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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:
= 3x(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

    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.

  • calculus -

    I took out the 3x which gave me 3x(x+6-9) but if that's'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

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