posted by help .
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:
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 + 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
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 )
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.
I took out the 3x which gave me 3x(x+6-9) 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