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Math Help - Slope by limit process

  1. #1
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    Slope by limit process

    Greetings,

    I need to use the limit definition of :

    m= \lim_{\Delta x\rightarrow 0} f(x) = (x+\Delta x)- f(x) / \Delta x

    ( / means all divided by \Delta x)

    I have to use the limit definition to find the slope of the tangent line to the graph of f given at these equations:

    f(x) = 4-x^2 at point (2,0)


    If someone could show me how to solve this step by step that would be cool.
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  2. #2
    MHF Contributor alexmahone's Avatar
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    f(x)=4-x^2
    f'(x)=-2x

    m=f'(x)=-4
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  3. #3
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    I think what he needs is to solve this using First Principle.

    f(x) = 4-x^2 at point (2,0), so x=2

    Now as you said:

    m= \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)- f(x)}{\Delta x}
    Sub x=2
    m=\lim_{\Delta x\rightarrow 0} \frac{f(2+\Delta x)- f(2)}{\Delta x}
    m=\lim_{\Delta x\rightarrow 0} \frac{4-(2+\Delta x)^2- (4-(2)^2)}{\Delta x}
    m=\lim_{\Delta x\rightarrow 0} \frac{4-(2^2+4\Delta x + (\Delta x)^2)- (4-(2)^2)}{\Delta x}
    m=\lim_{\Delta x\rightarrow 0} \frac{4-4-4\Delta x - (\Delta x)^2- 4+4}{\Delta x}

    We can do some cancelling
    m=\lim_{\Delta x\rightarrow 0} \frac{-4\Delta x - (\Delta x)^2}{\Delta x}

    We can do some more cancelling which leaves us with
    m=\lim_{\Delta x\rightarrow 0} -4 - \Delta x

    The limit is as \Delta x \rightarrow 0, so sub it in, which gives
    m= -4-0<br />
=-4

    So, at the point (2,0), m=-4


    You can also verify this by differentiating f(x) as alexmahone did:
    f' (x)=-2x
    f'(2)=-2(2) = -4

    Which is the same as what we obtained before, although this doesn't directly use the first principles or limits.
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