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Math Help - First derivative test

  1. #1
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    First derivative test

    Okay so here is the deal.

    I have the function.

    G(x)=(x^3-1)^3

    G'(x)= 3 (x^3-1)^2 * 3x

    And i need to find out where the function is increasing and decreasing. I know how to do that, I have to set it to zero to find the critical numbers. However, I have two questions I guess. One being is my first derv right, and then if it is....how to i solve for zero. Should I divide by 2x?? then 3?? I am confused.
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  2. #2
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    pickslides's Avatar
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    G'(x)= 3 (x^3-1)^2 \times  3x^2

    Finding where G'(x)=0 is a good start. This where the stationary points are. In between these are where the function is either increasing or decreasing, take some test points to find this out.
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  3. #3
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    e^(i*pi)'s Avatar
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    You forgot to take the derivative of x^3 properly when using the chain rule - you have 3x instead of 3x^2:

    g'(x) = 9x^2(x^3-1)^2 = 0

    You can use the same methods you do with equations like (x-a)(x-b) = 0 so either 9x^2 = 0 or (x^3-1)^2 = 0

    Then take some points either side to test
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