Find the local maximum and minimum values and saddle points of the function.


The objective of this question is to find the local minimum and maximum values and the saddle points of the given multi-variable function. For this purpose, a second derivative test is used.

A function of several variables, also known as a real multivariate function, is a function having more than one argument, all of which are real variables. A saddle point is a point on the surface of a function’s graph where the orthogonal slopes are all zero and the function does not have a local extremum.

A point $(x,y)$ on the graph of a function is said to be a local maximum if its $y$ coordinate is greater than all the other $y$ coordinates on the graph at the points near to $(x,y)$.  More accurately, we can say that $(x,f(x))$ will be a local maximum if $f(x)\geq f(z)$, $x,z\in(a,b)$ and $z\in$ domain of $f$. In a similar way, $(x,y)$ will be a local minimum if $y$ is the smallest coordinate locally, or $(x,f(x))$ will be a local minimum if $f(x)\leq f(z)$, $x,z\in(a,b)$ and $z\in$ domain of $f$.

Local maximum and minimum points on a function graph are quite distinguishable, and thus beneficial in recognizing the shape of the graph.

Expert Answer

The given function is $f(x,y)=y^4+4y^2-x^2$.

First, find the partial derivatives of the above function as:

$f_x(x,y)=-2x$  and $f_y(x,y)=4y^3+8y$

For critical points, let:

$-2x=0\implies x=0$

and $4y^3+8y=0\implies 4y(y^2+2)=0$

or  $y=0$

Hence, the function has critical points $(x,y)=(0,0)$.

Now for the discriminant $(D)$, we need to find the second order partial partial derivatives as:




And so:




Now at $(0,0)$:


Therefore, the function has a saddle point at $(0,0)$, and no local maximum or minimum.


Graph of $f(x,y)=y^4+4y^2-x^2$


Locate the saddle points, relative minimum or maximum, and the critical points of the function $f$ defined by:



Step 1



Step 2

$f_x=0\implies 2x+3y-3=0$  or  $2x+3y=3$                (1)

$f_y=0\implies 3x+8y=0$                             (2)

Simultaneous solution of (1) and (2) gives us:

$\left(\dfrac{24}{7},-\dfrac{9}{7}\right)$ as a critical point.

Step 3

For the discriminant $D$:







Since, $D>0$ and $f_{xx}\left(\dfrac{24}{7},-\dfrac{9}{7}\right)>0$, so by the second derivative test, the function has a local minimum at $\left(\dfrac{24}{7},-\dfrac{9}{7}\right)$.

 Images/mathematical drawings are created with GeoGebra.

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