calculus
posted by help 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^29x13 and find the local extremer.
So here's what I did:
f(x)=x^3+3x^29x13
F^1(x)=3x^2+6x9
= 3x(x3)
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?

your simplification of the derivative is not correct
F ' (x) = 3x^2 + 6x  9
now we set that equal to zero
3x^2 + 6x9 = 0
divide each term by 3
x^2 + 2x  3 = 0
which factors to
(x+3)(x1) = 0
so x = 3 or x = 1
if x=1, f(1) = 1+3913 = 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+6x9 to
3x(x3)
if you expand 3x(x3) you get 3x^2  9x , I don't see the 6x )
You should always check intermediate steps of your solution ) 
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 
I took out the 3x which gave me 3x(x+69) but if that's wrong...it's wrong :) thanks for the help!

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