Application of Rolle's Theorem

[mtayab1994]

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 .

 

[HallsofIvy]

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

 

[mtayab1994]

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

 

[mtayab1994]

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

\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}

and f'(0)=\lim_{h\rightarrow0}\frac{f(h)-f(0)}{h}=-\lim_{h'\rightarrow0^{+}}\frac{f(h)-f(0)}{h}
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