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Math Help - Difference quotient question

  1. #1
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    Difference quotient question

    Find the difference quotient of f

    f(x)=[7x^3]-6


    Not really sure where to go with this one, maybe cause its late I'm not thinking correctly. Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by Khaotical View Post
    Find the difference quotient of f

    f(x)=[7x^3]-6


    Not really sure where to go with this one, maybe cause its late I'm not thinking correctly. Any help is greatly appreciated.
    Is it

    f(x) = 7x^3 - 6?


    If so

    f(x + h) = 7(x + h)^2 - 6

     = 7(x^2 + 2xh + h^2) - 6

     = 7x^2 + 14xh + 7h^2 - 6.


    Therefore

    f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

     = \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}

     = \lim_{h \to 0}\frac{14xh + 7h^2}{h}

     = \lim_{h \to 0}(14x + 7h)

     = 14x.
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  3. #3
    Super Member Matt Westwood's Avatar
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    Quote Originally Posted by Prove It View Post
    Is it

    f(x) = 7x^3 - 6?


    If so

    f(x + h) = 7(x + h)^2 - 6

     = 7(x^2 + 2xh + h^2) - 6

     = 7x^2 + 14xh + 7h^2 - 6.


    Therefore

    f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

     = \lim_{h \to 0}\frac{7x^2 + 14xh + 7h^2 - 6 - (7x^2 - 6)}{h}

     = \lim_{h \to 0}\frac{14xh + 7h^2}{h}

     = \lim_{h \to 0}(14x + 7h)

     = 14x.
    It's cubed, not squared, in the initial problem. But the technique is okay.
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  4. #4
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    Quote Originally Posted by Matt Westwood View Post
    It's cubed, not squared, in the initial problem. But the technique is okay.
    Oh yeah... oops...

    f(x) = 7x^3 - 6.


    f(x + h) = 7(x + h)^3 - 6

    = 7(x^3 + 3x^2h + 3xh^2 + h^3) - 6

    = 7x^3 + 21x^2h + 21xh^2 + 7h^3 - 6.


    Now work out

    \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}.
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