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).


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