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Math Help - Evaluating a Function

  1. #1
    Senior Member vaironxxrd's Avatar
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    Evaluating a Function

    For the function f defined by f(x) = 2x^2-3x

    Evaluate: \frac{f(x+h)-f(x)}{h}

    I'm solving the problem using a book which actually gives me step-by-step instructions. Thing is I have a confusion with the following distribution, and the book is not showing the distribution step.

    \frac{f(x+h)-f(x)}{h}=\frac{[2(x+h)^2 -3(x+h)]-[2x^2-3x]}{h}

    When I do this step I receive the following:
    \frac{2x^2+2h^2-3x-3h-2x^2-3x}{h}


    What the book gives

    \frac{2(x^2+2xh+h^2)-3x-3h-2x^2+3x}{h}

    What am I doing wrong? Everything?
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Evaluating a Function

    Quote Originally Posted by vaironxxrd View Post
    For the function f defined by f(x) = 2x^2-3x

    Evaluate: \frac{f(x+h)-f(x)}{h}

    I'm solving the problem using a book which actually gives me step-by-step instructions. Thing is I have a confusion with the following distribution, and the book is not showing the distribution step.

    \frac{f(x+h)-f(x)}{h}=\frac{[2(x+h)^2 -3(x+h)]-[2x^2-3x]}{h}

    When I do this step I receive the following:
    \frac{2x^2+2h^2-3x-3h-2x^2-3x}{h}


    What the book gives

    \frac{2(x^2+2xh+h^2)-3x-3h-2x^2+3x}{h}

    What am I doing wrong? Everything?
    Try these corrections:
    1. (x + h)^2 = x^2 + 2xh + h^2, not (x + h)^2 = x^2  + h^2
    2. The sign on the last 3x is positive, not negative

    -Dan
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  3. #3
    Senior Member vaironxxrd's Avatar
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    Re: Evaluating a Function

    Quote Originally Posted by topsquark View Post
    Try these corrections:
    1. (x + h)^2 = x^2 + 2xh + h^2, not (x + h)^2 = x^2  + h^2
    2. The sign on the last 3x is positive, not negative

    -Dan
    I saw that before... a very simple mistake. Thank you, for correcting me.
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