If f : [– 5, 5] → R is a differentiable function and if f′(x) does not vanish anywhere, then prove that f (– 5) ≠ f (5).

Given: f : [-5, 5] → R is a differentiable function.

Mean Value Theorem states that for a function f : [a, b] → R, if

(a)f is continuous on [a, b]

(b)f is differentiable on (a, b)

Then there exists some c ∈ (a, b) such that

We know that a differentiable function is a continuous function.

So,

(a) f is continuous on [-5, 5]

(b) f is differentiable on (-5, 5)

∴ By Mean Value Theorem, there exists c ∈ (-5, 5) such that

⇒

⇒ 10 f'(c) = f(5) – f(-5)

It is given that f'(x) does not vanish anywhere.

∴ f'(c) ≠ 0

10 f'(c) ≠ 0

f(5) – f(-5) ≠0

f(5) ≠ f(-5)

Hence proved.

By Mean Value Theorem, it is proved that f(5) ≠ f(-5).