Functions of One Variable - II | Mathematics for Competitive Exams PDF Download

Rolle's Theorem : Statement

If a function f defined on [a, b] is such that it is :
[R₁] continuous in the closed interval [a, b]
[R₂] differentiable in the open interval (a, b)
and [R₃] f(a) = f(b),
then there exists atleast one point c ∈ (a, b) such that f’(c) = 0.

Proof : Since f is continuous in the closed interval [a, b], therefore by the property of continuity f will be bounded and will attain bounds in [a, b].
Let m and M be the infimum and supremum of f in [a, b], then there exist c and d in [a, b] such that f(c) = M and f(d) = m
Case 1 : When M = m, then
f(x) = M = m, ∀ x ∈ [a, b]
⇒ f’(x) = 0, ∀ x ∈ [a, b]
⇒ f’(c) = 0, where c ∈ (a, b)
Thus the theorem hold good at any point c of (a, b).
Case 2 : When M ≠ m, then by R₃, f(a) = f(b), and atleast one of the numbers M and m be different from f(a) and f(b) i.e.,
M = f(c) ≠ f(a), f(b)
⇒ c ≠ a, b ⇒ c ∈ (a, b)
Since the function f is differentiable in (a, b), then the derivative exist at a point c.
Therefore RHD and LHD are

both exist and each equal to f(c).
Since f(c) is the maximum value of f(x) in (a,b),
∴ f(c + h) < f(c) and f(c - h) < f(c) ...(1)
Now
 [Putting x = c + h]
≤ 0 [∵ by(1) f(c + h) - f(c) < 0]
and
 [Putting x = c - h]
≥ 0 [∵ by(1) f(c - h) - f(c) < 0]
Thus both the derivatives are of opposite signs.
Therefore both these derivatives will be equal only when f(c) = 0.

Geometrical Interpretation of Rolle’s Theorem

Statement : It a curve has a tangent at every point there of and the ordinates of its extremities A and B are equal i.e. f(a) = f(b); then there exists atleast one point P of the curve other than A and B, the tangent at which is parallel to x-axis.

Case 1 : When f(x) = k (constant),
Then the graph of this function will be a straight line parallel to the x-axis whose equation is y = k and the derivative
f’(x) = 0, ∀ x ∈ [a, b]
Case 2 : When f(x) ≠ k (constant),
Then the graph of this function is a continuous curve of the following type in [a, b] as shown below:

From these figures it is quite intuitively that there must be atleast one point P(x = c) (may be more) of the curve other than A and B, the tangent at which is parallel to x-axis and at these points f’(c) = 0.

Case of Failure : The theorem will not hold good if any of the three conditions fails to hold good i.e.,

1. the function f is discontinuous at x = c.
2. the derivative does not exist at x = c.
3. f(a) ≠ f(b).

The following figures will illustrate these cases of failure:

Algebraic interpretation of Rolle’s theorem:

Statement : If f(x) be a polynomial in x and x = a and x = b are any two roots of the equation f(x) = 0, then there exists at least one root of the equation f’(x) = 0, which lies between a and b.
Remarks : The Rolle’s theorem is not applicable for those functions which do not satisfy even one condition of Roll’s theorem.

Another useful form of Rolle’s Theorem:
If a function f defined on [a, a + h] is such that it is:
[R₁] continuous in [a, a + h],
[R₂] differentiable in (a, a + h) and
[R₃] f(a) = f(a + h),
then ∃ θ ∈ (0, 1) such that f (a + θh) = 0.

Example: Verify Rolle’s theorem for the function in the interval mentioned against:
f(x) = x(x + 3)e^-x/2, x ∈ [-3, 0]

R₁. Since the given function x(x + 3), x is a polynomial of x, therefore it is continuous for every value of x and e^-x/2 is also continuous for every value of x.
Therefore their product is also continuous for every value of x
Hence the function is continuous, particularly in [-3, 0].
R₂. f(x ) = (2x + 3) e^-x/2 + x(x + 3) • e^-x/2(-1/2) = (1/2)e^-x/2 (6 + x - x²)
which is not infinite or indeterminate at any point of (-3, 0).
Thus f(x) is differentiable in (-3, 0).
R₃. f(-3) = 0 = f(0)
Thus the given function f satisfies all the three conditions of Roll’s theorem.
Hence Verified.
Therefore there must be atleast one point c in (-3, 0), where

Since e^-c/2 is not zero for any finite value c, therefore 6 + c - c² = 0 ⇒ c = - 2, 3
One value of c = -2 ∈ (-3, 0) Hence Verified.

Example: Verify Rolle’s Theorem for

The given function being an algebraic function of x is continuous in [- 1, 1].
Now
So f(x) is derivable in ] - 1, 1[. Also f(1) = f(-1) = 0.
Thus f(x) satisfies all the conditions of Rolle’s Theorem.
So there exists a point c ∈ ] - 1, 1[ such that f'(c) = 0
Clearly f’(c) = 0 for c = 0 ∈ ] - 1, 1[.
Hence the given function satisfies the hypothesis and the conclusion of Rolle’s Theorem.

Example: Examine the validity of the hypotheses and the conclusion of Rolle’s theorem for the function f(x) = 1 - (x - 1)^2/3 on [0, 2].

The given function, being an algebraic function of x is continuous in [0, 2].
Now
⇒ f’(x) does not exist at x = 1 ⇒ f is not derivable at x = 1
⇒ f is not derivable in ]0, 2[.
But f(0) = f(2) = 0.
Hence f does not satisfy all the conditions of Rolle’s theorem. Clearly, the conclusion of Rolle’s theorem is not valid for the given function.

Example: Show that between any two roots of e^x cos x = 1, there exists at least one root of e^x sin x - 1 = 0.

Let α and β be any two distinct roots of e^x cos x = 1. ...(1)
∴ e^α cos α = 1 and eβ cos β = 1.
Define a function f as follows :
f(x) = e^-x - cos x, ∀ x ∈ [α, β]. ...(2)
Obviously f is continuous in [α, β] and f is derivable in ]α, β[.
Indeed, f’(x) = - e^-x + sin x ∀ x ∈ ]α, β[.
From (2),
Similarly, f(β) = 0 and so f(α) = f(β).
Thus f satisfies all the conditions of Rolle's Theorem in [α, β] and so there exists some γ ∈ ]α, β[ such that f(γ) = 0.
⇒ sin γ - e^-γ = 0 ⇒ e^γ sin γ - 1 = 0, α < γ < β.
Hence γ is a root of e^x sin x - 1 = 0, α < γ < β.

Example: Prove that between any two real roots of e^x sin x = x, there is at least one root of cos x + sin x = e^-x.

Let x = α, x = β be any two distinct roots of e^x sin x = x.
∴ e^α sin α = α and e^β sin β = β. ...(1)
Define a function f on [α, β] as follows :
f(x) = e^x sin x - x ∀ x ∈ [α, β]. ...(2)
Clearly, (i) f continuous in [α, β], (ii) f is derivable in ]α, β[ and f(α) = 0 = f(β),. by (1). Hence by Rolle’s theorem, there exists some γ ∈ ]α, β[ i.e.,α  < γ < β such that f’(γ) = 0. ...(3)
From (2), f’(x) = e^x sin x + e^x cos x - 1
⇒ f’(g) = e^γ (sin γ + cos γ) - 1
∴ e^γ (sin γ + cos γ) - 1 = 0, using (3)
or sin γ + cos γ = 1/e^γ = e^-γ, α < γ < β.
Hence γ is a root of sin x + cos x = e^-x, where α < γ < β.

Example: If p(x) is polynomial and k ∈ R, prove that between any two real roots of p(x) = 0, there is a roots of p’(x) + k p(x) = 0.

Let f(x) = e^kx p(x), x ∈ R. ...(1)
Let α and β be any two real roots of p(x), where α < β. Then p(α) = 0 = p(β) f(α) = 0 = f(β), by (1).
Obviously, (i) f is continuous in [α, β], (ii) f is derivable in ]α, β[ and (iii) f(α) = f(β). Hence, by Rolle’s theorem, there exists some γ ∈ ]α, β[ such that f’(γ) = 0.
From (1), f(x) = k e^kx p(x) + e^kx p’(x) = e^kx [p’ (x) + kp (x)].
∴ 0 = f (γ) = e^kγ [p' (γ) + k p (γ)]
⇒ p’(γ) + k p(γ) = 0, as e^kγ ≠ 0.
Hence γ is a root of p’(x) + kp (x) = 0, where α < γ < β.

Mean Value Theorem

Lagrange’s Mean Value Theorem

Statement : If a function f with domain [a, b] is such that it is
[L₁] continuous in [a, b], and  
[L₂] differentiable in (a, b),
then ∃ c ∈ (a, b) such that:

Proof : Let us define a new function φ with domain [a, b] involving the given function f(x) as follows :
φ(x) = f(x) + Ax ...(1)
where A is a constant to be determined such that φ(a) = φ(b) ...(2)
⇒ f(a) + Aa = f(b) + Ab
⇒  ...(3)
We observe that the function φ is also defined on [a, b] and is the sum of two continuous functions on [a, b] and differentiable in (a, b), therefore φ is :
R₁. φ,[a, b] is continuous in [a, b]
R₂. φ,(a, b) is differentiable in (a, b)
and R₃. φ,(a) = φ,(b) [by the condition of constant]
Thus φ satisfies all the three conditions of Rolle’s theorem, therefore accordingly, there must be atleast one point c in (a, b) such that
φ'(c) = 0
⇒ f'(c) + A = 0 ⇒ f ’(c) = - A
⇒  a < c < b [by (3)]

Important Particular Case : If f(a) = f(b), then it reduces to Rolle’s theorem.

Geometrical Interpretation of Lagrange’s MV Theorem:

If a curve y = f(x) is continuous between two given points whose abscissae are x = a and x = b respectively and a tangent can be drawn to the curve at every point, then there exists atleast one point x = c, c ∈ (a, b) such that the tangent there at is parallel to the chord joining the two given end points.
Let the arc APB represents the graph of the function y = f(x) and a and b be the ordinates of A and B respectively. Join AB. Draw perpendiculars from A and B on x-axis. Let the chord AB makes an angle ψ with the x-axis, then from the right angled triangle ACB,

Another Useful Form

Statement : A function f defined on [a, a + h] is such that it is :

1. Continuous in [a, a + h];
2. Differentiable in (a, a + h)

then ∃ θ ∈ (0, 1) such that :
f(a + h) - f(a) = h f' (a + θh).
If we put b - a = h or b = a + h in Lagrange’s MV theorem, then any point c can be taken as c = a + θ h between a and b, where 0 < θ < 1.
Therefore by Lagrange MVT, we have

⇒ f(a + h) - f(a) = h f(a + θh), 0 < θ < 1.

Important Deduction from Lagrange’s MV Theorem :

Theorem : If a function f is

1. continuous in [a, b]
2. differentiable in (a, b), ∀ x ∈ (a, b), then
   (a) f’(x) = 0 ⇒ f is a constant function in (a, b).
   (b) f’(x) > 0 ⇒ f is strictly increasing in [a, b].
   (c) f’(x) < 0 ⇒ f is strictly decreasing in [a, b].

If the derivatives of two functions at every point of the interval (a, b) is same, then the functions differ by a constant.

Example: Verify Lagrange’s Mean Value Theorem for the function f(x) =  in [2, 4].

The function f(x) =  s an algebraic function of x, so f is continuous in [2, 4]. Also f is derivable in ]2, 4[.
Now
Since f satisfies all the conditions of Lagrange’s mean value theorem, therefore, there exists some c ∈ ]2, 4[ such that

⇒
⇒
Obviously, c = √6 ∈ ]2, 4[.

Hence the Lagrange’s mean value theorem is verified.

Example: Verify Lagrange’s Mean Value Theorem for the function f(x) = x (x -1)(x -2) in [0, 1/2]

Since f is a polynomial, f is continuous in [0, 1/2] and derivable in ]0, 1/2[. Thus there exists c ∈ ]0, 1/2[ such that
 ...(1)
Now f’(x) = (x - 1) (x - 2) + x(x - 1) + x(x - 2)
or f ’(x) = 3x² - 6x + 2. Now f(0) = 0, f(1/2) = 3/8.
From (1), 3/8 = 1/2(3c² - 6c + 2)
or 12c² - 24c + 5 = 0 or c =
Now
Thus  and the theorem is verified.

Example: Let f be defined and continuous in [a - h, a + h] and derivable in ]a - h, a + h[. Prove that there is a real number θ between 0 and 1 such that f(a + h) - f(a - h) = h[f’(a + θh) + f’(a - θh)].

Define φ(x) = f(a + hx) - f(a - hx) on [0, 1]. ...(1)
∴ φ’(x) = hf' (a + hx) + hf' (a - hx). ...(2)
As x varies over [0, 1], a + hx varies over [a, a + h] and a - hx varies over [a - h, a]. Thus

1. φ is continuous in [0, 1].
2. φ is derivable in ]0, 1[.

By Lagrange's mean value theorem, there exists θ, 0 < θ < 1, satisfying
 φ’(θ) or φ(1) - φ(0) = φ’(θ).
Hence f(a + h) - f(a - h) = h [f' (a + θh) + f’(a - θh)], by (1) and (2).

Example: Let f be defined and continuous on [a - h, a + h] and derivable on ]a - h, a + h[. Prove that there is real number θ between 0 and 1 for which f(a + h) - 2f(a) + f(a - h) = h[f(a + θh) - f(a - θh)].

Consider φ(x) = f(a + hx) + f(a - hx) on [0, 1].
Then (i) φ is continuous in [0, 1], (ii) φ is derivable in ]0, 1[. So by Lagrange’s mean value theorem.
 for some θ ∈ ]0, 1[ 

or φ(1) - φ(0) = φ(θ)
or [f(a + h) + f(a - h)] - [f(a) + f(a)] = h[f'(a + θh) - f'(a - θh)].
Hence f(a + h) - 2f(a) + f(a - h) = h[f'(a + θh) - f'(a - θh)].

Example: Prove that if f be defined for all real x such that | f(x) - f(y) | < (x - y)² for all real x and y, then f is constant.

Let c ∈ R. For x ∈ c, we have
 using (1).
If ε > 0, then on choosing δ = ε, we obtain
 when |x - c| < δ.
∴
Thus f’(x) = 0 ∀ x ∈ R.
Hence f is a constant function.

Example: Use mean value theorem to prove that 1 + x < e^x < 1 + xe^x, ∀ x > 0.

We shall apply Lagrange’s mean value theorem to the function f(x) = e^x in the interval [0, x]. The function f(x) = e^x is derivable in [0, x] and, therefore, there exists some c, 0 < c < x, such that
 ...(1)
Since 0 < c < x, so 1 = e° < e^c < e^x. ...(2)
From (1) and (2), 1 < (1/x) (e^x - 1) < e^x or x < e^x - 1 < xe^x. (∵ x > 0)
Hence 1 + x < e^x < 1 + xe^x ∀ x > 0.

Example: Show that  tan^-1 v - tan^-1 u  if 0 < u < v, and deduce that

Applying Lagrange’s mean value theorem to the function
f(x ) = tan^-1 x in [u, v], we obtain
 for some c ∈ ] u, v [
or  for u < c < v. ...(1)
Now c > u ⇒ 1 + c² > 1 + u² ⇒  ...(2)
Again c < v ⇒ 1 + c² < 1 + v² ⇒  ...(3)
From (1), (2), (3) ; we obtain

Hence  ...(4) (∵ u < v ⇒ v - u > 0)
Taking u = 1 and v = 4/3 in (4), we obtain

or
Hence

Example: Prove that |tan^-1 x - tan^-1 y| ≤ |x - y| ∀ x, y ∈ R.

Let f(t) = tan^-1 t in [x, y], where x < y.
By Lagrange’s mean value theorem, there exists some c ∈ ]x, y[ such that f(y) - f(x) = (y -x) f'(x).
or f(y) - f (x) = (y - x). 1/(1+c²)
∴ f(y) - f(x) ≤ (y - x).
Similarly, f(x) - f(y) ≤ (x - y), when y < x.
∴ |f(x) - f(y)| = |x - y|.
Hence, |tan^-1 x - tan^-1 y|≤|x - y| ∀ x, y ∈ R.

Example: Let f be differentiable on an interval I and suppose that fs is bounded on I. Prove that f is uniformly continuous on I.

Let x₁, x₂ be any two arbitrary points of I with x₁ < x₂. Suppose f’(x) is bounded on I, there exists some k > 0 such that
|f'(x)| ≤ k ∀ x ∈ |. ...(1)
Applying Lagrange’s mean value theorem to f on [x₁, x₂], we get
 for some c ∈ ]x₁, x₂[
or | f(x₂) - 1(x₁) | = | x₂ - x₁ | | f’(c) | ≤ k I x₂ - x₁ I, by (1).
Let ε > 0 be given and let δ = ε/k > 0. Then | f(x₂) - f(x₁) | < ε, when | x₂ - x₁ | < δ, ∀ x₁, x₂ ∈ |.
Hence f is uniformly continuous on I.

Second Mean Value Theorem :

Cauchy’s Mean Value Theorem : Statement:
If two functions f and g with domain [a, b] are such that both are
[C₁] continuous in [a, b]
[C₂] differentiable in (a, b), and
[C₃] g’(x) ≠ 0 ∀ x ∈(a, b) then ∃ c ∈ (a, b) such that:

Proof. First we notice that g(b) ≠ g(a)
Let, if g(b) = g(a), then g would satisfy all the conditions of Rolle’s theorem, particularly [R₃], then by the theorem, ∃ c ∈(a, b), where g’(c) = 0 which contradicts the condition [C₃].
Therefore our assumption g(a) = g(b) is false.
Hence g(a) ≠ g(b) ...(1)
Now define a new function φ, with domain [a, b] involving the given functions f and g as follows: φ(x) = f(x) + A g(x), ...(2)
where A is a constant to be determined such that φ(a) = φ(b)
or, f(a) + A f(b) = g(a) + A g(b) [by (2)]
⇒  ...(3)
which exists because g(b) ≠ g(a).
Hence the function φ is expressed as sum of two continuous and derivable functions, therefore φ is :

1. continuous on [a, b]
2. differentiable on (a, b) and
3. φ(a) = φ(b) [by A of (3)]

Thus the given function φ satisfies all the three conditions of Rolle’s theorem. Therefore Rolle’s theorem is applicable and accordingly there must be atleast one point x = c in (a, b), where φ'(c) = 0.
∴ φ’(c) = f’(c) + A g'(c) = 0
⇒  ...(4)
By (3) and (4),  Hence Proved.

Important Special Case

When g(x) = x, it reduces to Lagrange’s MV Theorem.

Geometrical interpretation of Cauchy’s Mean value Theorem

Statement : There is an ordinate x = c between x = a and x = b such that the tangents at the points where x = c cut the graph of the function f(x) and  are mutually parallel.

Another useful form of Cauchy’s Mean Value Theorem:

Statement: If two functions f(x) and g(x) with domain [a, a + h] are such that both are:

1. Continuous in [a, a + h],
2. Differentiable in (a, a + h), and
3. g’(x) ≠ 0, x ∈ (a, a + h)

then ∃ θ ∈ (0, 1) such that :

If we put b = a + h, then any point c = a + θh , 0 < θ < 1 may be taken in the interval [a, a + h].

Generalised Mean Value Theorem

If three functions f, g and h defined in [a, b] are such that

1. f, g and h are continuous in [a, b]
2. f, g and h are derivable in ]a, b[.

Then there exists a real number c ∈ ]a, b[ such that

Proof. Define a function ψ on [a, b] as follows :

or y(x) = A f(x) + B g(x) + C h(x), ...(iii)
where A = g(a) h(b) - h(a) g(b) etc., B and C are constants.

Applying given conditions (i) and (ii) in (iii), it follows that y is continuous in [a, b] and derivable in ]a, b[.
Also ψ(a) = 0 = ψ(b). (∵ two rows in ψ(x) become identical by putting x = a and x = b)
So ψ(a) = φ(b).
Thus ψ satisfies all the conditions of Rolle’s theorem and, therefore, there exists some c ∈ ]a, b[ such that ψ’(c) = 0.
Now
Hence

Remarks 1. If we take g(x) = x and h(x) = 1 in the Generalised Mean Value Theorem, we obtain Lagrange’s mean value theorem.
2. If we take h(x) = 1 in the Generalised Mean Value Theorem we obtain Cauchy’s mean value theorem.

Example : Verify the Cauchy’s mean value theorem for :
(i) f(x) = x², g(x) = x³ in [1, 2].
(ii) f(x) = sin x, g(x) = cos x in [-π/2, 0]
(iii) f(x) = e^x, g(x) = e^-x in [0, 1].

(i) Clearly f(x) = x², g(x) = x³ are continuous in [1, 2] and derivable in ]1 , 2[. Further g’(x) = 3x² ≠ 0 ∀ x ∈]1, 2[. Thus the conditions of the Cuachy’s mean value theorem are satisfied and so there exists some point c ∈]1, 2[ such that

or
or
Hence Cauchy’s mean value theorem is verified.

(ii) Clearly f(x) = sin x, g(x) = cos x are continuous in [-π/2, 0] and derivable in ] -π/2, 0[. Further g’(x) = - sin x ≠ 0 for all x ∈ ] -π/2, 0 [. Thus the conditions of the Cauchy’s mean value theorem are satisfied and so there exists some point c ∈ ] - π/2, 0 [ such that

or tan c = -1, which gives c = - π/4 ∈ ] - π/2, 0[.
Hence Cauchy's mean value theorem is verified.

(iii) Clearly f(x) = e^x and g(x) = e^-x satisfy the conditions of the Cauchy’s mean value theorem and so there exists some c ∈ ]0, 1[ such that
or
or 2c = 1 or c = 1/2 ∈]0, 1[.
Hence Cauchy's mean value theorem is verified.

Example : Show that

Let f(x) = sin x, g(x) = cos x ∀ x ∈ [α, β] and derivable in ]α, β[.
Further g’(x) = - sin x ≠ 0 ∀ x ∈ ]a, b[ ∈ ]0, π/2[.
By Cauchy’s mean value theorem,

i e.,
Hence

Example : Let the function f be continuous in [a, b] and derivable in ]a, b[. Show that there exists a number c in ]a, b[ such that 2c [f(a) - f(b)] = f’(c) [a² - b²].

Let g(x) = x² ∀ x ∈ [a, b], a < b.
Then f and g are continuous in [a, b] and derivable in ]a, b[. Further g’(x) = 2x ∉ ]a, b[. By Cauchy’s mean value theorem,

i.e.,
Hence 2c [f(a) - f(b)] = f(c) [a² - b²], c ∈ ]a, b[

Example : If f and g’ exist for all x ∈ [a, b] and if g’(x) ≠ 0 ∀ x ∈ ]a, b[, then prove that for some c ∈ ]a, b[

Define a function ψ on [a, b] as follows :
ψ(x) = f(x) g(x) - f(a) g(x) - g(b) f(x), ∀ x ∈ ]a, b[.
Since f and g are derivable in [a, b], so ψ is continuous in [a, b] and derivable in ]a, b[.
Also ψ(a) = ψ(b) = -f(a) g(b).
Since ψ satisfied all the conditions of Rolle’s Theorem, therefore, there exists some c ∈ ]a, b[ such that ψ’(c) = 0. We have
ψ’(x) = f'(x) g(x) + f(x) g’(x) - f(a) g’(x) - g(b) f'(x).
∴ ψ'(c) = 0 gives us
f’(c) g(c) + f(c) g’(c) - f(a) g’(c) - g(b) f(c) = 0
or g’(c) (f(c) - f(a} = f’(c) (g(b) - g(c)}.
Hence

Example : If a function f is such that its derivative f’ is continuous on [a, b] and derivable on ]a, b[, then show that there exists a number c between a and b such that f(b) = f(a) + (b - a) f'(a) + 1/2(b - a)2f"(c).

Define a function φ on [a, b] as follows :
φ(x) = f(b) - f(x) - (b - x) f'(x) - (b - x)² A, ...(1)
where A is a constant to be determined by
φ(a) = φ(b) ...(2)
i.e., f(b) - f(a) - (b - a) f (a) - (b - a)² A = 0
i.e., f(b) = f(a) + (b - a) f (a) + (b - a)² A. ...(3)
Since f is continuous on [a, b], so f is also continuous on [a, b].
Also (b - x), (b - x)² are continuous on [a, b].
Thus by (1), φ is continuous.
Thus φ satisfies the conditions of Rolle’s Theorem and so there exists some point c ∈ ]a, b[ such that φ’(c) = 0. ...(4)
From (1), φ’(x) = - f'(x) - {-f(x) + (b - x) f"(x)} + 2(b - x)A
or φ’(x) = - (b - x) f”(x) + 2(b - x) A. ...(5)
∴ f ’(c) = 0 ⇒ - (b - c) f'’(c) + 2(b - c) A = 0
⇒ A = 1/2f"(c), since a < c < b ⇒ b - c ≠ 0.
Substituting this value of A in (3), the result is proved.

Example : If f is continuous on [a, a + h] and derivable on ] a, a + h [, then prove that there exists a real number c between a and a + h such that f(a + h) = f(a) + hf’(a) +h²/2 f" (c).

Define a function φ on [a, a + h] as follows:
φ(x) = f(x) + (a + h - x) f'(x) + 1/2(a + h - x)² A. ...(1)
where A is a constant to be determined by φ(a) = φ(a + h)
i.e., f(a) + hf’(a) + h²/2 A = f(a + h). ...(2)
Since f is continuous on [a, a + h], f and F are continuous on [a, a + h]. Further (a + h - x), (a + h - x)² are also continuous on [a, a + h] and so by (1), φ is continuous on [a, a + h]. Since f is derivable on ]a, a + h[, so by (1), φ is derivable on ]a, a + h[. Thus φ satisfies all the conditions of Rolle’s Theorem and so there exists some c ∈ ] a, a + h [ such that
∴ f’(c) = 0 ...(3)
From (1), φ’(x) = f'(x) - f'(x) + (a + h - x) f"(x) - (a + h - x) A.
∴ 0 = φ’(c) = (a + h - c) [f’(c) - A] ...(4)
⇒ f"(c) - A = 0, since c ∈ ]a, a + h[ means a + h - c ≠ 0.
Putting A = f”(c) in (2), we obtain
f(a + h) = f(a) + h f'(a) + 1/2 h²f"(c).

Example : Let f and g be two functions defined and continuous on [a, b] and derivable on (a, b). Show that there exists some c ∈ (a, b) such that

We define a function φ on [a, b] as follows :
...(1)
By the given hypothesis, φ is continuous on [a, b] and derivable on (a, b). Also φ(a) = 0 = φ(b). Thus φ satisfies the conditions of Rolle’s theorem and so there exists some point c ∈ (a, b) such that φ’(c) = 0.
From (1), we have

∴
Hence

Example : If f” be continuous on [a, b] and derivable on ]a, b[, then prove that

 for some real number d between a and b

Define a function g on [a, b] as follows ;
g (x ) = f(x) - f(a) - (1/2)(x - a) {f'(a) + f'(x)} + A (x - a)3,

where A is a constant to be suitably chosen.
Since f” is continuous on [a, b], so f and f are continuous on [a, b].
Thus g is continuous on [a, b].
Similarly, g is derivable on ]a, b[.
Let A be chosen such that g(a) = g(b).
∴ 0 = f(b) - f(a) - 1/2(b-a) {f'(a) + f(b)} + A (b - a)³. ...(1)
Since the function g satisfies all the conditions of Rolle’s theorem [a, b], therefore, there exists a real number c between a and b such that
g’(c) = 0
We have g'(x) = f'(x) - (1/2){f'(a) + f'(x)} - 1/2(x-a)f"(x) + 3A (x-a)²
or g'x = 1/2{f'(x) - f'(a)} - 1/2 (x - a)f"(c) + 3A(c - a)² = 0.
Thus g’(c) = 0 implies that
1/2{f'(c) - f'(a)} - 1/2(c-a)f"(c) + 3A(c-a)² = 0. ...(2)
Let h be a function defined on [a, b] as follows :
 ...(3)
Then (i) h is continuous on [a, c],
(ii) h is derivable on ]a, c[,
(iii) h(c) = 0 and h(a) = 0, by (2) and (3); so that h(a) = h(c).
Since the function h satisfies all the conditions of Rolle’s theorem it [a, c], therefore, there exists a real number d(a < d < c < b), such that
h'(d) = 0.
From (3),
∴ 0 = h'(d) = -(1/2)(d-a)f"(d) + 6A (d-a).
i.e., A = f”’(d)/12, since (d - a) ≠ 0. ...(4)
From (1) and (4), we have

Example : Assuming that f”(x) exists for all x in [a, b], show that  ...(1)
where c and ξ both lie in [a, b].

The equation (1) can be rewritten as
f(c)(b-a) -f(a)(b-c) -f(b)(c-a) - 1/2(b-a)(c-a)(c-b)f"ξ = 0
or f(a)(c-b) -f(c-a) -f(c)(b-a) - 1/2(a-b)(b-c)(c-a)f"ξ = 0
or  ...(2)
We shall prove (2) instead of (1).
The relation (2) helps us to define a function F(x) on [a, b] as follows:
 ...(3)
where A is a constant such that F(c) = 0. ...(4)
From (3), F(a) = 0, F(b) = 0.
Thus F(a) = F(c) and F(c) = F(b), a < c < b. ...(5)
Since f” exists in [a, b], therefore, f and f are derivable and continuous in [a, b]l an hence derivable on continuous in each of the intervals [a, c] and [c, b]. From (3) and (5), it follows that F satisfies the conditions of Rolle’s theorem in each of the intervals [a, c] and [c, b]. Consequently,

and
From (3),  ...(6)
As stated earlier, f is derivable and continuous in [a, b]. So by (6), F’ is derivable and continuous in [ξ₁, ξ₂] ⊂ [a, b] and F’(ξ₁) = F’(ξ₂). Thus F’ satisfies the conditions of Rolle’s theorem in [ξ₁, ξ₂] and so there exists x ∈ ]x₁, x₂[ c: [a, b] such that
F"ξ = 0
From (6), F”(x) =
= f”(x)(b - a) - 2A (b - a)
∴ F"(ξ) = 0 ⇒ f"(ξ)-2A = 0 ⇒ A = -(1/2)f"(ξ). (∵ a≠b)
Substituting A = 1/2f"(ξ) in F(c) = 0 and using (3), we obtain

which proves (2).

Example : If f(0) = 0 and f”(x) exists on [0, ∞[, show that
 ...(1)
and deduce that if f (x) is positive for positive values of x, then f(x)/x strictly increases in ]0, ∞[.

The relation (1) can be written as
f(x) - xf'(x) + (1/2)x²f"(ξ) = 0.
Define a function F as follows :
F(x) = f(x) - x f'(x) + (1/2)Ax², ...(2)
where A is a constant such that F(c) = 0, where c > 0.
We have F(0) = f(0) = 0 and F(c) = 0. ∴ F(0) = F(c).
Thus F satisfies the conditions of Rolle’s theorem on [0, c].
Therefore, there exists ξ, such that 0 < ξ < c and F’(ξ) = 0.
From (2) F’(x) = - x f"(x) + Ax.
∴ F’(x ) = 0 ⇒ A = f"(ξ).
From (2), F(c) = 0 ⇒ f(c) - cf’(c) + (1/2)Ac² = 0
⇒
Hence
(ii) Let G(x) = f(x)/x, whenever x > 0.
∴
Thus G’(x) > 0 ∀ x > 0. ...(3)
If x₁ and x₂ be any two positive real numbers such that x₂ > x₁, then by applying Lagrange’s mean value theorem to G in [x₁, x₂], we get
G(x₂) - G(x₁) = (x₂ - x₁) G’(θ), θ ∈ ]x₁, x₂[
Since G’ (θ) > 0, by (3), and x₂ - x, > 0, so
G(x₂) - G(x₁) > 0 i.e ., G(x₂) > G (x₁) for x₂ > x₁ > 0.
Hence G(x) = f(x)/x is strictly increasing in ]0, ∞[.

Example : A twice differentiable function f is such that f(a) = f(b) = 0 and f(c) > 0 for a < c < b. Prove that there is at least one value ξ between a and b for which f"(ξ) < 0.

Since f” exists in [a, b), therefore f and f both exist and are continuous in [a, b] and as well as in [a, c] and [c, b], since a < c < b. Applying Lagrange’s mean value theorem to f on [a, c] and [a, b], we get

and  respectively.
Since f(a) = f(b) = 0, the above relationship become
 ...(1)
Since f ’ is continuous and derivable on [ξ₁, ξ₂] so on applying Lagrange’s mean value theorem off’ on [ξ₁, ξ₂], we obtain
 ...(2)
From (1) and (2), we get

since f(c) > 0, a < c < b and ξ₁ < ξ₂.
Hence f"( ξ) < 0 for some a <  ξ < b.

Example : If f ” be defined on [a, b] and if |f ’(x)| ≤ M for all in [a, b], then prove that |f(b) - f(a) - 1/2 (b - a) (f(a) + f(b)}| ≤ (1/2)(b-a)²M.

Define a function φ on [a, b] as follows :
φ(x) = f(x) - f(a) - (1/2)x-a {f'(a) + f'(x)} + A(x-a)², ...(1)
where A is a constant to be chosen such that φ(b) = 0 ...(2)
i.e., f(b) - f(a) -(1/2)(b-a){f'(a) + f'(b)} + A(b-a)² = 0. ...(3)
Since f” is defined on [a, b], so f and f are both derivable on [a, b] i.e., φ is derivable on [a, b] and so φ is continuous on [a, b].
From (1) and (2), φ(a) = 0, φ(b) = 0 and so φ(a) = φ(b)
By Rolle’s Theorem, there exists some c ∈ ]a, b[ such that
φ'c = 0
From (1)
∴ φ’(c) = 0 ⇒ f’(c) - f'(a) - (c - a) f"(c) + 4 A(c - a) = 0. ...(4)
Applying Lagrange’s Mean Value Theorem to f on [a, c], we obtain
f(c) - f(a) = (c - a) f"(d) for some d ∈ ]a, c[.
Substituting in (4), we obtain
f”(d) - f"(c) + 4 A = 0 ⇒ A = 1/4[f"(c) - f"(d)]/
Putting this value of A in (3), we obtain
f(b) - f(a) - 1/2(b-a){f'(a) + f'(b)} = 1/4(b-a)²[f"(d)-f"(c)].
Hence | f(b) - f(a) - (1/2)(b-a){f'(a)+f'(b)}|
≤ (1/4)(b-a)²[|f"(d)| + |f"(c)|]
≤ (1/4)(b-a)² (M+M) {∵ | f"(x)| ≤ M ∀ x ∈ [a, b]}
= (1/4) b-a)²M.

Example : Show that ‘0’ which occurs in the Lagrange's mean value theorem tends to the limit 1/2 as h → 0, provided f” is continuous.

We may take f” to be continuous in [a, a + h].
So by Example we have
f(a + h) = f(a) + hf’(a) + (h²/2)f"(a + θ₁h), where 0 < θ₁ < 1. ...(1)
Applying Lagrange’ mean value theorem to f in [a, a + h], we get
f(a + h) - f(a) = hf'(a + θh), 0 < θ < 1. ...(2)
From (1) and (2), we obtain
 ...(3)
Applying Lagrange’s mean value theorem to the function f in [a, a + θh], we obtain
f’(a + θh) - f'(a) = θh. f ’’(a + θ₂θh), 0 < θ₂ < 1. ...(4)
From (3) and (4), we get

⇒
∴  since f" continuous in [a, a + h].
Hence θ → 1/2 as h → 0.

Example : If f”(x) exists ∀ x ∈ [a, b] and  where c ∈ ]a, b[. ...(1)
Then there exists some ξ ∈ ]a, b[ such that f’’(ξ) = 0.

Since f ’(x) exists ∀ x [a, b], therefore, f and f exist in [a, b] and are continuous in [a, b].
Applying Lagrange’s Mean Value Theorem to the function f on each of the interval [a, c] and [c, d], we get
 for some ξ₁ ∈ ]a, b[,
and  for some ξ₂ ∈ ]c, b[,
Using (1), it follows that f’(ξ₁) = f’(ξ₂). Further.
(i) f' is continuous in [ξ_1, ξ₂],(ii) f’ derivable in ]ξ_1, ξ₂[and (iii) f'(ξ₁) = f’(ξ₂). So by Rolle’s Theorem as applied to f’ in [ξ_1, ξ₂], we obtain
f"(ξ) = 0 for some ξ ∈ ]ξ_1, ξ₂[ ⊂ ]a, b [.

Example : If f”(x) > 0 for all x ∈ R, then show that
 ...(1)
for every pair of real numbers x₁ and x₂.

Let x₁, x₂ ∈ R be arbitrary. If x₁ = x₂, then the given relation (1) because equality and so the result follows. We may suppose that x₁ < x₂. Since f"(x) exists ∀ x ∈ R, if satisfies the conditions of Lagrange's mean value theorem in each of the intervals  Consequently,

and
On subtracting the corresponding sides of these equations, we get
 ...(2)
Since f”(x) exists ∀ x ∈ R, f satisfies the conditions of Lagrange’s mean value theorem in the interval [c₁, c₂] and so f’(c₂) - f'(c₁) = (c₂ - c₁) f”(t), for some t ∈ ]c₁, c₂[.
As given f”(t) > 0 and c₂ > c_1, so (c₂ - c₁) f"(t) > 0.
Thus f'(c₂) - f'(c₁) > 0 i.e., f’(c₁) - f'(c₂) < 0.
Also x₂ > x₁ ⇒ x₂ - x₁ > 0.
Using these inequalities in (2), we get

Hence

Example : If f', g ’ are continuous on [a - h, a + h] and derivable on ] a - h, a + h [, then prove that  for some d ∈ ] a - h, a + h [, provided g (a + h) - 2g (a) + g (a - h) ≠ 0 and g”(t) ≠ 0 for each t ∈ ] a - h, a + h [.

We define a function φ on [a - h, a + h] as follows :
φ(x) = f(x) + Ag(x) + Bx + C, ...(1)
where A, B, C are constants to be determined by
φ(a - h) = φ(a) = φ(a + h) = 0. ...(2)
Since f and g are continuous on [a - h, a + h] and Bx + C is also continuous on [a - h, a + h], so by (1), φ is continuous on [a - h, a + h]. Similarly, φ is derivable on ]a - h, a + h[. By virtue of (2), φ satisfies all the conditions of Rolle’s theorem in each of the intervals [a - h, a] and [a, a + h]. Consequently.
φ’(t₁) = 0, for some t₁ ∈ ]a - h, a[,
and φ’(t₂) = 0, for some t₂ ∈ ]a, a + h[.
For any x ∈ ]a - h, a + h [, we have by (1),
φ’(x) = f(x) + Ag’(x) + B. ...(3)
Since f' and g’ are continuous on [a - h, a + h], so by (3), φ’ is continuous on [a - h, a + h]. Similarly, 'φ' is derivable on ] a - h, a + h [. Also φ'(t₁) = φ'(t₂) = 0. Applying Rolle’s theorem to φ’ on [ t₁, t₂], there exists some d ∈ ]t₁, t₂[ such that φ”(d) = 0 for some d ∈ ] t₁, t₂ [. Using (3),
f'’(d) + Ag”(d) = 0 ⇒ A = -f'(d)/g” (d). ...(4)
From (2), φ(a+h) -2φ(a) + φ(a-h) = 0
or [f(a + h) - 2 f(a) + f(a - h)] + A[g(a + h) - 2g(a) + g(a - h)] + [B(a + h) + C - 2 (Ba + C) + B(a - h) + C] = 0
or  ...(5)
From (4) and (5), we get the desired result.

Example : Show that
(a) (x/1+x) < log (1 + x) < x, x > 0.
(b) (x - (x²/2)) < log (1 + x) < x - (x²/(2(1+x)), x > 0.
(c) (x²/2(1+x)) < x - log (1 + x ) < (x²/2), x > 0.

Let f(x) = log (1 + x)-(x/(1+x), so that f(0) = 0.
We shall show that f(x) > 0 for x > 0.
Now,
Thus f(x) is an increasing function ∀ x > 0
⇒ f(x) > f(0), for x > 0
⇒ f(x) > 0, for x > 0 (∵ f(0) = 0)
⇒ log(1 + x) - (x/1+x) >, 0, x > 0
∴ (x/1+x)< log (1 + x), x > 0.
Let φ(x) = x - log (1 + x), so that φ(0) = 0.
∴  for x > 0.
Thus φ(x) is an increasing function of x, for x > 0
⇒ φ(x) > f(0), for x > 0
⇒ φ(x) > 0 (∵ φ(0) = 0)
⇒ x - log (1 + x) > 0, for x > 0
⇒ log (1 + x) < x, for x > 0.
Hence (x/1+x) < log (1 + x) < x, x > 0
(b) Let f(x) = log (1 + x) - (x - (x2/2)), f (0) = 0.
∴  (∵ x > 0)
⇒ f(x) is an increasing function of x, for x > 0 ...(2)
⇒ f(x) > f(0), for x > 0 ⇒ f(x) > 0 (∵ f(0) = 0)
⇒ log (1 + x) -  for x > 0
∴ (x - (x²/2)) < log (1 + x), x > 0.
Let  - log(1 + x), so that φ(0) = 0.
∴
 (∵ x > 0)
⇒ φ(x) is an increasing function, for x > 0
⇒ φ(x) > φ(0), for x > 0
⇒ f(x) > 0 (∵ f(0) = 0)
⇒  - log (1 + x) > 0, for x > 0
∴ log(1 + x) < x- (x2/(2(1+x))), for x > 0
Hence x - (x²/2) < log (1 + x) < - x²(2(1+x), for x > 0
(c) By part (b), we have

or
Hence

Example : Prove that x < sin-1 x <  if 0 < x < 1.

Let f(x) = sin^-1 x - x, so that f(0) = 0.
∴
⇒ f’(x) > 0, for 0 < x < 1
⇒ f is strictly increasing in 0 < x < 1
⇒ f(x) > f(0), for 0 < x < 1 (∵ x > 0)
⇒ f(x) > 0, for 0 < x < 1.
⇒ sin^-1 x - x > 0, for 0 < x < 1
⇒ sin^-1 x > x, for 0 < x < 1
⇒  x < sin^-1 x , for 0 < x < 1.
Let g(x) = - sin^-1 x, so that g(0) = 0.
∴

⇒ g’(x) > 0, for 0 < x < 1
⇒ g is strictly increasing, for 0 < x < 1
⇒  g(x) > g(0), for 0 < x < 1 (∵ x > 0)
⇒  sin^-1 x, for 0 < x < 1
⇒ sin^-1 x <, for 0 < x < 1.

Example : Using Lagrange's mean value theorem, show that x/(1+x) < log (1 + x) < x, x > 0.

Let f(x) = log (1 + x) in [0, x ], so that f'(x) = 1/(1+x).
Since f is continuous in [0, x] and derivable in ]0, x[, so by Lagrange’s mean value theorem, there exists some 0, 0 < 0 < 1 such that
 (∵ ]0, x [ = ]0 x , 1 x [, c = θx)
or log (1 + x) = x/(1+θx). [∵ f(0) = 0] ...(1)
Now 0 < c < 1 and x > 0 ⇒ θx < x
⇒ 1 + 1 + θx < 1 + x ⇒  
⇒  ...(2)
Again 0 < θ < 1 and x > 0 ⇒ y < 1 + θx
⇒  ...(3)
From (2) and (3), we obtain
 ...(4)
From (1) and (4), we obtain
x/(1+x) < log (1 + x) < x.

Example : Use the mean value theorem to prove  < tan^-1 x < x, if x > 0.

Let f(x) = tan^-1 x in [0, x]
Then f satisfies the conditions of Lagrange’s mean value theorem in [0, x]. Consequently, there exists some θ satisfying 0 < θ < 1 such that 
or tan^-1 x =  ...(1)
Now 0 < θ < 1 and x > 0 ⇒ θ² x² < x² ⇒  1 + θ² x² < 1 + x²
⇒  ...(2)
Again 0 < θ < 1 and x > 0 ⇒ 1 < 1 + θ² x²
⇒  ...(3)
From (2) and (3), we obtain
 ...(4)
From (1) and (4), we obtain
x/(1+x²) < tan^-1x < x , x > 0.

Example : Show that

Let f(x ) = x - log (1 + x ) - (1/2)x², so that f(0) =0.
∴  ...(1)
Now -1 < x < 0 ⇒ x + 1 > 0 . Thus, by (1),
f'(x) < 0, for -1 < x < 0
⇒ f is decreasing for - 1 < x < 0
⇒ f(x) > f(0), for -1 < x < 0
⇒ f(x) > 0 , for - 1 < x < 0
⇒ 1/2(x²) < x- log (1 + x), for - 1 < x < 0.
Let  so that g(0) = 0.
∴
⇒ g’(x) < 0, for -1 < x < 0
⇒ g is decreasing, for -1 < x < 0
⇒ g(x) > g(0), for -1 < x < 0
⇒ g(x) > 0, for -1 < x < 0
⇒ x - log(1 + x)   -1 < x < 0

Example : Prove that

Let f(x) = log(1 + x) -
∴
⇒ f’(x) > 0, for x > 0
⇒ f is increasing, for x > 0
⇒ f(x) > f(0), for x > 0
⇒ f(x) > 0 (∵ f(0) = 0)
⇒
Let
∴ g’(x) = 1 - x + x² -
⇒ g’(x) > 0, for x > 0
⇒ g is increasing, for x > 0
⇒ g(x) > g(0), for x > 0
⇒ g (x) > 0 (∵ g(0) = 0)
⇒ log (1 + x) < x -

Example: If 0 < x < 1, show that 2x < log

Let f(x) =
∴
Clearly, f(x) > 0, for 0 < x < 1 ⇒ f is increasing, for 0 < x < 1.
In particular, f(x) > f(0) = 0 as x > 0
∴  for 0 < x < 1.
Hence  for 0 < x < 1.
Let g(x) = 2x
∴

Thus g(x) is increasing, for 0 < x < 1. In particular, g(x) > g(0) = 0 ⇒ g(x) > 0
Hence

Example : Prove that tan x > x, whenever 0 < x < π/2.

Let c be any real number such that 0 < c < π/2.
Let f(x) = tan x - x for all x ∈ [0, c].
Then f is continuous as well as derivable on [0, c].
Now f’(x) = sec² x - 1 = tan² x > 0, for 0 < x < c.
Thus f is strictly increasing in [0, c]
⇒ f(c) > f(0), for c > 0.
But f(0) = 0. Therefore f(c) > 0 ⇒ tan c - c > 0.
Since c is any real number such that 0 < c < π/2, therefore, tan x > x, whenever 0 < x < π/2.

Example : Show that

Let f be defined as
 ...(1)
Then f is continuous in [0, π/2] and derivable in ]0, π/2[.
We have  ...(2)
Let g(x) = x cos x - sin x in [0, π/2]
∴ g’(x) = cos x - x sin x - cos x
= - x sin x < 0 in ]0, π/2[.
⇒ g is strictly decreasing in [0, π/2).
⇒ (x) < g(0) = 0 in [0, π/2] (∵ x > 0)
Using in (2), it follows that f’(x) < 0 in ]0, π/2[
⇒  is strictly decreasing in [0, π/2]
⇒ f(0) > f(x) > f(π/2), for 0 < x < π/2
⇒  using (1)
Hence

Example: Prove that

Since x sin x > 0 for all x in ]0, π/2[, therefore, we need to show that tan x sin x - x² > 0, ∀ x ∈]0, π/2[.
Let c any real number in ]0, π/2[.
Let f(x) = tan x sin x - x² ∀ x ∈ [0, c]
Then f is continuous as well as derivable in [0, c].
Now f’(x) = sec² x sin x + tan x cos x - 2x
= sin x (sec² x + 1) - 2x.
The form of f’(x) is such that we cannot decide about its sign.
Let g’(x) = cos x (sec²x + 1) + sin x . 2 sec²x tan x - 2 + 2 sin² x sec³ x.
Since g’(x) > 0 for all x ∈ ]0, c[⇒ g is strictly increasing in [0, c]
⇒ g(x) > g(0), whenever 0 < x < c. Since g(0) = 0, this means that g(x) > 0, whenever 0 < x < c ⇒ f’(x) > 0, whenever 0 < x < c ⇒ f is strictly increasing in [0, c] ⇒ f(c) > f(0) = 0 ⇒ f(c) > 0.
⇒ tan c sin c - c²> 0
⇒
Since c is any point of ]0, π/2[, it follows that  whenever 0 < x < π/2.

Example: Verify Lagranges MV Theorem for the function f(x) = |x| in the interval [-1 , 2]. 

L₁. The given function is continuous in [-1, 2], because

and
L₂. At the point x = 0 of the open interval (-1, 2)

Thus f’(0 + 0) ≠ f'(0 - 0)
Therefore the given function is not derivable in (-1, 2).
Hence Lagrange's theorem is not applicable for the given function.

Taylor's Theorem

Taylor’s theorem with Lagrange’s form of remainder;

Statement : If a function f with domain [a, a + h] is such that:
(i) f^n-1 is continuous in [a, a + h]
(ii) fⁿ exists in (a, a + h)
then ∃ θ ∈ (0, 1) such that :
f(a + h) = f(a) + hf'(a) +

Proof : Let us define a new function φ with domain [a, a + h] involving the derivatives of f as follows:
 ...(1)
where A is a constant to be determined such that
φ (a + h) = φ (a) ...(2)
From (1), φ (a + h) = f (a + h) ...(3)
and  ...(4)
Using (3) and (4) in (2), we obtain
 ...(5)
The value of A is given by (5).
Now by first condition, f(x), f’(x), f"(x)...f^n-1(x) are continuous in [a, a + h] and their derivatives exist and are finite in (a, a + h).
Polynomials of x, (a + h - x), are continuous and derivable for all values of x.
Therefore the function is also continuous in [a, a + h] and derivable in (a, a + h).
Thus the function φ is :
R₁. Continuous in [a, a + h]
R₂. Derivable in (a, a + h) and
R₃. φ(a) = φ(a + h) [by (2)]
Therefore the function φ satisfies all the three conditions of Rolle’s theorem and accordingly, there exists at least one positive number θ, 0 < θ < 1,
where
φ’(a + θh) = 0, 0 < θ < 1 ...(6)
Differentiating (1),
φ‘(x) = f'(x) + [(a + h - x) f"(x) - f'(x)]

The terms cancel in pairs on RHS, therefore

Therefore,
or,  [by (6)]

therefore
or  [by(6)]
⇒ A = f’(a + θh) [∵ h ≠ 0, (1 - θ) ≠ 0] ...(7)
Substituting the value of A from (7) in (5), we obtain the required result.
Hence Proved.
Remainder after n terms [R_n] :

This is also known as Lagrange’s form of Remainder.

Taylor’s theorem with Cauchy’s form of remainder :

Statement: If a function f with domain [a, a + h] is such that
(i) f^n-1 is continuous in [a, a + h]
(ii) fⁿ exist in (a, a + h)
then ∃ θ ∈ (0, 1) such that :
f (a + h) = f(a) + h f'(a) +

Proof. Consider a function φ defined as follows :
φ(x) = f(x) + (a + h - x) f'(x) + (a + h - x) A ...(1)
where A is a constant to be determined such that
φ(a + h) = φ(a) ...(2)
From (1), φ(a + h) = f(a + h) ...(3)
and  ...(4)
Using (3) and (4) in (2),
 ...(5)
The value of A is given by (5).
Now by the first condition.
f(x), f’(x), f’'(x), .... f^n-1 (x) are continuous in [a, a + h] and their derivable f'(x), f'’(x), fⁿ(x) exist and are finite in (a, a + h).
and the polynomials (a + h - x),  are continuous and derivable for all values of x. Therefore these are continuous in [a, a + h] and derivable in (a, a + h).
Thus the function φ is : R₁. continuous in [a, a + h]
R₂, derivable in (a, a + h)
and R₃, φ(a) = φ(b)
Thus the function φ satisfies all the three conditions of Rolle’s theorem, accordingly there exists atleast one positive number θ, 0 < θ < 1, where
φ'(a + θh) = 0, 0 < θ < 1 ...(6)
Differentiating (1), we get
φ’(x) = f'(x) + [(a + h - x) f"(x) - f(x)]

The terms cancel in pairs on RHS, therefore

Therefore φ’(a + θh)
[by (6)]
⇒  ...(7)
Substituting the value of A from (7) in (5), we obtain the required result

Another form :
Replacing h = b - a in the theorem, we get the following use fulform :
f(b) = f(a) + (b - a) f'(a) +  

f(a+h) = f(a) + (h)f'(a) +  

Remainder after n terms [R_n] :

This is known as Cauchy's form of Remainder.

Taylor’s theorem with Schlomitch and Roche form of Remainder.

Statement : If a function f defined in [a, a + h] is such that :
(i) its all derivative up to order (n - 1) i.e. f^{n - 1} are continuous in [a, a + h]
(ii) fⁿ exists in (a, a + h)
(iii) p ∈ N
then ∃ θ ∈ (0, 1) such that :

Proof. Consider a function φ defined as follows :
  ...(1)
where A is a constant to be determined such that
φ(a + h) = φ(a) ...(2)
From (1), φ(a + h) = f(a + h) ...(3)
and   ...(4)
Using (3) and (4) in (2), we get

f(a + h) = f(a) + hf’(a) +  ...(5)
The value of A is given by (5).
Now the function φ is :
(R₁) continuous in [a, a + h], because f, f', f", .... f^n-1, (a - h - x)^p are continuous.
(R₂) derivable in (a, a + h), because f, f', f" ......f^n-1, (a - h - x)^p are derivable.
(R₃) φ(a + h) = φ(a)
“Thus the function f satisfies all the three conditions of Rolle’s theorem, according ∃θ ∈ (0, 1), where
φ(a + θh) = 0 ...(6)
But
⇒
⇒ φ’(a + θh) = 0
⇒  ...(7)
Substituting the value of A from (7) in (5), we obtain the required result

Important Special Case :

1. When p = n, then R_n(x) = (hⁿ/n!) fⁿ(a+θh),
   which is called Lagrange’s form of remainder and we get
   Taylor’s theorem with Lagrange’s form of remainder.
2. When p = 1, then R_n(x) =
   which is called Cauchy’s form of remainder and we get Taylor’s theorem with Cauchy’s form of remainder.

Another form :

Replacing a + h = x or h = x - a, we get the following form of Taylor’s theorem :
f(x) = f(a) + (x - a) f(a) +
where  ...(8)
is called the remainder after n terms.

Maclaurin’s theorem : Statement :

If a function f with domain [0, x] be such that

1. all of its derivatives upto order (n - 1 ) are continuous in [0, x]
2. fⁿ exists in (0, x)
3.  Then ∃ θ ∈ (0, 1) such that

where

Proof. First of all we observe that condition (i) in the statement of theorem implies that f, f', f", f^n-1 are all defined (i.e., exist) and continuous on [0, x].
Consider the function φ defined on [0, x] as
φ(t) = f(t) + (x - t) f'(t) +
where A is a constant to be determined such that φ(0) = φ(x).
But
and φ(x) = f(x)
∴ φ(0) = φ(x)
⇒ f(x) = f(0) + xf’(0)  ...(1)
Now, (i) Since f, f', f", ... f^n-1 are all continuous on [0, x ] and (x - t)^r, r ∈ N is continuous on R.
∴ φ is continuous on [0, x].
(ii) Since f, f’, f” ...... f^n-1 are all derivable on (0, x) and (x - t)^r, r ∈ N is derivable on R.
∴ φ is derivable on (0, x).
(iii) Also φ(0) = φ(x)
Thus the function φ satisfies all the three conditions of Rolle’s Theorem on [0, x] and, hence, there exists a real number φ ∈ (0, 1) such that φ’(θx) = 0.
But φ’(t) = f'(t) - f'(t) + (x - t) f"(t) - (x - t) f"(t) + ...
 [other terms cancel in pairs]
⇒

∴ φ’(θx) = 0
⇒
⇒
Putting this value of A in (1), we have

The term  which occurs after n terms in known as Schlomilch and Roche’s form of remainder.

Note. (i) For p = 1, we get R_n =  called Cauchy’s form of remainder.
(ii) For p = n, we get R_n = (xn/n!)(fⁿ(θx) called Lagrange’s form of remainder.

Example : Prove that the number ‘θ’ which occurs in the Taylor’s theorem with Lagrange’s form of remainder after n terms approaches the limit 1/(n+1) as h approaches zero provided that fⁿ⁺¹ (x) is continuous and different from zero at x = a.

By Taylor’s theorem with Lagrange' s form of remainder after n terms and (n + 1) terms successively, we have
f(a + h) = f(a) + hf’(a) + ... +  where 0 < θ < 1.
f(a + h) = f(a) + hf’(a) + . . . +  where 0 < θ₁ < 1
On subtraction, we get 0

or  ...(1)
Using Lagrange’s mean value theorem for fⁿ(x) on [a, a + θh], we have
fⁿ(a + θh) = fⁿ(a) + 0hfⁿ⁺¹ (a + θ₂θh) where 0 < θ2 < 1
or fⁿ(a + θh) - fⁿ(a) = θhfⁿ⁺¹ (a + θ₂θh) ...(2)
From (1) and (2), we have

∴
= 1/(n+1). [∵ fⁿ⁺¹ (a) ≠ 0]

Ex : Assuming the derivatives which occur are continuous, apply the mean value theorem to prove that φ'(x) = F'{f(x)} f'(x) where φ(x) = F {f(x)}.

Let f(x) = t so that φ(x) = F(t)
Now
 [∵ f(x + h) = f(x) + hf'(x + θ₁h) by Mean Value Theorem]

[∵ F(t + H) = F(t) + HF’(t + θ₂H) by Mean Value Theorem]


= f'(x) F'(f(x)) [∵ f and F’ are continuous]
= F’{f(x)} f'(x).

Example : Using Taylor’s theorem, show that
(i) cos x ≥ 1 - (x²/2) ∀ x ∈ R
(ii) 1 + x + (x²/2) < e^x < 1 + x + (x²/2)e^x, x > 0
(iii) x - (x³/3!) < sin x < x, x > 0
(iv)

(i) Case 1. Let x = 0
The cos x = 1, 1 - (x²/2) = 1 ∴ cos x = 1 - x²/2.
Case 2. Let x > 0 and f(x) = cos x
Then f'(x) = - sin x, f"(x) = - cos x
Since f(x) = f(0) + xf’(0) + (x²/2!) f”(θx) where 0 < θ < 1.
∴ cos x = 1 - (x²/2) cos θx
But cos θ < 1 θx, x > 0
∴
Case 3. Let x < 0
Put y = -x so that y > 0
By case 2, cos y > 1 - (y²/2) ⇒
Combining all cases, cos x ≥ 1 - x²/2 ∀ x ∈ R.
(ii) Let f(x) = e^x, e > 0, then f’(x) = f'’(x) = e^x
Since f(x) = f(0) + xf’(0) + (x²/2!) f”(θx) where 0 < θ < 1
∴ e^x = 1 + x + x²/2 (e^θx)
Now 0 < θ < 1 and x > 0 ⇒ 0 < θx < x
⇒ e⁰ < e^θx < e^x [∵ e^x is an increasing function]
⇒
⇒ 1 + x + x²/2 < e^x < 1 + x + x²/2 (e^x).
(iii) Let f(x) = sin x, x > 0, then f’(x) = cos x, f"(x) = - sin x
f”(x) = - cos x
Since f(x) = f(0) + xf’(0) + where 0 < θ < 1
∴ sin x = x - (x³/3!)cos θx
But cos θx < 1 ∀θx, x > 0
∴ x - (x³/3!) cosθ x > x - x³/3!
⇒ sin x > x - x³/3! ...(1)
Also f(x) = f(0) + xf’(θx) where 0 < θ < 1
∴ sin x = x cos θx
But cos θx < 1 ∀ θx, x > 0
∴ x cos θx < x ⇒ sin x < x ...(2)
Combining (1) and (2), we get x - (x³/3!) < sin x < x.
(iv) Case 1. Let x = 0
Then
Case 2. Let x > 0 and f(x) = sin x
Then f'(x) = cos x, f'’(x) = - sin x, f”(x) = - cos x,
f^iv(x) = sin x, f^v(x) = cos x
Since f(x) = f(0) + x f’(0) +
∴ sin x = x - (x³/3!) cos θx
But cos θx < 1 ∀ θx, x > 0
∴
⇒ sin x > x - (x³/3!) ...(1)
Also f(x) = f(0) + xf’(0) +
∴ Sin x = x -
But cos θx < 1 ∀ θx, x > 0
∴
⇒ sin x < x -  ...(2)
Combining (1) and (2),
Combining the two cases, we have

Taylor’s Series :

If a function f is defined in the interval [a + ah] and satisfy all conditions of Taylor’s theorem, then by Taylor’s theorem,
f(a + h) = f(a) + hf '(a) +
or, f(a + h) = S_r + R_n, where S_n represents the sum of first n terms and R_n is called the Taylor’s remainder after n terms.

or, f(a + h) =
Therefore if the function f is expressible in an Infinite series, then clearly

1. fⁿ (a) exists ∀ n ∈ N and
2. for the convergence of the Taylor’s series,
   Taylor’s remainder after n terms, R_n → 0 when n → ∞, then
   f(a + h) = f(a) + h f'(a) +

The infinite series on RHS is known as Taylor’s Series.

Maclaurin’s Series :
When a = 0 and replace h by x, the n the Taylor’s series reduces to the following form :
f(x) = f(0) + xf’(0) +
This series is called Maclaurin’s Series and is very useful in the expansion of functions. 

Example : Expand e^x as an infinite series.

Let f(x) = e^x so that fⁿ(x) = e^x and fⁿ(0) = e⁰ = 1 ∀ n ∈ N
Clearly, f and all its derivatives exist and are continuous for every real value of x.
Lagrange’s form of remainder is

Let
so that
⇒  
⇒
∴
Thus the conditions of Maclaurin's infinite expansion are satisfied.
∴ For all x ∈ R, e^x = f(x)

Example : Expand sin x as an infinite series.

Let f(x) = sin x so that fⁿ(x) = sin
and fⁿ(0) = sin (nπ/2) ∀ n ∈ N
⇒
⇒ f”(0) = f^iv(0) = ... = 0, f’(0) = 1, f”(0) = -1, f(0) = 1, ....
Clearly, f and all its derivatives exist and are continuous for every real value of x.
Lagrange’s form of remainder is

∴
 
But
∴
Thus the conditions of Maclaurin’s infinite expansion are satisfied.
∴ For all x ∈ R, sin x = f(x) = f(0) + xf’(0) +

Example : Expand (1 + x)^m, m ∈ R.

Two cases arise according as m is not a positive integer.
Case 1. When m is a positive integer.
Let f(x) = (1 + x)^m, x ∈ R
Then fⁿ(x) exists for all x and all n.
In fact, if 1 ≤ n ≤ m, then fⁿ(x) = m(m - 1)(m - 2) ... (m - n + 1) (1 + x)^m-n
so that f^m(x) = m ! and fⁿ(x) = 0, if n > m
⇒
Since fⁿ(x) = 0 for all n > m, it follows that R_n → 0 as n → ∞. Thus the conditions of Maclaurin’s expansion are satisfied.
∴ (1 + x)^m = f(x) = f(0) + xf’(0)

Case 2. When m is not a positive integer.
Let f(x) = (1 + x)^m, x ≠ - 1
Taking Cauchy form of remainder, we have



Now let
so that
∴

It follows that if |-x| = |x| < 1, then
⇒  ...(1)
Since 0 < θ < 1 and -1 < x < 1
∴ -θ < θx < θ ⇒ 1 - θ < 1 + θx < 1 + θ
⇒ 0 < 1 - θ < 1 + θx ⇒
Consequently, ...(2)
Also, since -
⇒
 ...(3)
From (1), (2) and (3), we find that for I x I < 1,
Thus the conditions of Maclaurin's infinite expansion are satisfied and

Example : Expand log (1 + x) as an infinite series.

Let f(x) = log (1 + x) where 1 + x > 0 i.e., x > -1
Then
Case 1. When 0 ≤ x ≤ 1
Writing Lagrange’s remainder after n terms, we have


If x = 1, then |R_n| =
∴
If 0 ≤ x < 1, then since 0 < θ < 1
∴ 0 ≤ x < 1 + θx
⇒  ⇒
∴
∴
Case 2. When -1 < x < 0
Since in this case, need not be less than unity, therefore, it may not be easily shown that R_n → 0 as n → ∞ by considering Lagrange’s remainder.
∴ Writing Cauchy’s remainder after n terms, we have


Now -1 < x < 0 and 0 < θ < 1 ⇒ - θ < θx
⇒ 1 - θ < 1 + θx ⇒
Also, -|x| ≤ x ⇒ - θ |x| ≤ θx
⇒ 1 - |x| < 1 + θx ⇒
Consequently,
and |x|ⁿ → 0 as n → ∞ (since |x| < 1)
∴
Thus, we find that if -1 < x ≤ 1, then
Hence log(1 + x) = f(x) = f(0) + x f’(0) +
= log 1 + x.1 +  

MAXIMA AND MINIMA OF ONE VARIABLE

Let I be an interval.
A function f : I → R is said to have a global maxima (or an absolute maximum) on I if there exists a point c ∈ I such that f(c) ≥ f(x) for all x ∈ I. c is said to be a point of global maxima for f on I.
f is said to have a global minima (or an absolute minimum) on I if there exists a point c ∈ I such that f(c) ≤ f(x) for all x ∈ I. c is said to be a point of global minima for f on I.
A function f : I → R is said to have a local maximum (or a relative maximum) at a point c ∈ I if there exists a neighbourhood N (c, δ) of c such that f(c) ≥ f(x) for all x ∈ N (c, δ) ∩ I.
f is said to have a local minimum (or a relative minimum) at a point c e I, If there exists a neighbourhood N(c, δ) of c such that f(c) ≤ f(x) for all x ∈ N(c, δ) ∩ I.
We say that f has a local extremum (or a relative extremum) at a point c ∈ I, if f has either a local maximum or a local minimum at c.

Note : If f : I → R has a local maximum (a local minimum) at a point c ∈ I then c is a point of global maxima (a point of global minima) for f on N(c, δ) ∩ I for some suitable δ > 0.
N (c, δ) = (c - δ, c + δ)

Theorem : Let f : I → R be such that f has a local extremum at an interior point c of I. If f’(c) exists then f’(c) = 0.
Proof. We prove that theorem for the case when f has a local maximum at c. The proof of the other case is similar.
Since f’(c) exists, either f'(c) > 0, or f'(c) < 0, or f'(c) = 0.
Let f'(c) > 0. Then
Therefore there exists a positive δ such that
Let c < x < c + δ. Then x - c > 0 and therefore f(x) > f(c) for all x ∈ (c, c + δ). This contradicts that f has a local maximum at c.
Consequently, f’(c) ≥ 0 ....... (i)
Let f ’(c) < 0. Then
Therefore there exists a positive δ such that
Let c - δ < x < c. Then x - c < 0 and therefore f(x) > f(c) for all x ∈ (c - δ, c). This contradicts that f has a local maximum at c.
Consequently, f’(c) ≤ 0 ....... (ii)
From (i) and (ii) we have f'(c) = 0.
This completes the proof.

Corollary : Let f : I  and c be an interior point of I, where f has a local extremum. Then either f’(c) does not exist, or f'(c) = 0.
Note 1. The theorem says that if the derivative f(c) exists at an interior point c of local extremum, f’(c) must be 0. A function may, however have a local extremum at an interior point c of its domain without being differentiable at c. For example, the function defined by f(x) = |x|, x ∈ R has a local minimum at 0 but f'(0) does not exist.
Note 2. The condition f’(c) = 0 (when f'(c) exists) is only a necessary condition for an interior point c to be a point of local extremum of the function f.
For example, for the function f defined by f(x) = x³, x ∈ R, 0 is an interior point of the domain of f. f’(0) = 0 but 0 is neither a point of local maximum nor a point of local minimum of the function f.
Note 3. The theorem holds if c is an interior point of I.
Let a function f be defined on [0, 1] by f(x) = x, x ∈ [0, 1]. Then f has a local maximum at 1 (not an interior point of I), f is differentiable at 1, but f’(1) ≠ 0.

Theorem : (First derivative test for extrema)

Let f be continuous on I = [a, b] and c be an interior point of I. Let f be differentiable on (a, c) and (c, b).

1. If there exists a neighbourhood (c - δ, c + δ) ⊂ I such that f’(x) ≥ 0 for x ∈ (c - δ,c) and f'(x) ≤ 0 for x ∈ (c, c + δ), then f has a local maximum at c.
2. If there exists a neighbourhood (c - δ, c + δ ) ⊂ I such that f’(x) ≤ 0 for x ∈ (c - δ,c) and f(x) ≥ 0 for x ∈ (c, c + δ), then f has a local minimum at c.
3. If f (x) keeps the same sign on (c - δ.c) and (c .c + δ), then f has no extremum at c.

Proof. 1. Let x ∈ (c - δ, c). Applying Mean value theorem to the function f on [x, c], we have f(c) - f(x ) = (c - x) f'(ξ) for some x ∈ (x, c).
Since f'(ξ) ≥ 0, we have f(x) ≤ f(c) for x ∈ (c - δ, c).
Let x ∈ (c, c + δ). Applying Mean value theorem to the function f on [c, x], we have f(x) - f(c) = (x - c)f'(η) for some η ∈ (c, x).
Since f’(η) ≤ 0. we have f(x) ≤ f(c) for x ∈ (c, c + δ).
If follow that f(c) ≥ f(x) for all x ∈ N(c, δ) ∩ I.
Therefore f has a local maximum at c.
2. Similar proof.
3. Let f’(x) > 0 for x ∈(c - δ, c) and for x ∈ (c, c + δ).
Then f(x) < f(c) for x ∈ (c - 8, c) and f(c) < f(x) for x ∈ (c, c + δ).
Therefore f has neither a maximum nor a minimum at c.
Similar proof if f ’(x) < 0 for x ∈ (c - δ, c) and for (c, c + δ).
Note. The converse of the theorem is not true.
For example, let f(x) = 2x² + x² sin(1/x), x ≠ 0
= 0, x = 0.
Then f has a local minimum at 0.
f'(x) = 4x + 2x sin
= 0, x = 0. f' takes both positive and negative values on both sides of 0 (in the immediate neighbourhood).

Examples:

1. Let f(x) = I x I, x ∈
   f is continuous on . f is not differentiable at 0.
   f’(x) < 0 for x ∈ (-δ, 0) and f’(x) > 0 for x ∈ (0, δ) for some δ > 0.
   Therefore f has a local minimum at 0.
2. Let f(x) = |x - 1| + |x - 2|, x ∈ [0, 3]
   Then f(x) = 3 - 2x, if 0 ≤ x < 1
   = 1 , if 1 ≤ x ≤ 2
   = 2x - 3, if 2 < x ≤ 3.
   f is continuous on [0, 3]. f is not differentiable at 1 and 2.
   f’(x) < 0 for x ∈ (1 - δ, 1), f'(x) = 0 for x ∈ (1, 1 + 8) for some δ satisfying 0 < δ < 1. Therefore f has a local minimum at 1.
   f’(x) = 0 for x ∈ (2 - 8, 2), f’(x) > 0 for x ∈ (2 , 2 + 8) for some δ satisfying 0 < δ < 1.
   Therefore f has a local minimum at 2.
3. f(x) = (x - 1)² (x - 3)³, x ∈ R.
   f'(x) = 2(x - 1)(x - 3)³ + 3(x - 1)² (x - 3)²
   = (x - 1)(x - 3)² (5x - 9), x ∈ R.
   f is continuous on R. f(x) = 0 at the points 1, 3, 9/5.
   f’(x) > 0 for x ∈ (1 - δ, 1) and f'(x) < 0 for x ∈ (1, 1 + δ) for some δ > 0. Therefore f has a local maximum at 1.
   f’(x) > 0 for x ∈ ( 3 - δ, 3) and f'(x) > 0 for x ∈ (3, 3 + δ) for some δ > 0. Therefore f has neither a maximum nor a minimum at 3.
   f’(x) < 0 for x ∈  and f’(x) > 0 for x ∈  for some δ > 0. Therefore f has a local minimum at 9/5.

Theorem : (Higher order derivative test for extrema)

Let f : I → R and c be an interior point of I.
If f'(c) = f(c) = ... = f^n-1 (c) = 0 and fⁿ(c) ≠ 0, then f has
(i) no extremum at c if n be odd, and
(ii) a local extremum at c if n be even:
a local maximum if fⁿ(c) < 0, a local minimum if fⁿ(c) > 0.

Example :

1. f(x) = x⁵ - 5x⁴ + 5x³ + 10, x ∈ R.
   Show that f has a maximum at 1 and a minimum at 3 and f has neither a maximum nor a minimum at 0.
   For an extremum f’(x) = 0. f'(x) = 0 at x = 1, 3, 0.
   f”(x) = 20x³ - 60x² + 30x. Therefore f”(1) < 0, f"(3) > 0, f"(0) = 0.
   Since f’(1) = 0 and f"(1) < 0, f has a local maximum at 1.
   Since f’(3) = 0 and f"(3) > 0, f has a local minimum at 3.
   Since f’(0) and f"(0) = 0, in order to decide the nature of f at 0, we are to examine derivative of higher order at 0.
   f”(x) = 60x² - 120x + 30. f”(0) = 30 ≠ 0.
   Therefore f has neither a maximum nor a minimum at 0.
2. Find the local extremum point of the function f(x)
   f’(x) = 0 at x = -2, 0.
   Let h be an arbitrarily small positive number.
   f’( - 2 - h) > 0, f'(-2) = 0, f'(-2 + h) < 0.
   f’(0 - h) < 0, f'(0) = 0, f'(0 + h) > 0.
   f is continuous at - 2. f'(x) > 0 for x ∈ (- 2 - δ, 2) and f (x) < 0 for x ∈ (-2, -2 + δ) for some δ > 0.
   f is continuous at 0.f’(x) < 0 for x ∈ (-δ, 0) and f’(x) > 0 for x ∈ (0, δ) for some δ > 0. Hence f has a local maximum at -2 and a local minimum at 0.
3. Find the global maximum and the global minimum of the function f on R, where f(x) =
   
   f’(x) = 0 at x = ±2. f'(x) < 0 for |x| < 2 and f'(x) > 0 for |x| > 2.
   f is continuous at 2. f’(2 + h) > 0 and f'(2 - h) < 0 for sufficiently small h > 0. Therefore f has a local minimum at 2 and f(2) = 1/3.
   f is continuous at - 2. f'(-2 + h) < 0 and f‘(- 2 - h) > 0 for sufficiently small h > 0. Therefore f has a local maximum at -2 and f(-2) = 3.
   As f'(x) > 0 for x > 2 and f is continuous at 2, f is an increasing function [2, ∞) and  f(x) = 1.
   Therefore
   As f’(x) > 0 for x < - 2 and f is continuous at - 2, f is an increasing function on (-∞, - 2] and
   Therefore
   
   Therefore

Example : Find the maximum and minimum points of the function f given by
f(x) = (x - 1) (x - 2) (x - 3).

f(x) = x³ - 6x² + 11 x - 6
f’(x) = 3x² - 12x + 11
f”(x) = 6x - 12
For maximum or minimum,
f'(x) = 0 ⇒ 3x² - 12x + 11 = 0
⇒  
At
⇒ f is minimum at x = 2
At x = 2 -
⇒ f is minimum at x = 2 -

Example : Show that x⁵ - 5x⁴ + 5x³ - 1 has a maximum when x = 1, a minimum when x = 3 and neither when x = 0.

Let f(x) = x⁵ - 5x⁴ + 5x³ - 1
f’(x) = 5x⁴ - 20x³ + 15x²
f”(x) = 20x³ - 60x² + 30x  
For maximum or minimum.
f’(x) = 0 ⇒ 5x²(x² - 4x + 3) = 0
⇒ x²(x - 1) (x - 3) = 0 x = 0, 1, 3
At x = 0, f”(x) = 0
Since f"(x) = 60x² - 120x + 30, f”(0) = 30 ≠ 0
⇒ f has neither a max. nor mini, when x = 0
At x = 1, f”(x) = 20 - 60 + 30 = -10 < 0
⇒ f has a maximum when x = 1
At x = 3, f”(x) = 20(3)³ - 60(3)² + 30(3) = 90 > 0
⇒ f has a minimum when = 3.

Example : Examine the following function for extreme values : (x - 3)⁵ (x + 1)⁴.

Let f(x) = (x - 3)⁵ (x + 1)⁴
then f(x) = (x - 3)⁵. 4(x + 1)³ + 5(x - 3)⁴ (x + 1)⁴
= (x - 3)⁴ (x + 1)³ [4(x - 3) + 5(x + 1)] = (x - 3)⁴ (x + 1)³ (9x - 7)
For maximum or minimum,
f(x) = 0 ⇒ x = 3, - 1, 7/9
Let us test values one by one.
(i) For x slightly < 3, f’(x) = (+)(+)(+) = +ve
For x slightly > 3, f’(x) = (+)(+)(+) = +ve
Since f’(x) does not changes sign as a passes through 3, f is neither maximum nor minimum at x = 3.

(ii) For x slightly < -1, f'(x) = (+)(-)(-) = +ve
For x slightly > - 1, f'(x) = (+)(+)(-) = - ve
Since f’(x) changes sign from +ve to -ve as x passes through -1, f is maximum at x = -1
f_max = f(-1) = 0
(iii) For x slightly < 7/9, f’(x) = (+)(+)(-) = -ve
For x slightly > 7/9, f’(x) = (+)(+)(+) = +ve
Since f’(x) changes sign from -ve to +ve as x passes through 7/9, f is minimum at x = 7/9

Example : Find the maximum and minimum values, if any, of the function (1 - x)² e^x.

Let f(x) = (1 - x)² e^x
f(x) = (1 - x)² . e^x - 2(1 - x)e^x
= (1 - x)(1 - x - 2) e^x = (1 - x)(-1 -x) e^x
= (x + 1)(x - 1) e^x = (x² - 1) e^x
f”(x) = (x² - 1).e^x + 2x . e^x = (x² + 2x - 1) e^x
For maximum or minimum, f'(x) = 0
⇒ (x² - 1)ex = 0 ⇒ x² - 1 = 0 [∵ e^x ≠ 0 for any x ∈ R]
∴ x = ± 1
When x = 1, f”(x) = 2e > 0 ⇒ f is minimum at x = 1
f_min = f(1) = 0
When x = - , f"(x) = -2e^-1 < 0 f is maximum at x = -1
f_max = f(-1) = 4e^-1 = 4/e.

Example : Find the maximum value of (logx/x), 0 < x < ∞.

Let f(x) = logx/x
then

For maximum or min., f(x) = 0 ⇒ 1 - log x = 0
⇒ log x = 1 = log e ⇒ x = e
When x = e,
⇒ f is maximum at x = e.

Example : Prove that the function (1/x)^x, x > 0 has a maximum at x = 1/e.

Let
∴
∴
For max. (or min.) f(x) = 0 ⇒ - f(x) [1 + log x] = 0
⇒ 1 + log x = 0 [∵ f(x) ≠ 0]
⇒ log x = - 1 ⇒ x = e^-1
Also f”(e^-1) = -f(e^-1) . e - f(e^-1)[1 + log e^-1] = - (e)^1/e. e - 0 < 0 [∵ f’(e^-1) = 0]
⇒ f is maximum at x = e^-1 = 1/e.

Example : Show that sin x(1 + cos x) is a maximum when x = π/3. 

Let f(x) = sin x(1 + cos x)
then f'(x) = sin x(- sin x) + cos x(1 + cos x)
= cos² x - sin² x + cos x = cos 2x + cos x
f”(x) = -2 sin 2x - sin x
For max. or min., f(x) = 0
⇒ cos 2x + cos x = 0 ⇒
⇒ either
Here we have to consider only the point x = π/3

⇒ f(x) has a maximum at x = π/3.

Example: Find the maximum and minimum values of the function

Let
then f'(x) = cos x + cos 2x + cos 3x
f"(x) = - [sin x + 2 sin 2x + 3 sin 3x]  
For max. or min., f’(x) = 0
⇒ cos x + cos 2x + cos 3x = 0 ⇒ (cos 3x + cos x) + cos 2x = 0
⇒ 2 cos 2x cos x + cos 2x = 0 ⇒ cos 2x(2 cos x + 1) = 0
⇒ either cos 2x = 0 or cos x = -(1/2)
⇒
When
⇒ f is maximum at x = π/4

When x =
⇒ f  is maximum at x = 2π/3

When

⇒ f is maximum at x = 3π/4

Example : Prove that the function f(0) = sin^p θ cos^q θ has a maximum at q = tan^-1

f(θ) = sin^p θ cos^q θ
f’(θ) = sin^p θ . θ cos^q-1 θ (- sin θ) + p sin^p-1 θ (cos θ) cos^q θ
= - q sin^p-1 θ cos^q-1 θ + p sin^{p - 1} θ cos ^q+1 θ
= sin^p-1 θ cos^q-1 θ(p cos² θ - q sin² θ)
For maxima or minima, f(θ) = 0
⇒ sin θ = 0 or cos θ = 0 or p = cos² θ = q sin² θ
⇒
Also f"(θ) = sin^P-1 θ cos^q-1 θ (p cos²θ - q sin²θ)

∴ f"(θ) = f(θ) [ - p cosec²θ - q sec²θ] + f’(θ) (p cot θ - q tan θ)
At θ  
∴ f”(θ) = -f(θ) [p cosec² θ + q sec² θ] at θ = tan^-1 which is negative
Hence f(θ) is maximum at θ =

Indeterminate forms and L'Hospital's Rule

Theorem : (L’Hospital’s Rule) : Suppose f and g are differentiable and g’(x) ≠ 0 near a (except possible at a). Suppose that
or that
(In other words, we have an indeterminate form of type ). Then 

 if the limit on the right side exists (or is ∞ or -∞).
Indeterminate Forms of Type

Example: Find

We have

In short,

Example : Find

We have
 

In short,

Example : Find

We have

Example: Find

We have

Example : Find

We have

Since  is an indeterminate form of type 0/0, we can use L’Hospital’s Rule again. But it is easier to do trigonometry instead. Note that 1 - sec² 5x = - tan² 5x. Therefore

Example : Find

One can show, however, that  does not exist. In fact, we first note that 1 + cos x and 1 - sin x may attain any value between 0 and 2. From this one can deduce that 

 attains any nonnegative value infinitely often as x → ∞. This means that 

 does not exist, so L’Hospital’s Rule can’t be applied here.

Example : Find

We have

The is WRONG. In fact, although the numerator sin x → 0 as x → π^-, notice that the denominator (1 - cos x) does not approach 0, so L’Hosptial Rule can’t be applied here. The required limit is easy to find, because the function is continuous at n and the denominator is nonzero here:

Example : Find

We have

Indeterminate Forms of Type ∞ - ∞ and 0 • ∞

Example : Find

We have
Note that
therefore

OR

Example : Find

We have

Example : Find

Note that  type of an indeterminate form. Put y = (tan x)^2x-n
then
In y = In ((tan x)^2x-π) = (2x - π) ln(tan x)

We have





Therefore

Example : Find

In short, y = x^x ⇒ In y = x In x =
We have
Therefore

Example : Find

Note that  (tan 5x)^x is 0° type of an indeterminate form. Put y = (tan 5x)^x
then
In y = ln(tan 5x)^x = x ln(tan 5x) =
We have

Therefore

Example : Find

Note that  is 0⁰ type of an indeterminate form. Put y = (sin 2x)^{tan 3x}
then In y = ln((sin 2x)^{tan 3x}) = tan 3x ln(sin 2x) =
We have

Therefore

Example : Find

Note that  is 1∞ type of an indeterminate form. Put y =  then
In y = In
We have




Therefore,

Example : Find  (1 + sin 7x)^cot.5x.

Note that  is 1^∞ type of an indeterminate form. Put y = (1 + sin 7x)^{cot 5x}
then
In y = ln((1 + sin 7x)^{cot 5x}) = cot 5x ln(1 + sin 7x) =
We have



Therefore