1
$\begingroup$

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!

$\endgroup$
  • 2
    $\begingroup$ Your first approach works just fine. $\endgroup$ – Parcly Taxel Mar 3 '18 at 18:50
  • $\begingroup$ 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. $\endgroup$ – Mehrdad Zandigohar Mar 3 '18 at 18:52
1
$\begingroup$

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 ...

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ OK, thank you! Is it even possible to use the generalized IVT theorem for this? $\endgroup$ – rose12 Mar 4 '18 at 9:50
  • $\begingroup$ @jonny1245: what's generalized IVT? My point is that there is no real need to invoke IVT. $\endgroup$ – Paramanand Singh Mar 4 '18 at 10:34
  • $\begingroup$ 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 $f(c)=r$ $\endgroup$ – rose12 Mar 4 '18 at 13:09
  • $\begingroup$ @jonny1245: for me this is same as IVT. And other answers have used it. $\endgroup$ – Paramanand Singh Mar 4 '18 at 14:17
1
$\begingroup$

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 $f'(c)=0$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.