Krdytkyu

Here is the theorem :

"Let $Isubseteq mathbb{R}$ an interval which contains $ain mathbb{R}$. Let $M$ and $m$ two functions defined on $I$, differentiable at $a$ and $f$ a function defined on $I$ which takes value in $mathbb{R}$.

If :

$i) forall x in I , m(x)le f(x)le M(x)$

$ii) m(a) = f(a) = M(a)$

$iii) m'(a)= M'(a)$

Then $f$ is differentiable at $a$ and we have $f'(a)=m'(a)=M'(a)$"

Here is my proof :

I consider the function : $xmapsto frac{f(x)-f(a)}{x-a}$ with $xne a$

Then $forall x in I setminus {a}$ by using $i)$ we have directly :

$frac{m(x)-f(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-f(a)}{x-a}$ if $x-a>0$ (for the other case we just have to change the sign of the inequality)

But by $ii)$ we have : $m(a)=f(a)=M(a)$ so we obtain :

$frac{m(x)-m(a)}{x-a} le frac{f(x)-f(a)}{x-a} le frac{M(x)-M(a)}{x-a}$

$Rightarrow$
$m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$

Then by using $iii)$ we have : $m'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le m'(a)$ or $M'(a) le lim limits_{xto a} frac{f(x)-f(a)}{x-a} le M'(a)$

So we prove that $f$ is differentiable at $a$ (by the Squeeze theorem the limit gives $m'(a)$ or $M'(a)$) and we have $f'(a)=m'(a)=M'(a)$

The theorem is proved.

Am I right ? Thanks in advance.

It's correct. just a formal thing: you can't move to the inequality
$$
m^prime(a)leqlim_{xrightarrow a}frac{f(x)-f(a)}{x-a}leq M^prime(a)
$$
Because you haven't proved that the limit exists. you need to move from the first inequality straight to the conclusion.