Application of Rolle's Theorem

In summary, if f is an even function and is differentiable on R, the equation f'(x)=0 has at least one solution in R. Additionally, if f is an even and differentiable function on R and f' is continuous at 0, then f'(0)=0. This can be proved using Rolle's theorem and the definition of differentiability.
  • #1
mtayab1994
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Homework Statement



1) Prove that if f is an even function and is differentiable on R then the equation f'(x)=0 has at least a solution in R.

2) Conclude that if f is an even and differentiable function on R and f' is continuous at 0 than f'(0)=0.



The Attempt at a Solution



1)We know that f is an even function so that means that f(x)=f(-x) and it's differentiable on R. So that means that there exists either an Mam or a min value in R such that f'(m)=0 or f'(M)=0, and by substituting x for m we get f'(x)=0.

2)We know that f is even and differentiable on R so f has an axis of symmetry or a point of symmetry at the point 0. So f'(0)=0. Can someone tell me if this correct? If not what can i do to fix it. Thank you .
 
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  • #2
"Max" or "min" is not relevant. If f is an even function, then f(1)= f(-1). Now apply Rolle's theorem.

For 2, you can go back to the definition of "differentiable".

If f is differentiable at 0, [itex]f'(0)= \lim_{h\to 0} (f(h)- f(0))/h[/itex]. Now look at the limits "from above" and "from below". From below, [itex]\lim_{h\to 0^-} (f(h)- f(0))/h[/itex]. Let h= -h' and that becomes [itex]\lim_{h'\to 0^+}(f(h')- f(0))/h= \lim_{h\to 0^+} (f(-h)- f(0))/(-h)[/itex]. Because f is even, that is the same as [itex]\lim_{h\to 0^+}(f(h)- f(0)/(-h)= -\lim_{h\to 0}(f(h)- f(0))/h[/itex]. Now compare that with the limit from above.
 
  • #3
HallsofIvy said:
"Max" or "min" is not relevant. If f is an even function, then f(1)= f(-1). Now apply Rolle's theorem.

For 2, you can go back to the definition of "differentiable".

If f is differentiable at 0, [itex]f'(0)= \lim_{h\to 0} (f(h)- f(0))/h[/itex]. Now look at the limits "from above" and "from below". From below, [itex]\lim_{h\to 0^-} (f(h)- f(0))/h[/itex]. Let h= -h' and that becomes [itex]\lim_{h'\to 0^+}(f(h')- f(0))/h= \lim_{h\to 0^+} (f(-h)- f(0))/(-h)[/itex]. Because f is even, that is the same as [itex]\lim_{h\to 0^+}(f(h)- f(0)/(-h)= -\lim_{h\to 0}(f(h)- f(0))/h[/itex]. Now compare that with the limit from above.

If f is continuous and differentiable on R then it's differentiable on [-1,1] so we get f(1)=f(-1) like you said. And rolle's theorem states that if f is continuous on [-1,1] and is differentiable on the open interval (-1,1) than there exists a c in (-1,1) such that f'(c)=0. Correct right?
 
  • #4
HallsofIvy said:
"Max" or "min" is not relevant. If f is an even function, then f(1)= f(-1). Now apply Rolle's theorem.

For 2, you can go back to the definition of "differentiable".

If f is differentiable at 0, [itex]f'(0)= \lim_{h\to 0} (f(h)- f(0))/h[/itex]. Now look at the limits "from above" and "from below". From below, [itex]\lim_{h\to 0^-} (f(h)- f(0))/h[/itex]. Let h= -h' and that becomes [itex]\lim_{h'\to 0^+}(f(h')- f(0))/h= \lim_{h\to 0^+} (f(-h)- f(0))/(-h)[/itex]. Because f is even, that is the same as [itex]\lim_{h\to 0^+}(f(h)- f(0)/(-h)= -\lim_{h\to 0}(f(h)- f(0))/h[/itex]. Now compare that with the limit from above.

[tex]\lim_{h\rightarrow0^{-}}\frac{f(h)-f(0)}{h}=\lim_{h'\rightarrow0^{+}}\frac{f(h)-f(0)}{-h}=-\lim_{h'\rightarrow0^{+}}\frac{f(h)-f(0)}{h}[/tex]

and [tex]f'(0)=\lim_{h\rightarrow0}\frac{f(h)-f(0)}{h}=-\lim_{h'\rightarrow0^{+}}\frac{f(h)-f(0)}{h}[/tex]
and those two are equal because we said h=-h' so the negative from h' will go away with the negative from the limit and we will get that the limit equals the limit. So therefore f'(0)=0
 

FAQ: Application of Rolle's Theorem

What is Rolle's Theorem?

Rolle's Theorem is a mathematical theorem that states that if a function is continuous on a closed interval and differentiable on the open interval, and if the function's values are equal at both endpoints of the interval, then there must be at least one point within the interval where the derivative of the function is equal to zero.

What is the significance of Rolle's Theorem?

Rolle's Theorem is significant because it provides a way to prove the existence of a point where the derivative of a function is equal to zero, which can have important applications in finding maximum and minimum values of a function.

How is Rolle's Theorem used in calculus?

Rolle's Theorem is used in calculus to prove the Mean Value Theorem, which states that if a function is continuous on a closed interval and differentiable on the open interval, then there must be at least one point within the interval where the slope of the tangent line is equal to the average rate of change of the function over the interval.

What are some real-world applications of Rolle's Theorem?

Rolle's Theorem has applications in various fields such as physics, engineering, and economics. For example, it can be used to find the maximum and minimum values of a function that represents the trajectory of a projectile, the temperature distribution in a metal rod, or the profit function of a company.

Are there any limitations to Rolle's Theorem?

Yes, Rolle's Theorem has some limitations. It can only be applied to functions that are continuous on a closed interval and differentiable on the open interval. Additionally, it does not provide any information about the location of additional points where the derivative is equal to zero, which can be important in some applications.

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