# calculus

posted by on .

hey I just wanted to make sure that I was doing this problem correctly.

The question says to graph the function f given by f(x)=x^3+3x^2-9x-13 and find the local extremer.

So here's what I did:
f(x)=x^3+3x^2-9x-13
F^1(x)=3x^2+6x-9
= 3x(x-3)
x=0
x=3
I put those two xes back into the original equation and got: (0,-13) (3,-4) and then graphed. Is this the proper way to look for local extremer?

• calculus - ,

your simplification of the derivative is not correct

F ' (x) = 3x^2 + 6x - 9
now we set that equal to zero

3x^2 + 6x-9 = 0
divide each term by 3
x^2 + 2x - 3 = 0
which factors to
(x+3)(x-1) = 0
so x = -3 or x = 1

if x=1, f(1) = 1+3-9-13 = -18
if x=-3, f(-3) = -27 + 27 + 27 - 13 = 14

so the points are (1,-18) and (-3,14)

(I don't understand how you went from
3x^2+6x-9 to
3x(x-3)

if you expand 3x(x-3) you get 3x^2 - 9x , I don't see the 6x )
You should always check intermediate steps of your solution )

• calculus - ,

yes, you have the max or min. These are the localized extremes. Look at x=+- inf? what happens there? (But they are not local extremes. Read this example to see the difference between local extrema and global extrema.

http://en.wikipedia.org/wiki/Maxima_and_minima

• calculus - ,

I took out the 3x which gave me 3x(x+6-9) but if that's wrong...it's wrong :) thanks for the help!

• calculus - ,

Andddd I just realized to take out the 3 out the 6 and 9 T_T opps. Thanks, now I feel dumb XD