# Using Exponentials

• Feb 21st 2011, 04:47 PM
zebra2147
Using Exponentials
I could use some help getting started on this proof. Any help would be appreciated.

Prove that if $f:[0,1]\rightarrow \mathbb{R}$ is differentiable and $f(0)=f(1)=0$, then for any $a\in \mathbb{R},\exists c\in (0,1)$ such that $f'(c)=af(c)$.

I was told a hint that I have to apply Rolle's Theorem to $f(x)e^{ax}$ but I'm not sure how to work that in yet...
• Feb 21st 2011, 04:52 PM
DrSteve
If $a\in \mathbb{R}$, let $b=-a$. By Rolle's Theorem, there is a c in (0,1) with $f'(c)e^{bc}+bf(c)e^{bc}=0$.
• Feb 21st 2011, 04:54 PM
Chris L T521
Quote:

Originally Posted by zebra2147
I could use some help getting started on this proof. Any help would be appreciated.

Prove that if $f:[0,1]\rightarrow \mathbb{R}$ is differentiable and $f(0)=f(1)=0$, then for any $a\in \mathbb{R},\exists c\in (0,1)$ such that $f'(c)=af(c)$.

I was told a hint that I have to apply Rolle's Theorem to $f(x)e^{ax}$ but I'm not sure how to work that in yet...

First note that if $g(x)=f(x)e^{ax}$, then $g(0)=f(0)=0$ and $g(1)=f(1)e^{a}=0$ since $f(0)=f(1)=0$. Now, by Rolle's theorem, $\exists c\in(0,1):g^{\prime}(c)=0$.

Note that $g^{\prime}(x)=f^{\prime}(x)e^{ax}+af(x)e^{ax}$. So $g^{\prime}(c)=0\implies\ldots$

Can you finish this problem?
• Feb 21st 2011, 05:19 PM
zebra2147
This implies that $f^{\prime}(x)e^{ax}+af(x)e^{ax}=0$. Therefore,
$f^{\prime}(x)e^{ax}=-af(x)e^{ax}$ Which implies $f^{\prime}(x)=-af(x)$
• Feb 22nd 2011, 09:11 AM
Plato
Quote:

Originally Posted by zebra2147
Prove that if $f:[0,1]\rightarrow \mathbb{R}$ is differentiable and $f(0)=f(1)=0$, then for any $a\in \mathbb{R},\exists c\in (0,1)$ such that $f'(c)=af(c)$.
I was told a hint that I have to apply Rolle's Theorem

Use $\displaystyle g(x)=f(x)\cdot e^{-ax}$