If f:[-5, 5] → R is a differentiable function and f’(x) does not vanish anywhere. Then, prove that f(-5) ≠ f(5).

Given: f:[-5, 5] → R is a differentiable function and f’(x) does not vanish anywhere

To prove: f(-5) ≠ f(5)

Since f(x) is differentiable function in [-5, 5]

We know, every differentiable function is continuous too

⇒ f(x) is continuous and differentiable function in [-5, 5]

Lagrange’s mean value theorem states that if a function f(x) is continuous on a closed interval [a,b] and differentiable on the open interval (a,b), then there is at least one point x=c on this interval, such that

f(b)−f(a)=f′(c)(b−a)

This theorem is also known as First Mean Value Theorem.

As, f’(x) does not vanish anywhere

⇒ f’(x) ≠ 0 for any value of x

Thus f’(c) ≠ 0

⇒ f(5) – f(-5) ≠ 0 × 10

⇒ f(5) – f(-5) ≠ 0

⇒ f(5) ≠ f(-5)

Hence Proved