Question

Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b). (Select all that apply.)

√[(2-x^(2/3)]^3 [-1,1]

1. None of these.
2. f(a) does not equal f(b) for all possible values of a and b in the interval [π, 3π].
3. There are points on the interval [a, b] where f is not continuous.
4. f '(a) does not equal f '(b) for any values in the interval [π, 3π].
5. There are points on the interval (a, b) where f is not differentiable.

Answer 1

Do you remember what conditions you need to have before you can apply Rolle's theorem?

Answer 2

I've never used Rolle's Theorem but I think there are 4 conditions

Answer 3

There's usually only three conditions. Rolle's theorem says that "if a real-valued function f is continuous on a closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists a c in the open interval (a, b) such that f'(c)=0."

So the three conditions are
1. Continuous on the closed interval [a,b]
2. Differentiable on the open interval (a,b)
3. f(a)=f(b).

Let's go through these one by one. First, is your function continuous on the interval [-1,1]?

Answer 4

yes

Answer 5

Good. Let's skip to the other easy property to check. Does f(-1)=f(1)?

Answer 6

no

Answer 7

I got √27

Answer 8

Just to make sure I'm thinking of the function correctly, the function is\[\Large\sqrt{(2-x^{2/3})^3}\]correct?

Answer 9

right

Answer 10

If we put \(x=1\) into the function, we get\[\Large\sqrt{(2-1^{2/3})^3}=\sqrt{(2-1)^3}=1\]If we put \(x=-1\), we get\[\Large\sqrt{(2-(-1)^{2/3})^3}=\sqrt{(2-1^{1/3})^3}=\sqrt{(2-1)^3}=1\]So they are equal.

Answer 11

ok

Answer 12

does that satisfy #3?

Answer 13

f(1)=1

Answer 14

Sorry for the delay, but yes. That means the functions satisfies the third condition. Now we just have to see if it's differentiable. Can you tell me what the derivative is?

Answer 15

√[2-x^(2/3)]
------------
3√x

Answer 16

- 4
-------------
[27√2-x^(8/27)] (x^(19/27)

Answer 17

The first one looked right to me. So using the first one, can you tell me where the derivative isn't defined?

Answer 18

at 0

Answer 19

Right. And 0 is inside the interval (-1,1). Therefore, the function is not differentiable on the interval (-1,1), and so it doesn't satisfy the second condition. And that means we can't use Rolle's theorem.

Answer 20

ok

Answer 21

choice 5 then

Answer 22

thanks

Answer 23

You're welcome.