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September 22, 2014

September 22, 2014

Posted by **help** on Thursday, July 5, 2012 at 8:25pm.

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 -
**Reiny**, Thursday, July 5, 2012 at 9:03pmyour 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 -
**bobpursley**, Thursday, July 5, 2012 at 9:06pmyes, 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 -
**help**, Thursday, July 5, 2012 at 10:37pmI took out the 3x which gave me 3x(x+6-9) but if that's wrong...it's wrong :) thanks for the help!

- calculus -
**help**, Thursday, July 5, 2012 at 10:39pmAndddd I just realized to take out the 3 out the 6 and 9 T_T opps. Thanks, now I feel dumb XD

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