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Math Help - Finding Values

  1. #1
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    Finding Values

    For the function f(x)= x^(6)(Aln(x)-1), where A is a constant.

    How would I find the value(s) of A if e^(5) is a critical point of f(x)?
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  2. #2
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    Quote Originally Posted by ctran View Post
    For the function f(x)= x^(6)(Aln(x)-1), where A is a constant.

    How would I find the value(s) of A if e^(5) is a critical point of f(x)?
    If e^5 is a critical point of f(x), then

    f'\left(e^5\right) = 0.

    Solve for A.
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  3. #3
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    Hello, ctran!

    Given the function: f(x)\:=\: x^6(A\ln x-1), where A is a constant.

    How would I find the value(s) of A if e^5 is a critical point of f(x) ?
    I assume you know what a critical point is . . .

    If x = e^5 is a critical point of f(x), then: . f'(e^5) \:=\:0


    Find f'(x)\!:\;\;f'(x) \;=\;x^6\cdot\frac{A}{x} + 6x^5(A\ln x - 1) \;=\;x^5(A + 6A\ln x - 6)


    Since f'(e^5) \,=\,0, we have:

    . f'(e^5) \;=\;(e^5)^5\bigg[A + 6A\ln(e^5) - 6\bigg] \:=\:0 \quad\Rightarrow\quad e^{25}\bigg[A + 6A\!\cdot\!5 -6\bigg] \:=\:0

    . . e^{25}\bigg[31A - 6\bigg] \:=\:0 \quad\Rightarrow\quad 31A - 6 \:=\:0 \quad\Rightarrow\quad 31A \:=\:6

    Therefore: . A \;=\;\frac{6}{31}

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