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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?

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

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

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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.

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