Please help: calculus prob

Question

If f(x) = |(x² - 4)(x² + 2)|, how many numbers in the interval [0, 1] satisfy the conclusion of the Mean Value Theorem?

Option:

1. None
2. 1
3. 2
4. 3

Answer 1

If the conditions of the mean value theorem are met, then there is a value, c, such that:

f'(c) = f(b) - f(a) / b-a

We cannot evaluate the derivate of an absolute function easily. However, we can break an absolute value equation down into different functions that are more easily differentiable. If If f(x) = |(x² - 4)(x² + 2)|, the piecewise function is:

(x² - 4)(x² + 2) for x< -2 and x> 2

-(x² - 4)(x² + 2) for -2<x<2

Since the problem asks you about the interval from 0 to 1, you'll be working with f(x)=-(x² - 4)(x² + 2).

First, determine if the conditions are met:

1: f(x) must be continuous on the closed interval [0,1]

2: f(x) must be differentiable on the open interval (0,1)

Since quadratic functions are continuous and differentiable, the conditions are met. (Note: If 2 or -2 were part of the interval, conditions would not be met because of the point created by the absolute value).

Find: f'(c) = f(b) - f(a) / b-a

[-(c² - 4)(c² + 2)] ' = (9-8) / (1-0)

You can simplify the left side of the equation before taking the derivative or use product rule to find the derivative of f(c). I'll simplify.

[-c⁴-2c²+4c²+8 ]' = 1

-4c³+4c = 1

-4c³+4c-1 = 0

c = .27, c = .838 (by graphing)

Since there are two c values, that would be the answer.