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Math Help - Very Simple Critical Points Problem with Answer (Need help understanding)

  1. #1
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    Very Simple Critical Points Problem with Answer (Need help understanding)

    Find the critical numbers of the function

    x^(4/5) * ((x-4))


    Answers: 0,4,8/7

    I used the chain rule. Got 8/5x^(-1/5) * (x-4).


    That would give me critical points of 0 and 4.

    Where does this 8/7 value come from?

    Thanks again guys!!

    Also I'm sure many of you are about to watch football so good luck!! (Go Skins!!)
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  2. #2
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    Re: Very Simple Critical Points Problem with Answer (Need help understanding)

    Using the product rule, we find:

    f'(x)=x^{\frac{4}{5}}(2(x-4))+\frac{4}{5}x^{-\frac{1}{5}}(x-4)^2=

    \frac{1}{5}x^{-\frac{1}{5}}(x-4)(10x+4(x-4))=\frac{2}{5}x^{-\frac{1}{5}}(x-4)(7x-8)
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    Re: Very Simple Critical Points Problem with Answer (Need help understanding)

    Quote Originally Posted by MarkFL2 View Post
    Using the product rule, we find:

    f'(x)=x^{\frac{4}{5}}(2(x-4))+\frac{4}{5}x^{-\frac{1}{5}}(x-4)^2=

    \frac{1}{5}x^{-\frac{1}{5}}(x-4)(10x+4(x-4))=\frac{2}{5}x^{-\frac{1}{5}}(x-4)(7x-8)

    I'm sooo confused. Why on Earth would you use the product rule? Maybe I wrote it wrong. it looks like

    X^4/5(x-4)

    That looks like classic chain rule? Why wouldn't you use the chain rule. (Sorry if I wrote it wrong!!)
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Very Simple Critical Points Problem with Answer (Need help understanding)

    You can use the product rule because the given function is the product of x^{\frac{4}{5}} and (x-4)^2.
    Thanks from skinsdomination09
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    Re: Very Simple Critical Points Problem with Answer (Need help understanding)

    To explain a little further, you use the chain rule for compositions - functions of the form f(g(x)), and the product rule for products - functions of the form f(x)g(x). For most functions you don't get to choose; you have to use whichever is appropriate.

    The function you have is a product, so you use the product rule:

    f(x)=(x^{\frac{4}{5}})((x-4)^2)
    f(x)=(x^{\frac{4}{5}})((x-4)^2)'+(x^{\frac{4}{5}})'((x-4)^2)

    You actually use the chain rule for (x-4)^2, though you probably do it without really thinking. It is the composition of f(y)=y^2 and g(x)=x-4. So f(g(x))=(x-4)^2 and by the chain rule, (f(g(x)))'=f'(g(x))g'(x)=(2(x-4))(1).

    and then what MarkFL2 said.

    - Hollywood
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