## Spring 2004 #5

Problem Statement: Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. Let $a,b\in \mathbb{R}$ such that $a and $f'(a)\neq f'(b)$. If $N$ is between $f'(a)$ and $f'(b)$ then there exists $M\in (a,b)$ such that $f'(M)=N$.

Proof: Wlog assume $f'(a). Fix $N\in (f'(a),f'(b))$. Define $g(x)=Nx-f(x)$. Then $g'(x)=N-f'(x)$. Note that $g'(a)=N-f'(a)>0$ and $g'(b)=N-f'(b)<0$ since $f'(a).

$g'(a)>0$ and so for $x$ sufficiently close to $a$, $g(x)>g(a)$ for $x>a$. Similarly, $g'(b)<0$ and so for $x$ sufficiently close to $x$, $g(x)>g(b)$ for $x

Since $g$ is continuous on $[a,b]$, which is a compact set, we know that $g$ attains both it’s max and min on $[a,b]$. We know from above that $g$  does not attain it’s max when $x=a$ or $x=b$. So, let $M\in (a,b)$ such that $g(M)\geq |g(x)|$ for every $x\in [a,b]$. Then $g(M)$ is the maximum, thus, $g'(M)=0$. Thus, $g'(M)=N-f'(M)=0$ and so it follows that $N=f'(M)$

$\Box$

Reflection: This one was not straightforward at all. I think that after doing this problem I would see the trick, but right away I didn’t see it. When I tried to attack it directly I didn’t make any real progress on the proof. So, maybe this is a good technique for me to have up my sleeve; when trying to prove something about the derivative create a function you know information about.