# Finding Critical Points and Local Maxima/Minima or Saddle Point

I need help to find critical points of the function: $$f(x,y)=\frac{-x^3}{3}+x-y^2$$ Then I have to classify these critical points as local maxima/minima or saddle points.

I thought that to find the critical points, I have to find the 1st derivative and to find local max/min or saddle, I have to use the second derivative test. I am having a little trouble both in finding first and second derivatives and how to use it to find the given above. Can someone help me?

Edit: I found the critical points to be $(1,0)$ and $(-1,0)$. Can someone verify this as well?

• Yepp, you are right. – user251257 Jul 13 '15 at 23:38
• @user251257 So how do I continue on with this? – Neel Jul 13 '15 at 23:39
• Like you said. You compute the 2nd derivative, the Hessian matrix at the critical points. If the Hessian is positive definite, you have a local minimum. If it is negative definite, you have a local maximum. If is is indefinite, you have a saddle point. – user251257 Jul 13 '15 at 23:41
• @user251257 I haven't seen the hessian matrix before. Can you please show me how this is done in this case? – Neel Jul 13 '15 at 23:41
• The Hessian is $$\nabla^2 f(x,y) = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2}(x,y) & \frac{\partial^2 f}{\partial y \partial x}(x,y) \\ \frac{\partial^2 f}{\partial x \partial y}(x,y) & \frac{\partial^2 f}{\partial y^2}(x,y) \end{bmatrix}.$$ For example $\frac{\partial^2}{\partial y^2}f(x,y) = \frac{d}{dy}(-2y) = -2$. – user251257 Jul 13 '15 at 23:48

You have $$\nabla f(x,y) = \begin{bmatrix} -x^2 + 1 \\ - 2y \end{bmatrix}.$$ So the critical points are $(-1,0)$ and $(1,0)$.
Now, the Hessian is $$\nabla^2 f(x,y) = \begin{bmatrix} -2x & 0 \\ 0 & -2 \end{bmatrix}.$$ The eigenvalues of $\nabla^2 f(-1,0)$ are $2$ and $-2$. Thus, $\nabla^2 f(-1,0)$ is indefinite and $(-1,0)$ is a saddle point. The eigenvalues of $\nabla^2 f(1,0)$ are $-2$ and $-2$. Thus, $\nabla^2 f(1,0)$ is negative definite and $(1,0)$ is a maximum.
• A (symmetric) matrix $A\in\mathbb R^{n\times n}$ is negative definite if and only if for every $x\in\mathbb R^n\setminus\{0\}$ it follows $x^T A x < 0$. See en.wikipedia.org/wiki/Positive-definite_matrix – user251257 Jul 14 '15 at 0:38
• @Neel $A$ represents a quadratic form; when $A$ is diagonal you get $x^T Ax=\sum\lambda_i x_i^2$, so when all $\lambda_i<0$ we get $x^T Ax=\sum\lambda_i x_i^2> 0$, when all $\lambda_i>0$ you have $x^T Ax\ge0$, and otherwise it's indefinite (note this is for nonzero $x$). – oldrinb Jul 14 '15 at 0:48