• cambo86
In summary, the conversation discusses a function f(x,y) and whether or not it has a saddle point at (0,0). The Hessian test is mentioned as a way to determine if a function has a saddle, min, or max at a critical point. The necessary and sufficient second-order conditions for a minimum are also mentioned, as well as the Hessian at (0,0) for f(x,y). The conclusion is that (0,0) is either a local minimum or a saddle point for f(x,y).
cambo86
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 $$g(x,y)=x^2-y^2$$

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.

cambo86 said:
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?

cambo86 said:
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 ).

Dick said:
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.

Dick said:
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'.

1. What is a saddle point?

A saddle point is a critical point on a surface where the tangent plane is horizontal in one direction and vertical in another, resembling the shape of a saddle. It is a point where the first partial derivatives of a function are equal to zero, but it is neither a maximum nor a minimum.

2. How do you find the saddle point of a function?

To find the saddle point, you need to calculate the first and second partial derivatives of the function. Then, set the first derivatives equal to zero and solve for the variables. If the second derivative is positive, the point is a local minimum, and if it is negative, the point is a local maximum. If the second derivative is zero, the point is a saddle point.

3. What is the significance of (0,0) in the saddle point at f(x,y)=x^2+y^3?

In this function, (0,0) represents the origin or the point where both x and y are equal to zero. This point is the critical point where the first partial derivatives are equal to zero, making it a potential saddle point.

4. How does the graph of f(x,y)=x^2+y^3 look like at the saddle point (0,0)?

The graph of this function at the saddle point (0,0) resembles a saddle, with a horizontal tangent plane in one direction and a vertical tangent plane in the other direction. The function increases in one direction from the saddle point and decreases in the other direction.

5. What is the significance of the second derivatives in determining a saddle point?

The second derivatives of a function at a critical point determine the concavity of the function. If the second derivatives are both positive or both negative, the point is a local minimum or maximum, respectively. However, if the second derivatives have different signs, the point is a saddle point.

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