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Math Help - Local extremes of a trig/ln function

  1. #1
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    Local extremes of a trig/ln function

    I am having trouble finding the local extremes for a particular function.

    f(xy)=e^(2x)cosy

    I have so far:
    f'(x)= 2e^(2x)cosy
    f'(y)=-e^(2x)cosy
    f''(x)=4e^(2x)cosy
    f''(y)=-e^(2x)cosy
    f''(xy)=-2e^(2x)siny

    Now I'm setting both f'(x) and f'(y) = 0, but this is where I'm having trouble. Does that mean that 2e^(2x)=0, cosy=0, -e^(2X)=0, and siny=0?

    Thanks for any help that can be provided.
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  2. #2
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Local extremes of a trig/ln function

    f(x,y) as a surface:
    as seen f(x,y)=0 can be considered as extreme value.



    Local extremes perhaps can have meaning for x=constant:

    Attached Thumbnails Attached Thumbnails Local extremes of a trig/ln function-local-extremes1.png   Local extremes of a trig/ln function-local-extremes3.png  
    Last edited by MaxJasper; October 13th 2012 at 12:30 PM.
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  3. #3
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    Re: Local extremes of a trig/ln function

    Thanks for the reply. I'm trying to understand what you mean. So the extremes exist @ x=n(pi)/2?
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  4. #4
    Senior Member MaxJasper's Avatar
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    Lightbulb Re: Local extremes of a trig/ln function



    As seen on the contours of f(x,y), there is no locality where f'x, f'y can be 0 at the same time ...except at x=-inf where the plane of f(x,y) is just a flat plane at z=0
    Attached Thumbnails Attached Thumbnails Local extremes of a trig/ln function-local-extremes2.png  
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  5. #5
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    Re: Local extremes of a trig/ln function

    Hello, MajMinor!

    \text{Find the local extremes: }\:f(x,y)\:=\:e^{2x}\cos y

    We have:
    . . \begin{Bmatrix}f_x &=& 2e^{2x}\cos y \\ f_y &=& \text{-}e^{2x}\sin y \end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}f_{xx} &=& 4e^{2x}\cos y \\ f_{yy} &=& \text{-}e^{2x}\cos y \\ f_{xy} &=& \text{-}2e^{2x}\sin y \end{Bmatrix}


    D \;=\;(f_{xx})(f_{yy}) - (f_{xy})^2

    . . =\;\left(4e^{2x}\cos y\right)\left(\text{-}e^{2x}\cos y\right) - \left(\text{-}2e^{2x}\sin y\right)^2

    . . =\;\text{-}4e^{4x}\cos^2\1y - 4e^{4x}\sin^2\!y

    . . =\;\text{-}4e^{4x}\underbrace{\left(\cos^2\!y + \sin^2\!y\right)}_{\text{This is 1}}
    . . =\;\text{-}4e^{4x}


    Since e^{4x} is always positive, D is always negative.
    . . Hence, all extremes are saddle points.

    \text{If }f_x=0\!:\;2e^{2x}\cos y \,=\,0 \quad\Rightarrow\quad \cos y \,=\,0 \quad\Rightarrow\quad y \,=\,\tfrac{\pi}{2} + \pi n

    \text{If }f_y = 0\!:\;\text{-}e^{2x}\sin y \,=\,0 \quad\Rightarrow\quad \sin y \,=\,0 \quad\Rightarrow\quad y \,=\,\pi n

    \text{There are saddle points when: }\:\begin{Bmatrix}y &=& \frac{\pi}{2}n \\ x &=& \text{any} \end{Bmatrix}
    Thanks from MajMinor
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  6. #6
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    Re: Local extremes of a trig/ln function

    Got it. I understand now. Thanks, guys!
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