points of inflection?

Set f"(x) = 0 then slove for values of X

0 = 12X + 12

X = -1 <-- this is a possible point of inflection

To check this point make a table to test the values of the function at certain points

- infinity -1 infinity

f"(x) - + <--- this is the value of f"(x)

between these values

If the signs are opposite, in this case which they are there is a POI.

So, at any point where f'(x) = 0, you have an inflection point, so factor the equation to find where it equals 0

0=(6x-6)(x+3), x=-3, 1

Also, from definition, a second derivative is negative at a max, and positive at a min. We can plug in the x-values of our inflection points to see if they are a max or min

12(-3)+12= negative (local max.)

12(1)+12=positive (local min)

We'll also need to plug our x-values back into the original eqn. to find their y components

for -3 -> 35, and for 1 -> -29

The answer is:

local max at (-3,35)

local min at (1,-29)