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Math Help - Calculus Problem

  1. #1
    Affinity
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    Smile Calculus Problem

    Find the point(s) on the graph of where the slope is 5. (Thanks!)
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  2. #2
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    Hello, Affinity!

    Find the point(s) on the graph of: f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}} where the slope is 5.

    We have: . f(x)\:=\:\frac{x - 3x^{\frac{3}{2}}}{x^{\frac{1}{2}}} \:=\:x^{\frac{1}{2}} - 3x

    Then: . f'(x)\;=\;\frac{1}{2}x^{-\frac{1}{2}} - 3 \;=\;5

    We have: . \frac{1}{2\sqrt{x}} \;=\;8\quad\Rightarrow\quad 16\sqrt{x}\;=\;1\quad\Rightarrow\quad \sqrt{x} \,=\,\frac{1}{16}\quad\Rightarrow\quad x\,=\,\frac{1}{256}

    Then: . f\left(\frac{1}{256}\right)\:=\:\sqrt{\frac{1}{256  }} - 3\left(\frac{1}{256}\right) \;=\;\frac{13}{256}


    Answer: . \left(\frac{1}{256},\,\frac{13}{256}\right)

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  3. #3
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    helo everybody
    siriban u have written a different function i think
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  4. #4
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    sorry thats soroban not siriban!!!
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  5. #5
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    Quote Originally Posted by Affinity View Post
    Find the point(s) on the graph of where the slope is 5. (Thanks!)
    We are looking for x values where the first derivative is equal to 5.

    y = \frac{x^2 + 3x - 1}{x}

    y' = \frac{(2x + 3)(x) - (x^2 + 3x - 1)(1)}{x^2}

    y' = \frac{2x^2 + 3x - x^2 - 3x + 1}{x^2}

    y' = \frac{x^2 + 1}{x^2}

    So:
    5 = \frac{x^2 + 1}{x^2}

    5x^2 = x^2 + 1

    4x^2 - 1 = 0

    (2x + 1)(2x - 1) = 0

    Thus x = \pm 1/2

    -Dan
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  6. #6
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    Help!
    Am I the only one seeing: . \boxed{f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}} ??

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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post
    Help!
    Am I the only one seeing: . \boxed{f(x) \:=\:\frac{x - 3x\sqrt{x}}{\sqrt{x}}} ??

    Interesting. He made an earlier post here with that function. Now as I look at the other post all I can see is the function for this one.

    An idea: perhaps both of us need to clear our browser's cache?

    -Dan
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  8. #8
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    Quote Originally Posted by Affinity View Post
    Find the point(s) on the graph of where the slope is 5. (Thanks!)
    Hello, Affinity,

    to avoid the quotient rule you can first do the division:

    y=\frac{x^2+3x-1}{x}=x+3-\frac{1}{x}

    Now derivate:

    y'=1+\frac{1}{x^2}

    If y' = 5 then you have the equation:

    1+\frac{1}{x^2}=5. Solve for x and you'll get the answer topsquark has already calculated.

    EB
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