Another local extreme problem

I am facing another problem that I can't compute the extremes for:(Doh)

f(xy)=e^{x}(x^{2}-y^{2})

so far I have:

fx= 2xe^{x }+ e^{x}(x^{2}-y^{2})

fy= -2ye^{x }fxx=2e^{x}+2xe^{x} + e^{x}(x^{2}-y^{2}) + 2xe^{x }

fyy= -2e^{x }fxy= -2ye^{x
How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.
Thanks for any help.}

Re: Another local extreme problem

Quote:

Originally Posted by

**MajMinor** I am facing another problem that I can't compute the extremes for:(Doh)

f(xy)=e^{x}(x^{2}-y^{2})

so far I have:

fx= 2xe^{x }+ e^{x}(x^{2}-y^{2})

fy= -2ye^{x }fxx=2e^{x}+2xe^{x} + e^{x}(x^{2}-y^{2}) + 2xe^{x }

fyy= -2e^{x }fxy= -2ye^{x
How do I find the extremes? I set fx and fy to 0, but I can't solve for x and y.
Thanks for any help.}

If you set the partial with respect to x equal to 0 you get

$\displaystyle 2xe^{x}+e^{x}(x^2-y^2)=0 \\ e^{x}(2x+x^2-y^2)=0$

Since the exponential is never zero his gives

$\displaystyle 2x+x^2-y^2=0 \iff y=\pm \sqrt{x^2+2x}$

If we plug this into the 2nd equation we get

$\displaystyle 2(\pm \sqrt{x^2+2x})e^{x}=0$

Again the exponential can never be zero so we get

$\displaystyle (\pm \sqrt{x^2+2x})=0 \iff x^2+2x=0^2 \iff x(x+2)=0$

So we get the two solutions

$\displaystyle x=0 \quad x=-2$

Can you finish from here?

Re: Another local extreme problem

Yes. Thank you very much! It's the little things that always trip me up.