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Math Help - Setting g'(x) to zero

  1. #1
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    Setting g'(x) to zero

    (I know this is the pre-cal forum, but the calculus isn't my problem here ... it's the basics that are dogging me.)
    In order to make a wiggle graph, I need to find where my slopes are zero. The formula of my derivative is:
    g'(x) = 6x^2 - 7x - 3

    So my question is, when g'(x) = 0, x = ?

    I don't think I'm just supposed to plug random numbers in and hope to hit 0, but simplifying has only gotten me a little closer. I think I can simplify to:

    x = 1/(2x) + 7/6

    It would be a little easier to just start plugging numbers in, but I think I'm supposed to actually solve this one before moving on. I'm stuck - how do I know what x is when g'(x) = 0?
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  2. #2
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    Quote Originally Posted by Boris B View Post
    (I know this is the pre-cal forum, but the calculus isn't my problem here ... it's the basics that are dogging me.)
    In order to make a wiggle graph, I need to find where my slopes are zero. The formula of my derivative is:
    g'(x) = 6x^2 - 7x - 3

    So my question is, when g'(x) = 0, x = ?

    I don't think I'm just supposed to plug random numbers in and hope to hit 0, but simplifying has only gotten me a little closer. I think I can simplify to:

    x = 1/(2x) + 7/6

    It would be a little easier to just start plugging numbers in, but I think I'm supposed to actually solve this one before moving on. I'm stuck - how do I know what x is when g'(x) = 0?
    You need to solve the quadratic equation

    0 = 6x^2 - 7x - 3 .....
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