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Thread: Linear Approximation

  1. #1
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    Linear Approximation

    The problem given was $\displaystyle f(x)=e^{x^2}$ near x = 5 solve for exact answer

    $\displaystyle e^{x^2}+2e^{x^2}(x-5)$

    $\displaystyle 3e^{x^2}(x-5)$ and then plug in 5 for x

    $\displaystyle 3e^{5^2}(x-5)$

    simplify

    $\displaystyle 3xe^{25}-15ex^25$


    is this fine?
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  2. #2
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    If the expression were $\displaystyle e^{x^2}(x-5)+2e^{x^2}(x-5)$, then you could derive $\displaystyle 3e^{x^2}(x-5)$, but not in this case.
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  3. #3
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    So is my answer correct?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    The problem given was $\displaystyle f(x)=e^{x^2}$ near x = 5 solve for exact answer

    $\displaystyle e^{x^2}+2e^{x^2}(x-5)$

    $\displaystyle 3e^{x^2}(x-5)$ and then plug in 5 for x

    $\displaystyle 3e^{5^2}(x-5)$

    simplify

    $\displaystyle 3xe^{25}-15ex^25$


    is this fine?
    First of all, the derivative of $\displaystyle e^{x^2}$ is $\displaystyle 2xe^{x^2}$, not $\displaystyle 2e^{x^2}$

    The formula you want is $\displaystyle f(5)+f'(5)(x-5) = e^{25}+ 10e^{25}(x-5) = 10e^{25}x-50e^{25}+e^{25} = \boxed{10e^{25}x-49e^{25}}$

    This won't be a great approximation though, since $\displaystyle e^{x^2}$ increases very rapidly.
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  5. #5
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    Thanks mate!

    would leaving

    $\displaystyle e^25+10e^25(x-5)$


    be sufficient?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by sk8erboyla2004 View Post
    Thanks mate!

    would leaving

    $\displaystyle e^{25}+10e^{25}(x-5)$


    be sufficient?
    I don't know how your grader will prefer it to be written, but it is still the correct answer; it's just up to personal preference how you write it. So I would say it's still fine.
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  7. #7
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    Thanks once again!
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