Question

How to find the maximum and minimum when given derivative?

Can someone please explain where I went wrong with 4g? I got max at -3 and min at 4 by drawing the sign graph, but the answer says the exact opposite.
Thanks!

Answer 1

We start by finding critical numbers:

`0 = -x^2 + x + 12`

`0 = x^2 - x - 12`

`0 = (x - 4)(x + 3)`

`x = 4 or -3`

So you are correct about the two turning points.

If we select a test point between the two turning points, say `x = 0` we get:

`y' = 0^2 + 0 + 12 = 12`

Since this is positive we know that the function is increasing on `(-3, 4)`. Then the function is decreasing on `(4, oo)` and thus `x = 4` will be a maximum and `x = -3` will be a minimum.

Hopefully this helps!