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

  • Thread starter HaLAA
  • Start date
  • #1
85
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.

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.
 

Answers and Replies

  • #2
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,928
1,492
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##.
 
  • #3
535
72
what is the sign of ##\frac{f(x) - f(0)}{x} ## for ##x## positive and negative ?
 
  • #4
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

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?
 
  • Like
Likes geoffrey159
  • #5
85
0
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
 
  • #6
85
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?
 
  • #7
andrewkirk
Science Advisor
Homework Helper
Insights Author
Gold Member
3,928
1,492
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.
 
  • #8
535
72
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
 
Last edited:

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

  • Last Post
Replies
2
Views
5K
Replies
1
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
13
Views
1K
Replies
5
Views
13K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
8
Views
2K
Replies
6
Views
6K
Replies
7
Views
1K
Top