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Math Help - rel extrema

  1. #1
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    rel extrema

    I'm having trouble finding the extremas of...

    I'm having trouble finding the crit # of this. Please show me how to do it...
    f(x)= e^x/1-x+x^2
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  2. #2
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    Quote Originally Posted by kwivo View Post
    I'm having trouble finding the extremas of...

    I'm having trouble finding the crit # of this. Please show me how to do it...
    f(x)= e^x/1-x+x^2
    This is f(x) = \frac{e^x}{1 - x + x^2} I presume? Please use parenthesis!

    f^{\prime}(x) = \frac{e^x(1 - x + x^2) - e^x(-1 + 2x)}{(1 - x + x^2)^2}

    f^{\prime}(x) = \frac{e^x(2 - 3x + x^2)}{(1 - x + x^2)^2}

    You will have critical numbers where f^{\prime}(x) = 0, so solve

    \frac{e^x(2 - 3x + x^2)}{(1 - x + x^2)^2} = 0

    e^x(x^2 - 3x + 2) = 0

    e^x(x - 1)(x - 2) = 0

    So either
    e^x = 0 <-- Doesn't happen
    or
    x - 1 = 0 \implies x = 1
    or
    x - 2 = 0 \implies x = 2

    So the critical points are x = 1 and x = 2.

    -Dan
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by kwivo View Post
    I'm having trouble finding the extremas of...

    I'm having trouble finding the crit # of this. Please show me how to do it...
    f(x)= e^x/1-x+x^2
    To find local extrema, it is when the derivative is equal to zero.

    f'(x)=\dfrac{e^x(1-x+x^2)-(-1+2x)(e^x)}{(1-x+x^2)^2}=\dfrac{e^x(x-1)(x-2)}{(1-x+x^2)^2}

    0=e^x(x-1)(x-2)
    therefore local max or min occur at x=1 and/or x=2

    Now look at the interval <1;at 1;1-2;at 2;>2

    you will see that at x=1 it is a local max and at x=2 it is a local min
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