Exercise applying mean value theorem and Intermediate value theorem let $f:[0,a]\to \mathbb{R}$ of class $C^1$ such that $f(0)=0$
show that 
$$\exists c\in ]0,a[, f'(c)=\frac{2f(a)+af'(a)}{3a}$$
First I apply mean value theorem then $\exists b\in]0,a[, f'(b)=\frac{f(a)-f(0)}{a-0}=\frac{f(a)}{a}$
then $f(a)=af'(b)$
so 
$\frac{2f(a)+af'(a)}{3a}=\frac23 f'(b)+\frac13 f'(a)$
How to prove that $\frac23 f'(b)+\frac13 f'(a)$ is between $f'(a)$ and $f'(b)$?
 A: In general, $\lambda x + (1 - \lambda)y$ lies between $x$ and $y$, when $\lambda \in [0, 1]$. In your case, $\lambda = 1/3$, $x = f'(a)$, and $y = f'(b)$.
To prove this, suppose that $x \le y$. Then,
$$\lambda x + (1 - \lambda)y - x = (\lambda - 1)x + (1 - \lambda)y = (y - x)(1 - \lambda).$$
Since $\lambda \le 1$ and $x \le y$, this is a product of positive numbers, hence it is positive. Thus,
$$\lambda x + (1 - \lambda)y - x \ge 0 \implies x \le \lambda x + (1 - \lambda)y.$$
Similarly (or symmetrically, if you switch the roles of $x$ and $y$, and swap $\lambda$ with $1 - \lambda$), if $y \le x$, then
$$y - \left(\lambda x + (1 - \lambda)y\right) = \lambda(y - x),$$
which is again positive. Thus,
$$\lambda x + (1 - \lambda)y \le y.$$
A: If $f'(b)=f'(a)$, it is easy. Otherwise, define
$$ g(x)=f'(x)-\bigg[\frac23 f'(b)+\frac13 f'(a)\bigg]. $$
Then $g(x)$ is continuous in $[b,a]$ and
$$ g(b)g(a)=-\frac{2}{9}(f'(b)-f'(a))^2<0. $$
By the IMVT, there exists $c\in(b,a)$ such that $g(c)=0$ or
$$ f'(c)=\frac23 f'(b)+\frac13 f'(a). $$
