## anonymous 5 years ago if f(x)=abs((x^2-12)(x^2+4)), how many numbers in the interval -2<x<3 satisfy the conclusion of the mean value theorem? thank you so much and please tell me how u got the answer.

1. anonymous

First, you need to put this function into something you can handle. You can use the definition of absolute value to do this:$f(x)=|(x^2-12)(x^2+4)|=\sqrt{[(x^2-12)(x^2+4)]^2}$so$f(x)=\sqrt{(x^2-12)^2(x^2+4)^2}$$=\sqrt{x^8-16 x^6-32 x^4+768 x^2+2304}$$=\sqrt{(x^4-8 x^2-48)^2}$Now, the mean value theorem says that if a function f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that$f'(c)=\frac{f(b)-f(a)}{b-a}$You need to find all those c's in (2,3) that satisfy,$f'(c)=\frac{f(3)-f(2)}{3-2}=f(3)-f(2)$...end part 1

2. nowhereman

only that its (-2, 3)

3. anonymous

lol I love how you always clean up the question.

4. anonymous

nowhereman, do you want to finish this? I want to go eat.

5. nowhereman

Ok, so you are looking for the points $c \in (-2, 3)$ so that $f'(c) = \frac{f(3) - f(-2)}{3-(-2)} = \frac{39-64}{5} = 5$ So we need to find the solutions of $\frac{2(x^4 - 8x^2-48)(4x^3 - 16x)}{2\sqrt{(x^4-8x^2-48)^2}} = 5$

6. nowhereman

Oh and we can do it a little bit easier, because $x \in [-2, 3] ⇒ x^2-12 < 0 ∧ x^2+4 > 0$$⇒ f(x) = |(x^2-12)(x^2+4)| = -(x^2-12)(x^2+4)$

7. nowhereman

So $f'(x) = -4x^3 + 16x$ and we need to solve $-4x^3 + 16x - 5 = 0$ by differentiating you get that the extrema of $-4x^3 + 16x - 5$ are at $\pm 2\sqrt{\frac{1}{3}}$

8. nowhereman

the function values are there $\pm32\frac{2}{3}\sqrt{\frac 1 3}-5$ one is bigger then zero and the other is smaller then zero. And so there must be 3 different solutions.

9. nowhereman

you should maybe draw the graph of f' or let some plotter do it for you, to get a better imagination of the situation.

10. anonymous

thank you both so much for providing such a detailed answer! u have saved my calc grade!

11. anonymous

You're welcome