# Thread: Use of the Signum Function

1. ## Use of the Signum Function

This is a tough one... And I really need someones help

Graph the following equation:

x(y-1/2 sgn(x)- 1/2 sgn (x-1)) (x-1) (y-2(x-3)^2) √(2y-y^2 )= 0

2. Originally Posted by FutureMD2be
This is a tough one... And I really need someones help
Graph the following equation:
$\displaystyle x(y-\frac{1}{2} sgn(x)- \frac{1}{2} sgn (x-1)) (x-1) (y-2(x-3)^2) \sqrt{(2y-y^2 )}= 0$
Have you considered the cases:

x = 0 , x = 1, and the intervals (-infinity, 0), (0, 1), and (1, infinity)?

at x = 0, for example, you get:
$\displaystyle -1 (y- 18) \sqrt{2y-y^2} = 0$

I tried for the first open interval but get a pretty messy implicit relationship. Are you asked to sketch the graph by hand?

Good luck!

3. Hello, FutureMD2be!

Graph the following equation:

. . $\displaystyle \underbrace{x}_{[1]}\cdot\underbrace{\left[y-\tfrac{1}{2}\text{sgn}(x)- \tfrac{1}{2}\text{sgn}(x-1)\right]}_{[2]} \cdot \underbrace{(x-1)}_{[3]} \cdot \underbrace{\left[y-2(x-3)^2\right]}_{[4]}\cdot \underbrace{\sqrt{2y-y^2}}_{[5]} \;=\; 0$
The function is a product of five factors set equal to zero.

We can consider each factor separately.

$\displaystyle [1]\;\,x = 0$ . . . vertical line (the $\displaystyle y$-axis)

$\displaystyle [3]\;\,x-1\:=\:0 \quad\Rightarrow\quad x \:=\:1$ . . . vertical line

$\displaystyle [4]\;\,y -2(x-3)^2\:=\:0 \quad\Rightarrow\quad y \:=\:2(x-3)^2$ . . . parabola, opens upward, vertex (3,0)

$\displaystyle [5]\;\;\sqrt{2y-y^2} \:=\:0 \quad\Rightarrow\quad y(2-y)\:=\:0\quad\Rightarrow\quad y\:=\;0,\:y\:=\:2$ . . . horizontal lines

$\displaystyle [2]\;\;y \:=\:\tfrac{1}{2}\text{sgn}(x) + \tfrac{1}{2}\text{sgn}(x-1)$
Recall the definition: .$\displaystyle \text{sgn}(x) \;=\;\begin{Bmatrix}1 & x > 0 \\ 0 & x = 0 \\ \text{-}1 & x < 0 \end{Bmatrix}$

We must consider five cases . . .

$\displaystyle x > 1\!:\;\;y \:=\:\tfrac{1}{2}(1) + \tfrac{1}{2}(1) \:=\:1$

$\displaystyle x = 1\!:\;\;y \:=\:\tfrac{1}{2}(1) + \tfrac{1}{2}(0) \:=\:\tfrac{1}{2}$

$\displaystyle 0 < x < 1\!:\;\;y \:=\:\tfrac{1}{2}(1) + \tfrac{1}{2}(\text{-}1) \:=\: 0$

$\displaystyle x = 0\!:\;\;y \:=\:\tfrac{1}{2}(0) + \tfrac{1}{2}(\text{-}1) \:=\:-\tfrac{1}{2}$

$\displaystyle x < 0\!:\;\;y \:=\:\tfrac{1}{2}(\text{-}1) + \tfrac{1}{2}(\text{-}1) \:=\:\text{-}1$

The graph looks like this:

Code:
                |
1+       o=======
|
+       *
|
- - - - - o=======o - - - -
|       1
*
|
=======o-1
|

Now plot all five graphs on the same set of coordinate axes.