# Math Help - saddle point

Find both stationary points of:
$f(x,y) = 2x^3 + 3x^2 + 6xy^2 + 9y^2 - 5$

and identify the nature of each one. Show on a diagram which points near the saddle point are such that
f > - 4 and which points are such that f < - 4 .

I'm having trouble understanding how to draw this diagram...the solution given by my lecturer is attached.

I've used Taylor series to examine the behaviour near the saddle point which led me to:

f(-1 + h, k) + 4 = -3h^2 + 3k^2 = 3(k + h)(k - h))[/tex]

So I see where k = h and k = -h comes from.

But I do not get, how to determine which areas are greater than or equal to 4 (but i know once you have found one they alternate in each section).

I tried sticking in (-1,1) into f(x,y) which gave me -1, so why does the diagram say f > 4??

2. Hello, hunkydory19!

a) Find both stationary points of: . $f(x,y) \:= \:2x^3 + 3x^2 + 6xy^2 + 9y^2 - 5$
and identify the nature of each one.

We have: . $\begin{array}{cccccc} f_x &=& 6x^2 + 6x + 6y^2 &=& 0 & {\color{blue}[1]} \\ f_y &=& 12xy + 18y &=&0 &{\color{blue}[2]} \end{array}$

These simplify to: . $\begin{array}{cccc}x^2+x + y^2 &=& 0 & {\color{blue}[3]} \\ 6y(2x+3) &=& 0 & {\color{blue}[4]} \end{array}$

From [2], we have: . $x = -\frac{3}{2}\:\text{ or }\;y = 0$

. . If $x = -\frac{3}{2}$, then [3] is: . $\frac{9}{4}-\frac{3}{2} + y^2 \:=\:0\quad\Rightarrow\quad y^2 \,=\,-\frac{3}{4}$ . . . no real roots

. . If $y = 0$, then [3] is: . $x^2+x \:=\:0\quad\Rightarrow\quad x \:=\:0,\,-1$

We have two stationary points: . $(0,0,-5)\,\text{ and }\,(-1,0,-4)$

And we can show that: . $\begin{Bmatrix}(0,0,-5)\text{ is a minimum} \\ (-1,0,-4)\text{ is a saddle point} \end{Bmatrix}$

Show on a diagram which points near the saddle point are such that
f > -4 and which points are such that f < -4.
Code:
                |
\  o  /  |
\   /   |
\ /    |
- o - - * - - o - -
/P\    |
/   \   |
/  o  \  |
|
Point $P(-1,0,-4)$ is a saddle point.

When $x = -1,\:y = 0$, the function is at $f = -4$

We investigate the value of $f$ in the four compass directions.

To the right: . $f(0,0) \:=\:2(0^3) + 3(0^2) + 6(0)(0^2) + 9(0^2) - 5 \;=\;-5$ . . . less than -4.
. . The surface is below $P$.

Above $P$: . $f(\text{-}1,1) \:=\:2(\text{-}1)^3 + 3(\text{-}1)^2 + 6(\text{-}1)(1^2) + 9(1^2) - 5 \;=\;-1$ . . . greater than -4.
. . The surface is above $P$.

Get the idea?