Module calculus problem

Question

Given f '(x) = (x + 1)(6 + 3x), find the x-coordinate for the relative minimum on the graph of f(x). 

Options:

1. 0
2. -1
3. -2
4. None of these

Answer 1

f '(x) = (x + 1)(6 + 3x)

f'(x) has roots of x = -1 and x = -2. These are either relative minimums OR maximums--we need to look at f(x) to see which:

f(x) is the integral of f'(x). To do the integration, first put f'(x) into standard polynomial form:

f'(x) = 3x² + 9x + 6

Then f(x) = x³ + 9/2 * x² + 6x The dominant shape is set by x³ : For large -x, y goes to -infinity, and for large +x, y goes to + infinity.

My first guess is that x = -2 is a local maximum, and x = -1 is a local minimum. To confirm, we need the 2nd derivative:

f"(x) = 6x + 9. Setting this = to 0, we find a root at -3/2. This is halfway between the 2 previous points, and appears to be an inflection.

Looking at f(x), we can also see that there is a root at x = 0. Factoring out the x, we have x2 +9x/2 + 6. This has no real roots, so there is only one crossing of the x axis (at 0,0)

So.....the local minimum is at x = -1

CHECK this by plotting on a graphing calculator or online tool