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Math Help - using limit derivative?

  1. #1
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    using limit derivative?

    how to solve using limit definition of the derivative for f(x)=2x^3+5x^2-7x-4

    my ans:

    6x^2+10x-7

    plse correct me...
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  2. #2
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    Quote Originally Posted by mathseek View Post
    how to solve using limit definition of the derivative for f(x)=2x^3+5x^2-7x-4

    my ans:

    6x^2+10x-7

    plse correct me...
    It's correct...
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  3. #3
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    limit def

    than what is limit definition??
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  4. #4
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    Definition of a derivative: f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
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    limit def

    so when i answer this question above....i just give my ans will do right?
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    Quote Originally Posted by mathseek View Post
    so when i answer this question above....i just give my ans will do right?
    No - I thought you had gone through the rigmarole of using the limit definition and gotten that answer.

    f(x) = 2x^3 + 5x^2 -7x - 4

    f(x + h) = 2(x + h)^3 + 5(x + h)^2 - 7(x + h) - 4

     = 2(x^3 + 3x^2h + 3xh^2 + h^3) + 5(x^2 + 2xh + h^2) - 7x -7h -4

     = 2x^3 + 6x^2h + 6xh^2 + 2h^3 + 5x^2 + 10xh + 5h^2 - 7x - 7h - 4


    So f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3 + 10xh + 5h^2 - 7h

     = h[6x^2 + 6xh + 2h^2 + 10x + 5h - 7].


    \frac{f(x+h) - f(x)}{h} = 6x^2 + 6xh + 2h^2 + 10x + 5h - 7


    \lim_{h \to 0}\frac{f(x+h) - f(x)}{h} = 6x^2 + 6x(0) + 2(0)^2 + 10x + 5(0) - 7

     = 6x^2 + 10x - 7.
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    Well you have to show your work probably and you do so by going through the grunt work of expanding and simplifying in order to evaluate the limit:

    \begin{aligned}f'(x) & = \lim_{h \to 0} \frac{2(x+h)^3+5(x+h)^2-7(x+h) - 4 \ - \ \left(2x^3 + 5x^2 - 7x - 4\right)}{h} \\ & = \lim_{h \to 0} \frac{2(x^3 + 2x^2h + 2xh^2 + h^3) +5(x^2 + 2xh + h^2) - \cdots}{h} \end{aligned}

    Eventually you can factor out an h from the top to cancel it with the bottom and you should get your answer. Like I said, a lot of grunt work.
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