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 $f'(c)=0$

Question

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!

Answer 1

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 ...

Answer 2

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.

A contradiction.

Hence there is a point

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

Answer 3

Due to the continuity of your function for any change of sign you get a zero and between two consecutive zeros you get at least one critical point.

You have two change of signs and your function is differentiable. Thus you have at least two zeros for your function and one zero for your derivative.

Answer 4

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.