# Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points

#### ewkimchi

Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points and which are saddle points.

I got to here:
Px = 1 -e^x
Py = 2y

But now I'm stuck.

I know I'm supposed to use this rule:

d=fxxfyy-(fxy)^2

If d>0:
fxx(a,b)>0, then f(a, b) is a rel min
fxx(a,b)<0, then f(a,b) is a rel. max

If d<0, it's a saddle point.

#### HallsofIvy

MHF Helper
Find the critical point(s) of f(x,y) = x + y^2 - e^x. Decide which are extreme points and which are saddle points.

I got to here:
Px = 1 -e^x
Py = 2y

But now I'm stuck.
At least on this one you have made an effort! The whole point of taking partial derivatives is that the critical points occur where the derivatives are 0!

Solve $$\displaystyle 1- e^x= 0$$ and 2y= 0.

I know I'm supposed to use this rule:

d=fxxfyy-(fxy)^2

If d>0:
fxx(a,b)>0, then f(a, b) is a rel min
fxx(a,b)<0, then f(a,b) is a rel. max
Yes, with a and b the values you got by solving those equations.

If d<0, it's a saddle point.

• ewkimchi

#### mr fantastic

MHF Hall of Fame
At least on this one you have made an effort! The whole point of taking partial derivatives is that the critical points occur where the derivatives are 0!

[snip]
I hope that ! is an exclamation mark (Rofl)

For the slow minded:

0! = 1 (Rofl)

#### ewkimchi

Thanks, but I what if fxx=0?

Which in this case, it does

#### General

I hope that ! is an exclamation mark (Rofl)

For the slow minded:

0! = 1 (Rofl)
+1
(Rofl)(Rofl)