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Math Help - solving problem using definition of derivative

  1. #1
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    solving problem using definition of derivative

    f(x) = x^3 + 2x

    Use the definition of the derivative as a limit to determine the slope of the
    tangent line to f(x) at x = 1

    I know the rule is limit as h goes to 0  f(x+h) - f(x)\over {h}

    But how do I rearrange this one so the answer isn't 0?
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  2. #2
    Super Member 11rdc11's Avatar
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    this is called finding the limit

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

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

    = \frac{3x^2h +3xh^2 +h^3 +2h}{h}

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

    = 3x^2 +3xh +h^2 +2

    Now plug in 0 for x and you get

    3x^2 +2

    Now plug in your x =1 for x and that is your answer
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  3. #3
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    Quote Originally Posted by 11rdc11 View Post
    this is called finding the limit

    ...
    = 3x^2 +3xh +h^2 +2

    Now plug in 0 for h and you get <=== tiny typo

    3x^2 +2

    ...
    Quote Originally Posted by smplease View Post
    f(x) = x^3 + 2x

    Use the definition of the derivative as a limit to determine the slope of the
    tangent line to f(x) at x = 1

    I know the rule is limit as h goes to 0  f(x+h) - f(x)\over {h}

    But how do I rearrange this one so the answer isn't 0?
    1. Using the formula you know that

    f'(x)=\lim_{h \to 0}\left(\dfrac{((x+h)^3+2(x+h))-(x^3+2x)}{h}\right)

    2. You are asked to calculate the slope of the graph at x = 1, that means:

    f'(1)=\lim_{h \to 0}\left(\dfrac{((1+h)^3+2(1+h))-(1^3+2\cdot 1)}{h}\right)

    3. Expand the brackets in the numerator:

    f'(1)=\lim_{h \to 0}\left(\dfrac{(1+3h+3h^2+h^3+2+2h)-(3)}{h}\right)

    4. Collect like terms and factor the numerator. Cancel the factor h. You'll get:

    f'(1)=\lim_{h \to 0}(h^2+3h+5) = 5
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