# Local extremes of a trig/ln function

• Oct 13th 2012, 10:20 AM
MajMinor
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.
• Oct 13th 2012, 10:44 AM
MaxJasper
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.

http://mathhelpforum.com/attachment....1&d=1350153665

Local extremes perhaps can have meaning for x=constant:

http://mathhelpforum.com/attachment....1&d=1350156578(Nerd)
• Oct 13th 2012, 10:57 AM
MajMinor
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?
• Oct 13th 2012, 11:10 AM
MaxJasper
Re: Local extremes of a trig/ln function
http://mathhelpforum.com/attachment....1&d=1350155212

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(Shake)
• Oct 13th 2012, 11:11 AM
Soroban
Re: Local extremes of a trig/ln function
Hello, MajMinor!

Quote:

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

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

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

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

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

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

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

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

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

$\displaystyle \text{There are saddle points when: }\:\begin{Bmatrix}y &=& \frac{\pi}{2}n \\ x &=& \text{any} \end{Bmatrix}$
• Oct 13th 2012, 11:24 AM
MajMinor
Re: Local extremes of a trig/ln function
Got it. I understand now. Thanks, guys!