trouble finding critical points on a function, R^2 -> R

**sgcb** · October 6th 2012, 05:53 PM

Q: Find the local maximum and minimum values and saddle point(s) of the function
$f(x, y) = (x^2 + y^2)e^{y^2 - x^2}$

I've found the the function's gradiant vector:

$\triangledown{f} = [-2xe^{y^2-x^2}(x^2+y^2-1), 2ye^{y^2-x^2}(x^2+y^2+1)]$

but I'm having trouble manipulating these equations to find where $\triangledown{f}=\vec{0}$ . BOB says these points are $(0, 0)$ and $(\pm{1}, 0)$ , but I'm not sure how to go about proving these are in fact the only points.

**HallsofIvy** · October 6th 2012, 06:43 PM

An exponential is never 0 so you can divide $\nabla f= \vec{0}$ by $e^{y^2- x^2}$ to get $\nabla f= [-2x(x^2+ y^2- 1), 2y(x^2+ y^2+ 1]= \vec{0}$ .

Since a vector is 0 if and only if each component is 0, we must have $-2x(x^2+ y^2- 1)= 0$ and $2y(x^2+ y^2- 1)= 0$ . And since ab= 0 if and only if either a= 0 or b= 0 we must have x=0, or $x^2+ y^2- 1= 0$ and y= 0 and $x^2+ y^2- 1= 0$ . What are the solutions to those equations? ( $(0, 0)$ and $(\pm1, 0)$ are solutions but there are two other solutions also.)

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