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    Hi,

    I need to differentiate f(x) = 10e^((x^2)+2).

    Is 20xe^((x^2)+2) correct?

    If so, I need to find the extreme value of the function...so I need to begin by finding the value of x where f'(x) = 0, right? How do I do this?

    Thanks a lot
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    (Original post by BJP)
    Hi,

    I need to differentiate f(x) = 10e^((x^2)+2).

    Is 20xe^((x^2)+2) correct?

    If so, I need to find the extreme value of the function...so I need to begin by finding the value of x where f'(x) = 0, right? How do I do this?

    Thanks a lot
    The differentiation is correct

    Now that you've got the derivative, and you know it equals zero, we can rewrite it as  \displaystyle \frac{dy}{dx} = 20xe^{x^2+2} = 0 . Now you solve the equation (note that e^anything cannot equal zero, which means that...? )

    By extreme values I assume you mean stationary points rather than global points.
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    (Original post by Goldfishy)
    The differentiation is correct

    Now that you've got the derivative, and you know it equals zero, we can rewrite it as  \displaystyle \frac{dy}{dx} = 20xe^{x^2+2} = 0 . Now you solve the equation (note that e^anything cannot equal zero, which means that...? )

    By extreme values I assume you mean stationary points rather than global points.
    Thanks for the confirmation

    So, does x = 0 at the stationary point? Thanks again
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    (Original post by BJP)
    Thanks for the confirmation

    So, does x = 0 at the stationary point? Thanks again
    No worries.

    Yep, x = 0 at the stationary point. To find out what type of stationary point is, you'd usually look at the 2nd derivative and see if it's +ve or -ve at that point, when x = 0 in this case (but y'' is zero). Plotting a graph always helps: Wolfram
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    extreme value of the function? Do you refer to local/global extrema?
 
 
 
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