Test: Lagrange's Mean Value Theorem - Civil Engineering (CE) MCQ

y = f(x) = 5x² + 10x in the internal x = (1, 2)
Since, the function y is continuous in the interval (1, 2), as well as is differentiable at each point so, from Lagrange mean value theorem there exist at least a point where:

Here, we have
a = 1, b = 2
So, for x = a = 1, we obtain:
y = f(a) = f(1) = 5(1)² + 10(1) = 15
and for x = b = 2
y = f(b) = f(2) = 5(2)² + 10(2) = 40
Therefore:

Newton–Gregory Forward Interpolation formula is given by
f(x) = y₀ + nΔy₀ + n(n-1)Δ²y₀/2! + n(n-1)(n-2) Δ³y₀ /3! + …..
This formula is obtained by the Newton’s Divided difference formula by substituting the intervals as h. This is done because we assume the intervals to be constant, that is, equally spaced.

Concept:
Lagrange MVT:
If a function f(x) is:
1. Continuous in closed interval a ≤ x ≤ b and
2. Differentiable in open interval (a, b) i.e, a < x < b.
Then there exist at most one value c of x lying in the open interval a < x < b such that, 
Calculation:
f(x) = x^1/3, x ∈ [-1, 1]
f(x) is continuous in [-1, 1]
Differentiating (1) with respect to x, we get

But, f'(0) → not defined
∴ f'(x) is not differentiable at every point in the -1 < x <1.
The condition for Langrage MVT is not satisfied and hence 'c' cannot be found.

The formula is
x = x₀ + nh.
Here x₀ is 0 as 0 is the first element and h is 1.
Since in the question it is given that we have to find f(0.2), x= 0.2.
So, substituting the values in the formula we get,
0.2 = 0 + n(1) .
Hence, n= 0.2.

Concept:
By lagrangian mean value theorem,

Calculation:
Given:
f(x) = x² – 2x + 2, x ϵ [1, 3]
By lagrangian mean value theorem

c = 2

Here, x₀ is 0, h is 0.5, x is 1.8.
Substituting the values in the formula
x = x₀ + nh,
1.8 = 0 + n(0.5)
n = 3.6.

Concept:
Let f(x) is a function define on [a ,b] such that, 

• f(x) is a Continuous on [a , b]
• f(x) is Differentiable on [a , b]

Then, there exist a real number C ∈ (a , b) such that, According to Lagrangian Mean Value Theorem,

Calculation:
Given:

The function satisfies Lagrange's Mean Value Theorem that means it satisfies two condition given above 1 and 2
Therefore for the value of C we can write down above formula 

Squaring on both sides,
3c² - 12 = c² 
2c² = 12
c² = 6
c = √6

Given
x₀ = 0.6
n = 2.6
h = 0.2
Substituting in the formula,
x = x₀ + nh
x = 0.6+(0.2)(2.6)
x = 1.12.

Concept:
Lagrange’s Mean Value Theorem:
If f(x) is real valued function such that –

• f(x) is continuous in the closed interval [a,b]
• (f(x) is differentiable in the open interval (a,b)
• f(a) ≠ f(b)

Then there exist at least one value x, c (a,b) such that –


Geometrical Interpretation:

• Between two points a and b, f(a) ≠ f(b) of the graph of f(x) then there exists one point where the tangent is parallel to the chord 

Explanation:
(a) is true with reference to geometrical interpretations.
(b) is false, if f(x) has same value for every value of x, it will violate f(a) ≠ f(b).

Given
x₀ = 0.75825
x = 0.759
h = 0.00005
Substituting in the formula,
x = x₀ + nh,
0.759 = 0.75825 + n(0.00005)
Therefore, n = 15.