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?

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 $$f'(\eta)=q(\xi)=c$$.

• Btw, you can refine this argument to a proof of the Darboux theorem mentioned in comments – Hagen von Eitzen May 27 at 13:08

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.

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.