Question

If f[-55]->R,: is a differentiable function and if f'(x) does not vanish anywhere, then prove that f(-5)≠f(5)

Answer 1

$\textbf{\underline{Mean Value Theorem states that for a function}}$
$\bf{f:[a,b]\rightarrow\mathbb{R}}$ if
(a)f is continuous on [a, b]
(b)f is differentiable on (a, b)
Then there exists some c ∈ (a, b) such that 
$\bf{f'(c)=\frac{f(b)-f(a)}{b-a}}$

given , f(x) is differentiable function.
we know if any function is differentiable then it means, function must be continuous.
so, f(x) is also continuous.
∴ By Mean Value Theorem, there exists c ∈ (-5, 5) such that f'(c) = {f(5) - f(-5)}/(5 + 5)
=> 10f'(c) = f(5) - f(-5) --------(1)
It is given that f'(x) does not vanish anywhere.
∴ f'(c) ≠ 0
=> f(5) - f(-5) ≠ 0 [from equation (1) ]
=> f(5) ≠ f(-5) $\textbf{\underline{hence proved}}$

Answer 2

Answer:

this is the correct answer