Prove that if $f$ is differential at $[0,1]$ and $0\le f'(x)\le 2$ for every $x\in [0,1]$ there exists $x\in [0,1]$ where $f'(x) = x^2+x$

Question

I want to prove that if $f$ is differential at $[0,1]$ and $0\le f'(x)\le 2$ for every $x\in [0,1]$ there exists $x\in [0,1]$ where $f'(x) = x^2+x$.

Steps:

1. define $g(x) = f(x)-\frac{1}{3}x^3-\frac{1}{2}x^2$. $g$ is differential at $[0,1]$ as the sum of a differential function and a polynom.
2. $g'(x) = f'(x) -(x^2+x)$.

3. $g'(0) = f'(0) \implies 0\le g'(0)\le 2$.

   $g'(1) = f'(1) - 2 \implies -2 \le g'(1) \le 0$

4. if $g'(0) = 0$ or $g'(1)=0 \implies $ done.

5. $g'(1) < 0 < g'(0)$, and according to Darboux's theorem, there exists $c\in (0,1)$ which holds: $g'(c) = 0$. $x=c \implies$ done.

Is this proof correct?