Posted by **Someone** on Wednesday, January 11, 2017 at 11:20am.

if f(x)= |(x^2-9)(x^2+1)| how many numbers in the interval [-1,1] satisfy the mean value theorem?

None

1

2

3

Will someone please explain in detail?

- Calculus -
**Steve**, Wednesday, January 11, 2017 at 12:00pm
absolute value problems can get tricky, because the functions are not differentiable at any cusps.

However, since (x^2-9)(x^2+1) is negative on [-1,1],

f(x) = -(x^2-9)(x^2+1)

This is a nice smooth polynomial, so it satisfies the requirements of the MVT on [-1,1].

Now take a look at the graph:

http://www.wolframalpha.com/input/?i=%7C(x%5E2-9)(x%5E2%2B1)%7C

Since f(-1) = f(1) we see that there is a single point where f'(c) = 0 on [-1,1].

To show that algebraically, we just go through the steps:

(f(1)-f(-1))/2 = 0

f'(x) = 4x(4-x^2)

f'(0) = 0

- Calculus -
**Ke$ha**, Wednesday, January 11, 2017 at 12:04pm
so the answer is none?

- Calculus -
**Steve**, Wednesday, January 11, 2017 at 12:08pm
Excuse me?

Did you look at the graph?

Did you actually read what I wrote?

I don't mind helping you arrive at a solution, but I do expect you to actually read it, rather than just hold out your hand and say "gimme the answer." In fact, I **did** give you the answer.

- Calculus -
**Ke$ha**, Wednesday, January 11, 2017 at 12:12pm
Yes I read it I still don't get it and 0 isn't an answer choice.

- Calculus -
**Steve**, Wednesday, January 11, 2017 at 12:19pm
That is because x=0 is the single point where f'(x) = 0

the number zero is a single choice, not an absence of choices!

The answer is 1

There is 1 (one) point on the interval [-1,1] where

f(1)-f(-1)

------------ = 0

1 - (-1)

That point is at x=0.

Just take a deep breath and read what is written.

- Calculus -
**Ke$ha**, Wednesday, January 11, 2017 at 12:20pm
wait its 3 right?

- Calculus -
**Ke$ha**, Wednesday, January 11, 2017 at 12:20pm
oh ok thank you!!!! sorry

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