Results 1 to 2 of 2

Math Help - max and min value

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    22

    Exclamation max and min value ( derivative)

    find the max and min : y = x^3 - 8x^2 +61

    max : dy/dx = 3x^2 - 16X ; dy/dx = 0

    0=x(3x-16)
    x = 5.33 y = -14.9

    min : d2y/dx2 = 9x - 16 ; d2y/dx2 = 0

    0=9x-16
    x= 1.78 y = 41.3


    its wrong ... im not sure what i did wrong though ( 3 significant digits btw)
    Last edited by rock candy; March 22nd 2009 at 09:18 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    Quote Originally Posted by rock candy View Post
    find the max and min : y = x^3 - 8x^2 +61

    max : dy/dx = 3x^2 - 16X ; dy/dx = 0

    0=x(3x-16)
    x = 5.33 y = -14.9
    How did you go from "x = 0 or 3x - 16 = 0" to "x = 5.33 and y = -14.9"?

    (Would it be correct to assume, in what follows, that "X" is meant to be the same as "x"?)

    Quote Originally Posted by rock candy View Post
    min : d2y/dx2 = 9x - 16 ; d2y/dx2 = 0
    How did you go from ``\, \frac{dy}{dx}\, =\, 3x^2\, -\, 16x\, " to ``\,\frac{d^2y}{dx^2}\, =\, 9x\, -\, 16\,"? (The derivative of 3x^2 is (2)(3x^(2-1)) = 2*3x = 6x.)

    Instead, try starting with a quick graph, so you have a good idea what you're working with. Then find the critical points (in exact form, rather than decimal approximations) of the function.

    Take the second derivative, find its zeroes, and then find the intervals of sign, which will prove what you already know from the graph.
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum