# Does f'(0)=0 where x in [-1,1] and f(x)<=f(0)

## Homework Statement

If the function f:ℝ→ℝ is differentiable and f(x)<=f(0) for all x ∈[-1,1], then f'(0)=0. True or False.

## The Attempt at a Solution

I think the statement is right. Since f(x)<= f(0) for all x in [-1,1], this tells us f is an even function or a symmetric function, then apply Rolle's theorem, we can prove the statement.

andrewkirk
Homework Helper
Gold Member
The function is not necessarily even or symmetric. Consider f(x) that is ##x^3## for ##x\leq 0## and ##-x^2## for ##x>0##.

what is the sign of ##\frac{f(x) - f(0)}{x} ## for ##x## positive and negative ?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

If the function f:ℝ→ℝ is differentiable and f(x)<=f(0) for all x ∈[-1,1], then f'(0)=0. True or False.

## The Attempt at a Solution

I think the statement is right. Since f(x)<= f(0) for all x in [-1,1], this tells us f is an even function or a symmetric function, then apply Rolle's theorem, we can prove the statement.

Your function is differentiable on [-1,1] and has a maximum over [-1,1] at the point x = 0. What does that tell you?

• geoffrey159
Your function is differentiable on [-1,1] and has a maximum over [-1,1] at the point x = 0. What does that tell you?
Your function is differentiable on [-1,1] and has a maximum over [-1,1] at the point x = 0. What does that tell you?
let h(x)=(f(x)-f(0))/x, when x>0, h(x)<=0; when x<0, h(x)=>0; f(x) is bounded above and the supremum is at x = 0, so f'(0)=0

The function is not necessarily even or symmetric. Consider f(x) that is ##x^3## for ##x\leq 0## and ##-x^2## for ##x>0##.

At x =0, we still have f'(0)=0?

andrewkirk
Homework Helper
Gold Member
At x =0, we still have f'(0)=0?
Yes, the conclusion still holds. In the OP you reached the correct conclusion based on an incorrect argument. Ray's and Geoffrey's posts indicate the direction that a valid argument would take.

let h(x)=(f(x)-f(0))/x, when x>0, h(x)<=0; when x<0, h(x)=>0; f(x) is bounded above and the supremum is at x = 0, so f'(0)=0

Can you clearly explain the link between ##h(x)## and ##f'(0) ## ?

EDIT: By the way, my hint is valid, but honestly, Mr Vickson's hint is more elegant

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