MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 August 2014

Lecture 16

Graph $f\left(x\right)=\frac{x^2-1}{x^3-4x}\text{.}$

Notes:

(a)	$y=\frac{x^2-1}{x^3-4x}=\frac{(x+1)(x-1)}{x(x^2-4)}=\frac{(x+1)(x-1)}{x(x+2)(x-2)}\text{.}$
(b)	If $x=1$ then $y=0\text{.}$
(c)	If $x=-1$ then $y=0\text{.}$
(d)	If $x=2^{+}$ then $y\to \infty \text{.}$ $\left(\frac{\text{pos}·\text{pos}}{\text{pos}·\text{pos}·\text{pos}}\right)$
(e)	If $x=2^{-}$ then $y\to -\infty \text{.}$ $\left(\frac{\text{pos}·\text{pos}}{\text{pos}·\text{pos}·\text{neg}}\right)$
(f)	If $x=0^{+}$ then $y\to \infty \text{.}$ $\left(\frac{\text{pos}·\text{neg}}{\text{pos}·\text{pos}·\text{neg}}\right)$
(g)	If $x=0^{-}$ then $y\to -\infty \text{.}$ $\left(\frac{\text{pos}·\text{neg}}{\text{neg}·\text{pos}·\text{neg}}\right)$
(h)	If $x=-2^{+}$ then $y\to \infty \text{.}$ $\left(\frac{\text{neg}·\text{neg}}{\text{neg}·\text{pos}·\text{neg}}\right)$
(i)	If $x=-2^{-}$ then $y\to -\infty \text{.}$ $\left(\frac{\text{neg}·\text{neg}}{\text{neg}·\text{neg}·\text{neg}}\right)$
(j)	$y=\frac{x^2-1}{x^3-4x}$ is the same as $(-y)=\frac{{(-x)}^2-1}{{(-x)}^3-4(-x)}$ so if we flip $y$ to $-y$ and $x$ to $-x$ the graph stays the same.
(k)	As $x\to \infty$ then $y\to 0^{+}\text{.}$
(l)	As $x\to -\infty$ then $y\to 0^{-}\text{.}$

$$\begin{array}{c} x y -1 -2 1 2 \end{array}$$

Graph $\sqrt{x}+\sqrt{y}=1\text{.}$

Notes:

(a)	If we switch $x$ and $y$ this graph stays the same.
(b)	If $x=0$ then $\sqrt{y}=1,$ so $y=1^2=1\text{.}$
(c)	If $y=0$ then $x=1\text{.}$
(d)	If $x=y$ then $\sqrt{x}+\sqrt{x}=1$ and $\sqrt{x}=\frac{1}{2},$ $x=\frac{1}{4}\text{.}$
(e)	This graph should be similar to $x^2+y^2=1$ or $x=y+1\text{.}$

$$\begin{array}{c} x y 1 1 22 22 x2+y2=1 x y 1 1 12 12 x+y=1 \\ x y 1 1 14 14 x+y=1 \end{array}$$

Graph $\frac{x^2-1}{x^2+1}=f\left(x\right)\text{.}$

Notes:

(a)	$y=\frac{x^2-1}{x^2+1}=\frac{x^2+1-2}{x^2+1}=1-\frac{2}{x^2+1}\text{.}$

$$\begin{array}{c} x y 1 y=1x2+1 \end{array}$$ Notes:

(a)	If $x=0,$ $y=\frac{1}{0^2+1}=\frac{1}{1}=1\text{.}$
(b)	If $x\to \infty$ then $y\to 0^{+}\text{.}$
(c)	If $x\to -\infty$ then $y\to 0^{+}\text{.}$
(d)	This graph stays the same if we flip $x$ to $-x,$ $y=\frac{1}{x^2+1}=\frac{1}{{(-x)}^2+1}\text{.}$

$$\begin{array}{c} x y -2 y=-2x2+1 x y -1 1 y=1-2x2+1 = x2-1 x2+1 \end{array}$$

Graph $\text{ln}(4-x^2)=f\left(x\right)\text{.}$

Notes:

(a)	$y=\text{ln}(4-x^2)=\text{ln}\left((2+x)(2-x)\right)=\text{ln}(2+x)+\text{ln}(2-x)\text{.}$
(b)	If we flip $x$ to $-x$ the graph stays the same since $y=\text{ln}(4-x^2)=\text{ln}(4-{(-x)}^2)\text{.}$

$$\begin{array}{c} x y 1 y=ln⁡x x y -1 y=ln⁡(-x) \\ x y -2 1 -1 ln 3 ln 2 y=ln⁡(x+2) x y -1 1 2 ln 3 ln 2 y=ln⁡(2-x) \\ x y -2 2 1 -1 ln 3 2ln 2 ln 2 \\ y=\text{ln}(4,x^2)=\text{ln}(2+x)+\text{ln}(2-x)\text{.}\end{array}$$

Graph $y=x^{\frac{2}{3}}{(6-x)}^{\frac{1}{3}}\text{.}$

Notes:

(a)	If $x=0$ then $y=0\text{.}$
(b)	If $x=6$ then $y=0\text{.}$
(c)	If $x\to \infty$ then $y\to x^{\frac{2}{3}}{(-x)}^{\frac{1}{3}}=-x\text{.}$
(d)	If $x\to -\infty$ then $y\to \infty$ $\text{(}y\to -x$ again).

Notes and References

These are a typed copy of Lecture 16 from a series of handwritten lecture notes for the class MATH 221 given on October 13, 2000.

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