Question

Given the function `f(x)= abs((x^2-12)(x^2+4))`, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-2,3] and find the c?

Answer 1

Here's a sketch or outline of the solution.

Explanation:

`f` is the composition of the absolute value and a polynomial, so `f` is continuous at every real number. In particular, `f` is continuous on `[-2,3]`. So `f` satisfies the first hypothesis on this interval..

In general `absu` is differentiable expect where `u` changes signs.

`(x^2 -12)` changes signs at `x= +- sqrt2`.

`(x^2+4)` deos not changes signs on the real numbers.

So `f` is non-differentiable ony at `+- sqrt12` .

Since `3 < sqrt12 < 4`, neither of `+- sqrt12` is in `(-2,3)`

So `f` is differentiable on `(-2,3)` And `f` satisfies the second hypothesis on this interval..

The Mean Value Theorem now assures us that there is a `c` in `(-2,3)` with `f'(c) = (f(3)-f(-2))/(3-(-2))`

Actually finding the `c` is tedious.

On the interval `[-2,3]`, the values of `(x^2-12)(x^2+4)` are negative, so

`f(x) = -(x^2-12)(x^2+4) = -x^4+8x^2+48`

And `f'(x) = -4x^3+16x`

`f(3) = 39` and `f(-2) = 64`, so we need to solve

`-4x^3 +16x = (39-64)/(3-(-2)`

`-4x^3+16x = -7`

`4x^3-16x-7=0`

Now use whatever techniques you have available to solve this third degree equation. (There are two solutions in `(-2,3)` and another outside the interval.)