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

  • Thread starter HaLAA
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In summary: Can you clearly explain the link between ##h(x)## and ##f'(0) ## ?Yes, the link between the two is clear.
  • #1
HaLAA
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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.
 
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  • #2
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
what is the sign of ##\frac{f(x) - f(0)}{x} ## for ##x## positive and negative ?
 
  • #4
HaLAA said:

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?
 
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  • #5
Ray Vickson said:
Your function is differentiable on [-1,1] and has a maximum over [-1,1] at the point x = 0. What does that tell you?
Ray Vickson said:
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
andrewkirk said:
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
HaLAA said:
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
HaLAA said:
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:

1. What does f'(0)=0 mean?

f'(0)=0 means that the derivative of the function f(x) at x=0 is equal to 0. This indicates that at x=0, the function has a horizontal tangent line and is neither increasing nor decreasing.

2. How do you determine if f'(0)=0 for a given function?

To determine if f'(0)=0, you can take the derivative of the function and set it equal to 0. If the resulting equation has a solution of x=0, then f'(0)=0. Alternatively, you can graph the function and visually inspect the slope at x=0 to see if it is equal to 0.

3. Does f'(0)=0 guarantee that f(x) is a maximum or minimum at x=0?

No, f'(0)=0 does not guarantee that f(x) is a maximum or minimum at x=0. It only indicates that the function has a horizontal tangent line at x=0. To determine if f(x) has a maximum or minimum at x=0, you would need to use the first or second derivative test.

4. Can f'(0)=0 for a function where f(x)<=f(0) for all x in [-1,1]?

Yes, it is possible for f'(0)=0 and f(x)<=f(0) for all x in [-1,1]. This would indicate that the function is constant on the interval [-1,1] and has a horizontal tangent line at x=0.

5. How does f'(0)=0 relate to the behavior of the function on the interval [-1,1]?

f'(0)=0 indicates that the function has a horizontal tangent line at x=0, but it does not necessarily indicate how the function behaves on the interval [-1,1]. To determine the behavior of the function, you would need to analyze the function on the entire interval using techniques such as the first and second derivative tests.

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