Rolle's Theorem and The Mean Value Theorem - CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

The reader must be familiar with the classical maxima and minima problems from calculus. For example, the graph of a differentiable function has a horizontal tangent at a maximum or minimum point. This is not quite accurate as we will see.
Definition : Let  : I → R, I an interval. A point x₀ ∈ I is a local maximum of  if there is a δ > 0 such that f (x) < f (x₀) whenever x ∈ I ∩ (x₀ - δ,x₀ + δ): Similarly, we can defiine local minimum.
Theorem 6.1 : Suppose  : [a; b] → R and suppose f has either a local maximum or a local minimum at x₀ ∈ (a; b). If  is differentiable at x₀ then f' (x₀) = 0.

Proof: Suppose f has a local maximum at x₀ ∈ (a; b). For small (enough) h, f (x₀ + h) < f (x₀): If h > 0 then 

Similarly, if h < 0, then 

By elementary properties of the limit, it follows that f' (x₀) = 0:

We remark that the previous theorem is not valid if x0 is a or b. For example, if we consider the function f : [0; 1] → R such that f (x) = x, then f has maximum at 1 but f '(x) = 1 for all x ∈ [0; 1]:

The following theorem is known as Rol le's theorem which is an application of the previous theorem.

Theorem 6.2 : Let f be continuous on [a; b], a < b, and differentiable on (a; b). Suppose f (a) = f (b). Then there exists c such that c ∈ (a; b) and f'(c) = 0.

Proof: If  is constant on [a; b] then f'(c) = 0 for all c ∈ [a; b]. Suppose there exists x ∈ (a; b) such that f (x) > f (a). (A similar argument can be given if f (x) < f (a)). Then there exists c ∈ (a; b) such that f (c) is a maximum. Hence by the previous theorem, we have f'(c) = 0.

Problem 1 : Show that the equation x¹³ + 7x³ - 5 = 0 has exactly one (real) root.

Solution : Let f (x) = x¹³ + 7x³ - 5. Then f(0) < 0 and f(1) > 0. By the IVP there is at least one positive root of f (x) = 0. If there are two distinct positive roots, then by Rolle's theorem there is some x₀ > 0 such that f '(x₀) = 0 which is not true. Moreover, observe that f (x) < 0 for x < 0.

Problem 2 : Let f and g be functions, continuous on [a; b]; differentiable on (a; b) and let f (a) = f (b) = 0: Prove that there is a point c ∈ (a; b) such that g'(c)f (c) + f'(c) = 0:

Solution : Define h(x) = f (x)e^g(x) . Here, h(x) is continuous on [a; b] and differentiable on (a; b). Since h(a) = h(b) = 0, by Rolle's theorem, there exists c ∈ (a; b) such that h'(c) = 0.

Since h'(x) = [f'(x) + g'(x)f (x)]e^g(x) and e^α ≠ 0 for any α ∈ R, we see that f'(c) + g'(c)f (c) = 0.

A geometric interpretation of the above theorem can be given as follows. If the values of a differentiable function f at the end points a and b are equal then somewhere between a and b there is a horizontal tangent. It is natural to ask the following question. If the value of f at the end points a and b are not the same, is it true that there is some c ∈ [a; b] such that the tangent line at c is parallel to the line connecting the endpoints of the curve? The answer is yes and this is essentially the Mean Value Theorem.

Theorem 6.3 : (Mean Value Theorem) Let  be continuous on [a; b] and differentiable on (a; b). Then there exists c ∈ (a, b) such that f (b) - f (a) = f'(c)(b - a):

Proof: Let

Then g(a) = g(b) = f (a). The result follows by applying Rolle's Theorem to g.

The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f'. For example, if we have a property of f' and we want to see the e®ect of this property on f, we usually try to apply the mean value theorem. Let us see some examples.

Example 1 : Let f : [a; b] → R be differentiable. Then f is constant if and only if f'(x) = 0 for every x ∈ [a; b].

Proof : Suppose that f is constant, then from the definition of f'(x) it is immediate that f'(x) = 0 for every x ∈ [a; b]

To prove the converse, let a < x < b. By the mean value theorem there exists c ∈ (a; x) such that f (x) - f (a) = f'(c)(x - a). Since f '(c) = 0, we conclude that f (x) = f (a), that is f is constant. (If we try to prove the converse directly from the definition of f ' (x) we wil l be in trouble.)

Example 2 : Suppose f is continuous on [a; b] and differentiable on (a; b).

(i) If f ' (x) ≠ 0 for al l x ∈ (a; b), then f is one-one (i.e, f (x) ≠ f (y) whenever x ≠ y).

(ii) If f ' (x) > 0 (resp. f ' (x) > 0) for all x ∈ (a, b) then f is increasing (resp. strictly increasing) on [a, b]. (We have a similar result for decreasing functions.)

Proof : Apply the mean value theorem as we did in the previous example. (Note that f can be one-one but f ' can be 0 at some point, for example take f (x) = x³ and x = 0.)

Problem 3 : Use the mean value theorem to prove that | sinx - siny | < | x - y | for all x, y ∈ R.

Solution : Let x; y ∈ R. By the mean value theorem sinx - siny = cosc (x - y) for some c between x and y. Hence | sinx - siny | < | x - y |.

Problem 4 : Let f be twice differentiable on [0; 2]. Show that if f (0) = 0, f (1) = 2 and f (2) = 4, then there is x₀ ∈ (0; 2) such that f '' (x₀) = 0.

Solution : By the mean value theorem there exist x1 2 (0; 1) and x2 2 (1; 2) such that

f '(x₁) = f (1) - f (0) = 2 and f '(x₂) = f (2) - f (1) = 2:

Apply Rolle's theorem to f ' on [x₁ , x₂].

Problem 5 : Let a > 0 and f : [-a; a] → R be continuous. Suppose f '(x) exists and f '(x) · 1 for all x ∈ (-a, a). If f (a) = a and f (-a) = -a, then show that f (x) = x for every x ∈ (-a, a).

Solution : Let g(x) = f (x) - x on [-a, a]. Note that g'(x) < 0 on (-a, a). Therefore, g is decreasing. Since g(a) = g(-a) = 0, we have g = 0. 

This problem can also be solved by applying the MVT for g on [-a, x] and [x, a].