## MATH 221 Lecture 17

Last update: 6 August 2012

## Graphing Examples

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}f\left(x\right)=3{x}^{2}-2x-1$

 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) $3{x}^{2}-2x-1=\left(x-1\right)\left(3x+1\right)$. We know $x-1$ should be a factor since when you plug in 1, $3·{1}^{2}-2·1-1=0$. (d) $f\left(x\right)=0\phantom{\rule{0.5em}{0ex}}\text{if}\phantom{\rule{0.5em}{0ex}}x=1\phantom{\rule{0.5em}{0ex}}\text{or if}\phantom{\rule{0.5em}{0ex}}x=-\frac{1}{3}$ x y -1 1 $\frac{1}{3}$ $-\frac{1}{3}$ $-\frac{4}{3}$ $y=f\left(x\right)$ (e) $\text{The minimum will be where}\phantom{\rule{0.5em}{0ex}}{\frac{df}{dx}|}_{x=a}\phantom{\rule{0.5em}{0ex}}\text{is 0.}$ ${\frac{df}{dx}|}_{x=a}={6x-2|}_{x=a}=6a-2\text{. This is 0 when}\phantom{\rule{0.5em}{0ex}}a=\frac{1}{3}\text{.}$ $f\left(\frac{1}{3}\right)=3{\left(\frac{1}{3}\right)}^{2}-2\left(\frac{1}{3}\right)-1=\frac{1}{3}-\frac{2}{3}-1=-\frac{4}{3}$

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}f\left(x\right)=2{x}^{3}-21{x}^{2}+36x-20$.

 Notes: (a) $\text{If}\phantom{\rule{0.5em}{0ex}}x\to \infty \phantom{\rule{0.5em}{0ex}}\text{then}\phantom{\rule{0.5em}{0ex}}f\left(x\right)\to \infty$. (b) $\text{If}\phantom{\rule{0.5em}{0ex}}x\to -\infty \phantom{\rule{0.5em}{0ex}}\text{then}\phantom{\rule{0.5em}{0ex}}f\left(x\right)\to -\infty$. (c) $\frac{df}{dx}=6{x}^{2}-42x+36=6\left({x}^{2}-7x+6\right)=6\left(x-6\right)\left(x-1\right)$. $\text{So}\phantom{\rule{0.5em}{0ex}}\frac{df}{dx}\phantom{\rule{0.5em}{0ex}}\text{is 0 when}\phantom{\rule{0.5em}{0ex}}x=6\phantom{\rule{0.5em}{0ex}}\text{and when}\phantom{\rule{0.5em}{0ex}}x=1$. $f\left(6\right)=2·{6}^{3}-21·{6}^{2}+36·6-20={6}^{2}\left(12-21+6\right)-20$ $\phantom{\rule{14.4em}{0ex}}={6}^{2}\left(-3\right)-20=-128$ $f\left(1\right)=2-21+36-20=38-41=-3$. (d) ${\frac{{d}^{2}f}{d{x}^{2}}|}_{x=6}={12x-42|}_{x=6}=72-42=30>0$. $\phantom{\rule{1em}{0ex}}$Concave up ${\frac{{d}^{2}f}{d{x}^{2}}|}_{x=1}={12x-42|}_{x=1}=12-42=-30<0$. $\phantom{\rule{1em}{0ex}}$Concave down

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}f\left(x\right)={x}^{3}-x+1$.

 Notes: (a) $\text{This is}\phantom{\rule{0.5em}{0ex}}{x}^{3}-x\phantom{\rule{0.5em}{0ex}}\text{shifted up by 1.}$ (b) ${x}^{3}-x=x\left({x}^{2}-1\right)=x\left(x+1\right)\left(x-1\right)$ (c) $\frac{d\left({x}^{3}-x\right)}{dx}=3{x}^{2}-1$ $\text{So}\phantom{\rule{0.5em}{0ex}}{\frac{d\left({x}^{3}-x\right)}{dx}|}_{x=a}\phantom{\rule{0.5em}{0ex}}\text{is}\phantom{\rule{0.5em}{0ex}}0\phantom{\rule{0.5em}{0ex}}\text{when}\phantom{\rule{0.5em}{0ex}}a=±\frac{1}{\sqrt{3}}$

So

Example:$\phantom{\rule{1em}{0ex}}$ $\text{Graph}\phantom{\rule{0.5em}{0ex}}x-{x}^{2}-27$

 Notes: (a) The ${x}^{2}$ indicates to us that this is a concave down parabola. (b)

So

Example:$\phantom{\rule{1em}{0ex}}$ For which values of $x$ is $\phantom{\rule{1em}{0ex}}f\left(x\right)=\left\{\begin{array}{ccc}\frac{|x-a|}{x-a},& & \text{if}\phantom{\rule{0.5em}{0ex}}x\ne a,\\ 1,& & \text{if}\phantom{\rule{0.5em}{0ex}}x=a,\end{array}\phantom{\rule{2em}{0ex}}$ continuous?

 Notes: $f\left(x\right)$ $=$ $\left\{\begin{array}{ccc}\frac{|x-a|}{x-a},& & \text{if}\phantom{\rule{0.5em}{0ex}}x\ne a,\\ 1,& & \text{if}\phantom{\rule{0.5em}{0ex}}x=a,\end{array}$ $=$ $\left\{\begin{array}{ccc}\frac{x-a}{x-a},& & \text{if}\phantom{\rule{0.5em}{0ex}}x\ne a\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}x-a>0,\\ -\frac{\left(x-a\right)}{x-a},& & \text{if}\phantom{\rule{0.5em}{0ex}}x\ne a\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}x-a<0,\\ 1,& & \text{if}\phantom{\rule{0.5em}{0ex}}x=a,\end{array}$ $=$ $\left\{\begin{array}{ccc}1,& & \text{if}\phantom{\rule{0.5em}{0ex}}x\ne a\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}x>a,\\ -1,& & \text{if}\phantom{\rule{0.5em}{0ex}}x\ne a\phantom{\rule{0.5em}{0ex}}\text{and}\phantom{\rule{0.5em}{0ex}}x

$f\left(x\right)$ has a jump at $x=a$
So $f\left(x\right)$ is not continuous at $x=a$

## Notes and References

These are a typed copy of lecture notes given by Arun Ram on October 16, 2000.