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 17

Graphing Examples

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

Notes:

(a)	The $x^2$ indicates this is a parabola.
(b)	Since the coefficient of $x^2$ is positive this is a concave up parabola.
(c)	$3x^2-2x-1=(x-1)(3x+1)\text{.}$ We know $x-1$ should be a factor since when you plug in $1,$ $3·1^2-2·1-1=0\text{.}$
(d)	$f\left(x\right)=0$ if $x=1$ or if $x=-\frac{1}{3}\text{.}$ $$\begin{array}{c} x y -1 1 13 -13 -43 y=f⁡(x) \end{array}$$
(e)	The minimum will be where $\frac{df}{dx}{|}_{x=a}$ is $0\text{.}$ $\frac{df}{dx}{|}_{x=a}=6x-2{|}_{x=a}=6a-2\text{.}$ This $0$ when $a=\frac{1}{3}\text{.}$ $$f\left(\frac{1}{3}\right)=3{\left(\frac{1}{3}\right)}^2-2\left(\frac{1}{3}\right)-1=\frac{1}{3}-\frac{2}{3}-1=-\frac{4}{3}\text{.}$$

Graph $f\left(x\right)=2x^3-21x^2+36x-20\text{.}$

Notes:

(a)	If $x\to \infty$ then $f\left(x\right)\to \infty \text{.}$
(b)	If $x\to -\infty$ then $f\left(x\right)\to -\infty \text{.}$
(c)	$\frac{df}{dx}=6x^2-42x+36=6(x^2-7x+6)=6(x-6)(x-1)\text{.}$ So $\frac{df}{dx}$ is $0$ when $x=6$ and when $x=1\text{.}$ $$\begin{array}{c}{\displaystyle f\left(6\right)=2·6^3-21·6^2+36·6-20=6^2(12-21+6)-20=6^2(-3)-20=-\mathrm{128,}}\\ {\displaystyle f\left(1\right)=2-21+36-20=38-41=-3\text{.}}\end{array}$$
(d)	$$\begin{array}{ccc}{\displaystyle \frac{d^{2}f}{d{x}^2}{|}_{x=6}}& =& {\displaystyle 12x-42{|}_{x=6}=72-42=30>0,\text{concave up,}}\\ {\displaystyle \frac{d^{2}f}{d{x}^2}{|}_{x=1}}& =& {\displaystyle 12x-42{|}_{x=1}=12-42=-30<0,\text{concave down.}}\end{array}$$

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

Notes:

(a)	This is $x^3-x$ shifted up by $1\text{.}$
(b)	$x^3-x=x(x^2-1)=x(x+1)(x-1)\text{.}$ $$\begin{array}{c} x y 1 -1 -13 13 y=x3-x \end{array}$$ $\frac{d(x^3-x)}{dx}=3x^2-1\text{.}$ So $\frac{d(x^3-x)}{dx}{|}_{x=a}$ is $0$ when $a=\pm \frac{1}{\sqrt{3}}\text{.}$

So $$\begin{array}{c} x y 1 -1 1 -13 13 y=x3-x+1 \end{array}$$

Graph $x-x^2-27\text{.}$

Notes:

(a)	The $-x^2$ indicates to us that this is a concave down parabola.
(b)	$$\begin{array}{ccc}{\displaystyle x-x^2-27}& =& {\displaystyle -(x^2-x+27)}\\ & =& {\displaystyle -(x^2-x+\frac{1}{4}-\frac{1}{4}+27)}\\ & =& {\displaystyle -({(x-\frac{1}{2})}^2+26\frac{3}{4})\text{.}}\end{array}$$

So

For which values of $x$ is $$f\left(x\right)=\{\begin{array}{cc}\frac{|x-a|}{x-a},& \text{if}\hspace{0.17em}x\ne a,\\ 1,& \text{if}\hspace{0.17em}x=a\end{array}$$ continuous?

Notes: $$\begin{array}{ccc}{\displaystyle f\left(x\right)}& =& {\displaystyle \{\begin{array}{cc}\frac{|x-a|}{x-a},& \text{if}\hspace{0.17em}x\ne a,\\ 1,& \text{if}\hspace{0.17em}x=a\end{array}}\\ & =& {\displaystyle \{\begin{array}{cc}\frac{x-a}{x-a},& \text{if}\hspace{0.17em}x\ne a\hspace{0.17em}\text{and}\hspace{0.17em}x-a>0,\\ -\frac{(x-a)}{x-a},& \text{if}\hspace{0.17em}x\ne a\hspace{0.17em}\text{and}\hspace{0.17em}x-a<0,\\ 1,& \text{if}\hspace{0.17em}x=a\end{array}}\\ & =& {\displaystyle \{\begin{array}{cc}1,& \text{if}\hspace{0.17em}x\ne a\hspace{0.17em}\text{and}\hspace{0.17em}x>a,\\ -1,& \text{if}\hspace{0.17em}x\ne a\hspace{0.17em}\text{and}\hspace{0.17em}x<a,\\ 1,& \text{if}\hspace{0.17em}x=a\text{.}\end{array}}\end{array}$$ $$\begin{array}{c} x y a -1 1 y=f⁡(x) \end{array}$$ $f\left(x\right)$ has a jump at $x=a\text{.}$ So $f\left(x\right)$ is not continuous at $x=a\text{.}$

Notes and References

These are a typed copy of Lecture 17 from a series of handwritten lecture notes for the class MATH 221 given on October 16, 2000.

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