# Multivariable critical points

• Oct 3rd 2012, 11:12 PM
cubejunkies
Multivariable critical points
Show that f(x,y) = $\sqrt{x^2+ y^2}$ has one critical point P and that f is nondifferentiable at P. Is P a saddle point, minimum, or maximum?

So I found fx = $\frac {2x} {\sqrt {x^2 + y^2}}$ and fy
= $\frac {2y} {\sqrt {x^2 + y^2}}$ and I tried some algebraic gymnastics to try and solve for critical points but unfortunately I'm just flat out stuck (Crying)

Thanks
Anthony
• Oct 3rd 2012, 11:16 PM
Prove It
Re: Multivariable critical points
Quote:

Originally Posted by cubejunkies
Show that f(x,y) = $\sqrt{x^2+ y^2}$ has one critical point P and that f is nondifferentiable at P. Is P a saddle point, minimum, or maximum?

So I found fx = $\frac {2x} {\sqrt {x^2 + y^2}}$ and fy
= $\frac {2y} {\sqrt {x^2 + y^2}}$ and I tried some algebraic gymnastics to try and solve for critical points but unfortunately I'm just flat out stuck (Crying)

Thanks
Anthony

At critical points, all partial derivatives equal to 0. Also, for a function to be differentiable at a point, it need to be defined at that point, continuous at that point, and have continuous partial derivatives at that point.
• Oct 3rd 2012, 11:34 PM
cubejunkies
Re: Multivariable critical points
But how do you solve the partial derivatives when they're set equal to 0? I always get (0,0) but the first partial derivatives don't exist at (0,0) so I can't figure out whether (0,0) if it even is a critical point, is a max min or saddle point because the second partial derivatives will not exist at (0,0) either.
• Oct 4th 2012, 05:07 AM
johnsomeone
Re: Multivariable critical points
Quote:

Originally Posted by cubejunkies
But how do you solve the partial derivatives when they're set equal to 0? I always get (0,0) but the first partial derivatives don't exist at (0,0)...

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Compare that with how you initially posed the question: "Show that f(x,y) = has one critical point P and that f is nondifferentiable at P..."
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Note that f is perfectly well defined at (0,0), even though its partial deriviatives aren't.

To decide about (0,0), you'll need some technique other than the standard one (standard one = examining the eigenvalues of the Hessian). A good first step is to see if you can approximately "see" what the graph of the function looks like.

To "see" the graph of the function, ask yourself about its "level curves". Think of it as graphed according to z = f(x, y). Then think of what points in the x-y plane correspond to z being a constant. Like z = r. This is like slicing the graph with a horizontal plane z = r. I don't know if that makes sense to you, but if it does, you should be able to "see" the graph, and then know the answer to your problem. Once you know the answer, it's pretty easy to "see" (and prove!) it algebraically too.