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Is it \[g(x,y) = \frac{ (x^2+y)e^y }{ 2 }\]

no just e^(y/2)

no y/2 is the exponent of e

\[g(x,y)= (x^2+y)e^{y/2}\]

yes

We have to do the second derivative test, you know it?

Lets make it easier on ourselves and expand it \[g(x,y) = x^2e^{y/2}+ye^{y/2}\]

ok

So find all the derivatives, find \[\huge f_x,f_y, f_{xx}, f_{yy}\]

But, lets just go over this and you will see, so find the partial derivatives as I've asked you to

fx= 2xe^(y/2)

fy=e^(y/2)(x^2 +y /2) + 1

fxx=2e^(y/2)

fyy=1/2 e^(y/2) +1/2 e^(y/2) (x^2 +y +2/2)

\[\large g_{y} = \frac{ 1 }{ 2 }e^{y/2} (x^2+y+2)\] this should be function with respect to y

Then \[\large g_{yy} = \frac{ 1 }{ 4 }e^{y/2}(x^2+y+4)\]

ok thanks

These are long problems..I know haha.

So for gx we are solving for y

\[2xe^{y/2}=0\]

xey/2=0

how do we solve for y

Mhm sec, I am getting - infinity for that

Oh oops, solve for x haha.

That makes more sense, and should be a lot more easier :)

so x=0

Right, now for gy solve for y

\[\large g_y = \frac{ 1 }{ 2 }e^{y/2}(x^2+y+2) = 0\]

i don't know how to solve it because of e^(y/2)

Get rid of it :P |dw:1438324656542:dw|

Now go ahead and solve for y

oh both e^(y/2) and (x2+y+2)=0 so y= -2 right?

so (0,-2) is the critical point

solving for y we get y = -x^2-2

So now if x = 0, y = y = -0^2 -2 = -2
and wow, that seems to be our only critical point

(0,-2)

Sorry I've been lagging a lot so it's hard to work, ok let me see...

no i really appreciate your work

So find \[g_{xy}\] first

so (2/e) (1/2e)- 0=1/e^2 right?

What's that?

\[g_{xy} = xe^{y/2}\]

\[D = g(0,-2) = g_{xx}(0,-2)g_{yy}(0,-2)-[g_{xy}(0,-2)]^2\]

D= (2/e) (1/2e)- 0=1/e^2

Oh ok, so you plugged it all in :)

So if 1/e^2 < 0 then we have a saddle point

but it is >0

so we see gxx

Good! Now is gxx >0 or <0

By the way, something to note, if D = 0, then the test fails

oh thanks

gxx >0 so will get min

Yes a local min :)

Awesome work @Astrophysics ^_^

Haha, thanks Zale, I just wish it wasn't as laggy

I can tell. Your icon does not appear on the top thread of this post

g(0,-2)= -2/e

is our local min value

Thanks a lot !!!

If gxx(0,2) < 0 then it would be local maximum
and np :)