Does $[−2, 3]\subset \operatorname{Im} f'$ for the defined function? I'm trying to prove that if a function
$$
f : [−1, 1] \rightarrow \mathbb{R}$$
 is continuous in 
$[−1, 1],\phantom{2}$ differentiable in
$(−1, 1)$ and verifies
$$
f(−1) = 1,\phantom{1} f(0) = −1, \phantom{1} f(1) = 2
$$
Then the interval $[−2, 3]$ is contained on the image of the derivative $f'(x)$.
I tried to solve it using Intermediate value theorem since 
$f(-1) \gt f(0) \Rightarrow \forall k \in (-1,1) \phantom{2}\exists c \in (-1,0):f(c) = k $ 
$f(0) \lt f(1) \Rightarrow \forall k \in (-1,2) \phantom{2}\exists c \in (0,1):f(c) = k $ 
But I didn't get nothing at all, any suggestions?
 A: Define 
$$ q(x)=f(x)-f(x-1)$$
on $[0,1]$. Then $q$ is continuous, $q(0)=-2$, $q(1)=3$. If $-2\le c\le 3$, IVT gives us $\xi\in[0,1]$ with $q(\xi)=c$. Then MVT gives us $\eta\in (\xi-1,\xi)$ with $f'(\eta)=q(\xi)=c$.
A: Well, as it is differentiable and the slope between $-1$ and $0$ is $-2$, you can say by the mean value theorem that there exists $c_1 \in (-1,0)$ for which $f'(c_1) = -2$. Same thing between $0$ and $2$ where the slope is $3$, there exists $c_2 \in (0,1à)$ for which $f'(c_2) = 3$.
You can conclude using Darboux's theorem, that states that the image of and interval by the derivative of a function is an interval. So $[-2,3] \subset f'\left((-1,1)\right)$. But it is not a trivial theorm!
If moreover $f$ is supposed to be $\mathcal{C}^1$, you can conclude directly because $f'$ will be continuous.
A: You are using IVP (intermediate value property) for $f$. However, it would be more beneficial to use it for $f'$.
However, note very carefully that $f$ being differentiable would not imply $f'$ is continuous and so, we cannot use IVP directly.
Fortunately, we have Darboux's Theorem which states that the derivative satisfies IVP anyway.

Now, note that by mean value theorem, $f'(x) = -2$ for some $x \in (-1, 0)$ and $f'(x) = 3$ for some $x \in (0, 1)$.
To see the above, just note that
$$\dfrac{f(0) - f(-1)}{0 - (-1)} = -2;\qquad \dfrac{f(1) - f(0)}{1 - 0} = 3.$$
Now, just use Darboux's theorem to conclude.
