# Let $f:[-1,1]\to\Bbb R$ be a continuous and differentiable function.prove that there exists a point $c\in(-1,1),$ such that $f'(c)=0$

Let $f:[-1,1]\to\Bbb R$ be a continuous and differentiable function. Assume that $f(-1)=\pi, f(0)= -3,f(1)=1.$ prove that there exists a point $c\in(-1,1),$ such that $f'(c)=0$  I know that $f(-1) \cdot f(0) <0$, and $f(0) \cdot f(1) <0$ , but I don't know if I can use the intermediate value theorem to say that there exists 2 points $-1<c<0<d<1$ s.t $f(c)=f(d)=0$ and use Rolle theorem to say that there exists $c \in (c,d)\subset(-1,1)$ s.t $f'(e)=0$  I also tried to use the generalized intemediate value theorem by : $f(-1)=\pi >1, f(0)=-3 <0$, but my interval is [-1,0] and $f(a)<r<f(b)\to f(-1)\not\lt 1 \not\lt f(0) = \pi\not\lt1\not\lt-3$.$$-$$

Thank you!

• Your first approach works just fine. – Parcly Taxel Mar 3 '18 at 18:50
• Why do you think you may not be able to use the intermediate value theorem? That's alright and your first method is the answer. – Mehrdad Zandigohar Mar 3 '18 at 18:52

As you said,

$f(-1) \cdot f(0) <0$, and $f(0) \cdot f(1) <0$

so the intermediate value theorem will gives you $a\in (-1,0)$ and $b\in (0,1)$ such that $f(a)=f(b)=0$ (with $a<b$).

You can then use Roll's theorem for $f:[a,b]\rightarrow \mathbb{R}$ ($f$ is continuous on $[a,b]$ and differentiable on $(a,b)$) which tells you the existence of $c$ ($\in (a,b) \subset(-1,1$)) such that $$f'(c)= 0.$$

So basically it is exactly what you said ...

This is where the proof (and not just the statement) of Rolle's theorem comes into play. Since $f(0)<f(-1)$ and $f(0)<f(1)$ it follows that $f$ attains its minimum value at some interior point $c\in(-1,1)$. Since $f$ is differentiable at $c$ by principle of minima we have $f'(c) =0$. There is no need to invoke additional theorems like IVT.

• OK, thank you! Is it even possible to use the generalized IVT theorem for this? – rose12 Mar 4 '18 at 9:50
• @jonny1245: what's generalized IVT? My point is that there is no real need to invoke IVT. – Paramanand Singh Mar 4 '18 at 10:34
• I understand. The generalized IVT claims that if $f: [a,b] \to \Bbb R$ a continuous function and there is $r\in \Bbb R$ such that $f(a)<r<f(b)$, then there exists a point $a<c<b$, for which $f(c)=r$ – rose12 Mar 4 '18 at 13:09
• @jonny1245: for me this is same as IVT. And other answers have used it. – Paramanand Singh Mar 4 '18 at 14:17

An option:

Assume there is no point

$c \in (-1,1)$ with $f'(c)=0$.

Then:

1) $f'(x) \gt 0$ for $x\in (-1,1)$, or

2) $f'(x) \lt 0$ for $x\in (-1,+1).$

1) $f$ is strictly increasing in $(-1,1)$ .

Ruled out, look at the given points.

2) $f$ is strictly decreasing in $(-1,1)$.

Ruled out, look at the given points.

$c \in (-1,1)$ with $f'(c)=0$.