let-fx-1-x23-show-that-f-1-f1-but-there-is-no-number-c-in-1-1-such-that-fc-0-why-does-this-not-contr

Name: let fx 1 x23 show that f 1 f1 but there is no number c in 1 1 such that fc 0 why does this not contr
Uploaded: 2017-12-19
Duration: 5 min 9 s
Channel: Numerade
Description: Let $ f(x) = 1 - x^{2/3} $. Show that $ f(-1) = f(1) $ but there is no number $ c $ in $ (-1, 1) $ such that $ f'(c) = 0 $. Why does this not contradict Rolle's Theorem?

Question

Let $ f(x) = 1 - x^{2/3} $. Show that $ f(-1) = f(1) $ but there is no number $ c $ in $ (-1, 1) $ such that $ f'(c) = 0 $. Why does this not contradict Rolle's Theorem?

Fahad Paryani · Accepted Answer

So now they&#x27;re the function affect because one minus X to the two third. And we&#x27;re being has to show that effort negative or is equal after one. But there&#x27;s no numbers. CNN won the one shot that promise city. And the question is, why does this not contradict rolls, too? So, um, just by looking at dysfunction, we know that dysfunction is continuous in this animal because this is a is actually a square root, some sort of brute function. And we can have values from negative one to one because this is a cube groups. The Coopers are defined in all on all real numbers. So we know that this is continuous, that&#x27;s for sure. Uh, we cannot assume different ability. This is a little tricky. We&#x27;re going to talk about this in the second, so we&#x27;re gonna come back to to oops. Sorry about that. Um and then we&#x27;re going to shank of after negative one equals that one right away. So if we plug in value of negative one a negative one into our functions or one minus negative. One two to third. Oh, and since since this two on top squares the inside and the coupe root of positive one is one. It&#x27;LL just be one minus one, which is Seo. And then half of one will be the same thing we wanted to third, which will be one minus one Should authors you got a zero. So now, now we know that this last condition is satisfied. So the defense ability over negative ones one is a little tricky in this case, we can assume it in this case right away. So why don&#x27;t we just first take the derivative? So after primary Rex. So now all we do is apply. Probably normal rules and justice. Looking at news, we&#x27;re going to spring this down, So this will be CEO, bring down to third, and then we subtract one. The two third minus one is just negative one third, which is this is the same thing as negative to remember. There&#x27;s a negative right here. I know this is Yeah, because we bring down to negative one toe. Important note eyes. This is the next to the one third. And now there&#x27;s after something very interesting going on in this function. Um, since this value of X, um, if you actually draw the function the cube root of ax. There is no value of excess that will give us zero. Um, because even if you think about that, just looking at this function negative too. Three x one third we can&#x27;t solve for X equals zero like we can&#x27;t find a value where X is equal to zero. Because when we multiply this, I&#x27;m three extra one third, we want to buy a zero and we and then we&#x27;re left with negative equals zero, which is just absolute. This is just nonsense doesn&#x27;t make sense. So we know that there&#x27;s no value of excess will give us the value that equals zero. So although there is a derivative, it is not defensible from negative one to one. So what&#x27;s actually happening here is that this second condition is not satisfied. And so initially seems like there is a It is defensible and we could take to evident, but there&#x27;s no value. There&#x27;s no point in which there it is equal to zero have no, um, there&#x27;s no f prime of C that equals zero. So this actually doesn&#x27;t contradict roasted because the second condition is not even satisfied because this tells this is asking us that it has to be defensible. I&#x27;m negative Wonder one. But there is no value from negative on the one. There&#x27;s no X equals zero has no value of effort. Zero that will give us this conclusion. So this Khun, this conditions cannot be satisfied. This is why does not contradict rule still.

Problem

Let $ f(x) = \tan x $. Show that $ f(0) = f(\pi) …

Problem 9 Medium Difficulty

Let $ f(x) = 1 - x^{2/3} $. Show that $ f(-1) = f(1) $ but there is no number $ c $ in $ (-1, 1) $ such that $ f'(c) = 0 $. Why does this not contradict Rolle's Theorem?

Answer

$f(x)=1-x^{2 / 3} . \quad f(-1)=1-(-1)^{2 / 3}=1-1=0=f(1) . \quad f^{\prime}(x)=-\frac{2}{3} x^{-1 / 3},$ so $f^{\prime}(c)=0$ has no solution. This
does not contradict Rolle's Theorem, since $f^{\prime}(0)$ does not exist, and so $f$ is not differentiable on (-1,1)

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Catherine R.

Anna Marie V.

Heather Z.

Michael J.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 4

Video Transcript

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Michael J.

Integration Review - Intro

Definite vs Indefinite

$f(x)=1-x^{2 / 3} . \quad f(-1)=1-(-1)^{2 / 3}=1-1=0=f(1) . \quad f^{\prime}(x)=-\frac{2}{3} x^{-1 / 3},$ so $f^{\prime}(c)=0$ has no solution. Thisdoes not contradict Rolle's Theorem, since $f^{\prime}(0)$ does not exist, and so $f$ is not differentiable on (-1,1)

Calculus: Early Transcendentals

Catherine R.

Anna Marie V.

Heather Z.

Michael J.

Integration Review - Intro

Definite vs Indefinite

Catherine R.

Anna Marie V.

Heather Z.

Michael J.

$f(x)=1-x^{2 / 3} . \quad f(-1)=1-(-1)^{2 / 3}=1-1=0=f(1) . \quad f^{\prime}(x)=-\frac{2}{3} x^{-1 / 3},$ so $f^{\prime}(c)=0$ has no solution. This
does not contradict Rolle's Theorem, since $f^{\prime}(0)$ does not exist, and so $f$ is not differentiable on (-1,1)