MATH 221 Lecture 17

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

Last update: 6 August 2012

Graphing Examples

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

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

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

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

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

Notes:
(a)	$\text{This is}{x}^3-x\text{shifted up by 1.}$
(b)	$x^3-x=x(x^2-1)=x(x+1)(x-1)$
(c)	${\displaystyle \frac{d(x^3-x)}{dx}=3x^2-1}$ ${\displaystyle \text{So}{\frac{d(x^3-x)}{dx}|}_{x=a}\text{is}0\text{when}a=\pm \frac{1}{\sqrt{3}}}$

So

Example:$$ $\text{Graph}x-x^2-27$

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

So

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

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

$f\left(x\right)$ 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.

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