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 15

A function $f\left(x\right)$ is continuous at $x=a$ if it doesn't jump at $x=a,$ i.e. if $\underset{x\to a}{\text{lim}}f\left(x\right)=f\left(a\right)\text{.}$ $$\begin{array}{c} a x y y=f⁡(x) \end{array}$$ Not continuous at $x=a\text{.}$ Think about $$\frac{df}{dx}{|}_{x=a}=\underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x}$$ in terms of the graph $$\begin{array}{c} x y a+Δx a f⁡(a+Δx) f⁡(a) y=f⁡(x) \end{array}$$ $$\begin{array}{ccc}{\displaystyle \frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x}}& =& {\displaystyle \frac{\text{change in}\hspace{0.17em}f}{\text{change in}\hspace{0.17em}x}}\\ & =& {\displaystyle \frac{\text{rise}}{\text{run}}}\\ & =& {\displaystyle \text{slope of line connecting}\hspace{0.17em}(a,f\left(a\right))\hspace{0.17em}\text{and}\hspace{0.17em}(a+\mathrm{\Delta }x,f(a+\mathrm{\Delta }x))\text{.}}\end{array}$$ $\underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(a+\mathrm{\Delta }x)-f\left(a\right)}{\mathrm{\Delta }x}=$ slope of $f$ at the point $x=a\text{.}$

A function $f\left(x\right)$ differentiable at $x=a$ if the derivative $\frac{df}{dx}{|}_{x=a}$ exists, i.e. if the slope of the graph of $f\left(x\right)$ at $x=a$ exists.

Graph $$f\left(x\right)=\left|x\right|=\{\begin{array}{cc}x,& \text{if}\hspace{0.17em}x\ge 0,\\ -x,& \text{if}\hspace{0.17em}x\le 0\text{.}\end{array}$$ $$\begin{array}{c} x y 1 1 -1 \end{array}$$ Then $$\frac{df}{dx}{|}_{x=a}=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}a>0,\\ -1,& \text{if}\hspace{0.17em}a<0,\\ \text{does not exist,}& \text{if}\hspace{0.17em}a=0\text{.}\end{array}$$ So $f$ is not differentiable at $x=0\text{.}$

Graph $y=x^{\frac{1}{3}}\text{.}$ $$\begin{array}{c} x y -1 1 -1 1 y=x3 x y -1 1 -1 1 y=x13 \end{array}$$ Notes:

(a)	$y=x^{\frac{1}{3}}$ is the same as $y^3=x\text{.}$

$$\frac{dy}{dx}=\frac{1}{3}{x}^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}}\text{.}$$ So $\frac{dy}{dx}{|}_{x=0}=\infty \text{.}$ So $f\left(x\right)$ is not differentiable at $x=0\text{.}$

A function $f\left(x\right)$ is increasing at $x=a$ if it is going up at $x=a,$ i.e. if $f(a+\mathrm{\Delta }x)>f\left(a\right)$ for all small $\mathrm{\Delta }x>0,$
i.e. if slope is positive,
i.e. if $\frac{df}{dx}{|}_{x=a}>0\text{.}$ $$\begin{array}{c} x y a+Δx a f⁡(a+Δx) f⁡(a) y=f⁡(x) increasing at x=a x y a+Δx a f⁡(a+Δx) f⁡(a) y=f⁡(x) decreasing at x=a \end{array}$$

A function $f\left(x\right)$ is decreasing at $x=a$ if it is going down at $x=a,$ i.e. if $f(a+\mathrm{\Delta }x)<f\left(a\right)$ for all small $\mathrm{\Delta }x>0,$
i.e. if the slope of $f\left(x\right)$ at $x=a$ is negative,
i.e. if $\frac{df}{dx}{|}_{x=a}<0\text{.}$

$f$ is concave up at $x=a$ if it is right side up bowl shaped $x=a,$ i.e. if the slope of $f$ is getting larger at $x=a,$
i.e. if $\frac{df}{dx}$ is increasing at $x=a,$
i.e. if $\frac{d^{2}f}{d{x}^2}{|}_{x=a}>0\text{.}$

$f$ is concave down at $x=a$ if it is upside down bowl shaped $x=a,$ i.e. if the slope of $f$ is getting smaller,
i.e. if $\frac{df}{dx}$ is decreasing at $x=a,$
i.e. if $\frac{d^{2}f}{d{x}^2}{|}_{x=a}<0\text{.}$

A point of inflection is a point where $f$ changes from concave up to concave down, or from concave down to concave up. $$\begin{array}{c} x y p a1 a2 concave up at x=a1 point of inflection concave down at x=a1 \end{array}$$

A local maximum is a point $x=a$ where $f\left(a\right)$ is bigger then the $f\left(x\right)$ around it.

A local minimum is a point $x=a$ where $f\left(a\right)$ is smaller then the $f\left(x\right)$ around it. i.e. $f\left(a\right)<f(a+\mathrm{\Delta }x)$ for small $\mathrm{\Delta }x\text{.}$ $$\begin{array}{c} x y a1 a2 a3 a4 local maximum at x=a1 local minimum at x=a2 local maximum at x=a3 local minimum at x=a4 \end{array}$$

A critical point is a point where a maximum or minimum might occur.

Note:

(1)	If $f\left(x\right)$ is continuous and differentiable and $x=a$ is a maximum then $$\frac{df}{dx}{|}_{x=a}=0\text{and}\frac{d^{2}f}{d{x}^2}{|}_{x=a}<0\text{.}$$
(2)	If $f\left(x\right)$ is continuous at $x=a,$ $f\left(x\right)$ is differentiable at $x=a,$ $$\frac{df}{dx}{|}_{x=a}=0\text{and}\frac{d^{2}f}{d{x}^2}{|}_{x=a}>0$$ then $x=a$ is a minimum.

Where can a maximum or minimum occur?

(a)	A point $x=a$ where $f\left(x\right)$ is differentiable and $\frac{df}{dx}{|}_{x=a}=0\text{.}$ $$\begin{array}{c} x y a \end{array}$$
(b)	A point $x=a$ where $f\left(x\right)$ is not continous. $$\begin{array}{c} x y 1 2 3 1 2 \end{array}$$ $$f\left(x\right)=\{\begin{array}{cc}{x}^2+1,& \text{if}\hspace{0.17em}0\le x\le 1,\\ 2-x,& \text{if}\hspace{0.17em}x>1\text{.}\end{array}$$ $x=1$ is a maximum.
(c)	A point $x=a$ on the boundary of where $f\left(x\right)$ is defined. $$\begin{array}{c} x y 1 2 3 1 2 \end{array}$$ $$f\left(x\right)=\{\begin{array}{cc}{x}^2+1,& \text{if}\hspace{0.17em}0\le x\le 1,\\ 2-x,& \text{if}\hspace{0.17em}x>1\text{.}\end{array}$$ $x=0$ is a minimum.

Notes and References

These are a typed copy of Lecture 15 from a series of handwritten lecture notes for the class MATH 221 given on October 11, 2000.

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