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

[HaLAA]

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.

Homework Equations

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]

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

 

[geoffrey159]

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

 

[Ray Vickson]

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.

Homework Equations

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?

 

[HaLAA]

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

 

[HaLAA]

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]

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.

 

[geoffrey159]

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