For $$f(x,y)=x^2+y^3$$
Is there a saddle point at (0,0) or does the function have to have 2 or more sides going down like $$g(x,y)=x^2-y^2$$

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 Okay so have you taken the required derivative and found your critical points?
 The critical point is (0,0). I'm just wondering if f(0,0) is a saddle point when the graph is shaped more like a chair than a saddle.

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 Quote by cambo86 The critical point is (0,0). I'm just wondering if f(0,0) is a saddle point when the graph is shaped more like a chair than a saddle.
The exact shape doesn't matter. A saddle point is a stationary point that isn't a local max or min. Is yours?

 Quote by cambo86 The critical point is (0,0). I'm just wondering if f(0,0) is a saddle point when the graph is shaped more like a chair than a saddle.
Are you familiar with the second derivative test? As in where you check :

$D = f_{xx}f_{yy} - f_{xy}^{2}$

This will allow you to see if your function has a saddle, min or max at a critical point ( Doesn't alwayyyyys work though ).

 Quote by Dick The exact shape doesn't matter. A saddle point is a stationary point that isn't a local max or min. Is yours?
Yes, thank you.

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 Quote by Dick The exact shape doesn't matter. A saddle point is a stationary point that isn't a local max or min. Is yours?
The Hessian test fails in this case. If H is the Hessian evaluated at the stationary point, the exact second-order conditions are: (i) a necessary condition for a minimum is that H is positive semi-definite; and (ii) a sufficient condition for a strict local minimum is that H is positive definite. For testing a max, just test -H for a min of -f (or replace positive (semi) definite by negative (semi) definite).

Note that there is a bit of mismatch between the necessary and sufficient second-order conditions: there is no sufficient condition for a non-strict local min, so functions like that of the OP slip through the cracks. This happens as well in one dimension, where testing functions like f(x) = x^3 for an optimum at x = 0 cannot just rely on the second derivative.

For the OP's function f(x,y), the Hessian at (0,0) is positive semi-definite, so (0,0) cannot be a local *maximum *; it must either be a local minimum or a saddle point. In fact, f(x,0) = x^2, so x = 0 is a minimum along the line (x,0), and f(0,y) = y^3, so y = 0 is an inflection point along the line (0,y). Thus, there are points near (0,0) giving f(x,y) > f(0,0) and other points near (0,0) giving f(x,y) < f(0,0); that is more-or-less what we mean by a 'saddle point'.