Precalculus Index Cards

FUNCTIONS

A FUNCTION is a rule that assigns to each input EXACTLY ONE output. If $\,x\,$ is the INPUT, then $\,f(x)\,$ is the corresponding output.

[labels on graphic] INPUT $\,x\,,$ NAME OF FCT $\,f\,,$ OUTPUT $\,f(x)$

What is $\,f(2)\,$? It is the OUTPUT from the function $\,f\,$ when the INPUT is $\,2\,.$

What is $\,g(x+h)\,$? It is the OUTPUT from the fct $\,g\,$ when the INPUT is $\,x+h\,.$

GRAPH OF A FUNCTION

The GRAPH OF A FCT $\,f\,$ is a picture, in the coordinate plane, of all its (INPUT,OUTPUT) pairs.

Precisely,

$$ \text{GRAPH OF $\,f$} = \{(x,f(x))\ |\ x\in\text{dom}(f)\} $$

Let

$$ \begin{align} \text{dom}(f) &= \text{DOMAIN OF $\,f$}\cr &= \text{set of allowable inputs} \end{align} $$

EX: $\,f(x) = \sqrt{x}$
$\text{dom}(f) = [0,\infty]$

$x$	$f(x)$
$0$	$0$
$1$	$1$
$4$	$2$
$9$	$3$

[labels on graph] point $\,(x,\sqrt x)\,,$ $\,y = \sqrt x\,,$ GRAPH OF $\,f\,,$ $\,y = f(x)$

SOLVING LINEAR INEQUALITIES

Remember: If $\,C\lt 0\,,$ then

$$A\le B\ \ \iff\ \ CA\ge CB$$

[labels on image] Why? $\,-1\ge -2\,,$ $\,1\lt 2\,,$ tick marks $\,-2\,,$ $\,-1\,,$ $\,0\,,$ $\,1\,,$ $\,2$

SOLVE:

$$ \begin{gather} 5-3x\le -16\cr \text{(subtract $5$)}\cr\cr -3x\le -21\cr \text{(divide by $-3$)}\cr\cr x\ge 7 \end{gather} $$

SOLN SET: $\,[7,\infty)$

‘SPOT CHECK’:

$$ \begin{gather} 5 - 3(8) \overset{?}{\le} -16\cr -19 \le -16\ \ \text{TRUE!}\cr\cr 5 - 3(0) \overset{?}{\le} -16\cr 5 \le -16\ \ \text{FALSE} \end{gather} $$

SOLVING NONLINEAR INEQUALITIES

Note: You can write ANY inequality in the form

$$ f(x)\ \substack{ \gt\strut \\ \ge\strut \\ \lt\strut \\ \le\strut }\ 0 $$

[label on graph] $\,y = f(x)$

SOLUTION SETS FOR:

$f(x) \gt 0$
value(s) of $\,x\,$ for which graph of $\,f\,$ is ABOVE $x$-axis
$(-\infty,1) \cup (2,3)$

$f(x) \ge 0$
on or above $x$-axis
$(-\infty,1] \cup (2,3)$

$f(x) \lt 0$
below $x$-axis
$(1,2) \cup [3,\infty)$

$f(x) \le 0$
on or below $x$-axis
$[1,2) \cup [3,\infty)$

KEY IDEA for solving NONLINEAR INEQUALITIES

There are only $\,2\,$ TYPES OF PLACES where the graph of a function $\,f\,$ can go from above to below the $x$-axis (or vice versa):

• where $\,f(x) = 0$
• where there's a BREAK in the graph of $\,f$

the TEST POINT METHOD for solving NONLINEAR INEQUALITIES

1. Get $\,0\,$ on one side; identify $\,f(x)\,.$
2. Where is $\,f(x) = 0\,$? Break(s) in graph? Mark these on a # line: $\,B\,,z$
3. TEST resulting subintervals. (MAY USE a graphing calculator—SHOW GRAPH!)
4. REPORT SOLUTION SET
5. ‘SPOT-CHECK’ in ORIGINAL SENTENCE

EXAMPLE: Using the TP METHOD

Solve: $\,x^2 + 5x \gt -6\,$ (ORIGINAL SENTENCE)

(1) $\,\underbrace{x^2 + 5x + 6}_{f(x)} \gt 0$

(2) $\,x^2 + 5x + 6 = 0$

$(x+2)(x+3) = 0$

(3) (test subintervals)
[labels on number line]
$T = -4\ \ (-)(-)$
$T = -2.5\ (-)(+)$
$T = 0\ \ (+)(+)$
zeros at $\,-3\,$ and $\,-2$

(4) SOLN SET to $\,f(x) \gt 0\,$: $(-\infty,-3) \cup (-2,\infty)$

(5) $$ \begin{gather} (-4)^2 + 5(-4) \overset{?}{\gt} -6\cr 16 - 20 \overset{?}{\gt} -6\cr -4 \gt -6\ \ \text{True!}\cr\cr (-2.5)^2 + 5(-2.5) \overset{?}{\gt} -6\cr -6.25 \gt -6\ \ \text{False!} \end{gather} $$

EXAMPLE: Using the TP METHOD

Solve: $\,x\le \frac{2}{x-1}\,$ (ORIGINAL SENTENCE)

(1) $$ \begin{gather} x - \frac 2{x-1}\le 0\cr\cr \frac{x(x-1)-2}{x-1}\le 0\cr\cr \frac{x^2 - x - 2}{x-1}\le 0\cr\cr \underbrace{\frac{(x-2)(x+1)}{x-1}}_{f(x)}\le 0 \end{gather} $$

(2) and (3)
[labels on number line] zero at $\,-1\,,$ break at $\,1\,,$ zero at $\,2$

(4) Soln set to $\,f(x) \le 0\,$: $(-\infty,-1] \cup (1,2]$

(5) $$ \begin{gather} 0 \overset{?}{\le} \frac 2{0-1}\cr\cr 0 \le -2\cr\ \text{False!}\cr\cr -2 \overset{?}{\le} \frac{2}{-2-1}\cr\cr -2 \overset{?}{\le} \frac{-2}3\cr \text{True!} \end{gather} $$

ABSOLUTE VALUE as DISTANCE FROM $\,0$

$|x| = \,$ distance between $\,x\,$ and $\,0\,$ on # line.

Solve:
$|x|\lt 3\,,$   $\,-3 \lt x \lt 3\,,$   Soln set: $\,(-3,3)$

Solve:
$|x|\ge 2\,,$   $\,x\ge 2\ \ \text{or}\ \ x\le -2\,,$   Soln set: $\,(-\infty,-2] \cup (2,\infty)$

$$ \begin{gather} |2x-3| \gt 1\cr 2x-3\gt 1\ \ \text{or}\ \ 2x-3 \lt -1\cr 2x \gt 4\ \ \text{or}\ \ 2x\lt 2\cr x\gt 2\ \ \text{or}\ \ x\lt 1 \end{gather} $$

SPOT CHECK!

Soln set $\,(-\infty,1)\cup (2,\infty)$

ABSOLUTE VALUE as DISTANCE BETWEEN $\,2\,$ NUMBERS

$|x-y| = \,$ distance between $\,x\,$ and $\,y$

Solve:
$|x-1| \lt 2\,,$   $\,-1 \lt x \lt 3\,,$   Soln set: $\,(-1,3)$

$$ \begin{gather} |x+3|\ge 1\cr |x-(-3)|\ge 1 \end{gather} $$

$x\le -4\ \ \text{or}\ \ x\ge -2\,,$   Soln set: $\,(-\infty,-4] \cup [-2,\infty)$

$0 \lt |x-2| \lt 1\,,$   Soln set: $\,(1,2)\cup (2,3)$

CIRCLES

Let $\,(x,y)\,$ be ANY point on the circle with center $\,(h,k)\,$ and radius $\,r\,.$ Then,

$$ \begin{align} &\text{distance from $\,(h,k)\,$ to $\,(x,y)\,$}\cr &\quad = \sqrt{(x-h)^2 + (y-k)^2}\cr &\quad = r \end{align} $$

EQUATION OF CIRCLE with CENTER $\,(h,k)\,$ and RADIUS $\,r$
$(x-h)^2 + (y-k)^2 = r^2$

If you multiply out:

• COEFFICIENTS of $\,x^2\,$ and $\,y^2\,$ are the SAME
• NO $\,xy\,$ TERM
• MAY HAVE $\,x\,,$ $\,y\,,$ CONSTANT TERMS

COMPLETING THE SQUARE TECHNIQUE

EXAMPLE:
GRAPH: $\,4x^2 + 4y^2 - 8x + 16y = -19$

$$ \begin{gather} x^2 + y^2 - 2x + 4y = -\frac{19}4\cr \text{(get coefficients of $\,x^2\,, y^2\,$ equal to $\,1\,$)}\cr\cr \bigl( x^2 - 2x + (\frac{-2}2)^2\bigr) + \bigl( y^2 + 4y + (\frac 42)^2\bigr) = -\frac{19}4 + 1 + 4\cr \text{(Complete the SQUARES!)}\cr\cr (x-1)^2 + (y+2)^2 = \frac 14\cr\cr (x-\underset{\substack{\uparrow\\ h=1}}{1})^2 + \bigl(y - \underset{\substack{\uparrow\\ k = -2}}{(-2)}\bigr)^2 = ( \underset{\substack{\uparrow\\ r = 1/2}}{\frac 12} )^2 \end{gather} $$

Circle with CENTER $\,(1,-2)\,,$ RADIUS $\,\frac 12$

SLOPES of LINES

SLOPE is a NUMBER that measures the “steepness” of a line;

$$ m = \text{SLOPE} = \frac{\text{rise}}{\text{run}} = \frac{y_2-y_1}{x_2-x_1} $$

HORIZONTAL LINES: SLOPE $= 0$

VERTICAL LINES: NO SLOPE

PARALLEL LINES: SAME SLOPE

PERPENDICULAR LINES: SLOPES ARE OPPOSITE RECIPROCALS

LARGE POSITIVE SLOPE; SMALL POSITIVE SLOPE; LARGE NEGATIVE SLOPE; SMALL NEGATIVE SLOPE

LINES

LINES have the property that EQUAL CHANGES in the INPUT give rise to EQUAL CHANGES in the OUTPUT:

$$ \begin{gather} \underbrace{f(x) = 2x - 3}_{\text{LINEAR FCT}}\cr\cr \frac{\Delta y}{\Delta x} = m = 2 \end{gather} $$

$0$ into $\,f\,$ box, $-3\,$ comes out
$3$ into $\,f\,$ box, $3\,$ comes out
$\Delta x = 3\,,$ $\,\Delta y = 6$

$100$ into $\,f\,$ box, $197\,$ comes out
$103$ into $\,f\,$ box, $203\,$ comes out
$\Delta x = 3\,,$ $\,\Delta y = 6$

$y = mx+b$

SLOPE-INTERCEPT FORM;
$m = $ SLOPE
$(0,b) = $ $y$-INTERCEPT

$y - y_1 = m(x - x_1)$

POINT-SLOPE FORM;
$m = $ SLOPE
$(x_1,y_1) = \text{POINT ON LINE}$

REVIEW of COORDINATE GEOMETRY

$$ \begin{gather} \text{DISTANCE BETWEEN}\ (x_1,y_1)\ \text{and}\ (x_2,y_2)\cr = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \end{gather} $$
$$ \begin{gather} \text{MIDPOINT of LINE SEGMENT}\cr \text{between} \ (x_1,y_1)\ \text{and}\ (x_2,y_2)\cr = \biggl( \underset{\substack{\uparrow \\ \text{AVERAGE}\\ x-\text{values}}}{\frac{x_1 + x_2}2} \ ,\ \underset{\substack{\uparrow \\ \text{AVERAGE}\\ y-\text{values}}}{\frac{y_1+y_2}2} \biggr) \end{gather} $$

SKETCHING REGIONS IN A PLANE

$$ \overset{\substack{\text{the set of all ...}}\\ \qquad \downarrow}{\bigl\{ } (x,y)\ \overset{\substack{\text{such that}\\ \downarrow}}{\ |}\ x\gt 1 \overset{\substack{\text{BRACES}\\ \downarrow}}{\ \bigr\} } $$
$$ \bigl\{ (x,y)\ |\ \underset{\substack{\uparrow\\ x\gt 0\text{ and } y\lt 0\\ \text{OR}\\ x\lt 0\text{ and } y\gt 0}}{xy \lt 0} \bigr\} $$

SYMMETRY

A GRAPH HAS:

symmetry about the $y$-AXIS

iff [if and only if]

whenever $\,(a,b)\,$ is on the graph, so is $\,(-a,b)$

symmetry about the $x$-AXIS

iff

whenever $\,(a,b)\,$ is on the graph, so is $\,(a,-b)$

symmetry about the ORIGIN

iff

whenever $\,(a,b)\,$ is on the graph, so is $\,(-a,-b)$

$y$-AXIS SYMMETRY, $x$-AXIS SYMMETRY, ORIGIN SYMMETRY

TESTING EQUATIONS FOR SYMMETRY (an example)

Test $\,x^2y^3 - x^4y = y\,$ for symmetry about $x$-AXIS, $y$-AXIS, ORIGIN.

Spse [suppose] $\,(a,b)\,$ is on graph, so $\,a^2b^3 - a^4b = b\,.$

$y$-axis:

Is $\,(-a,b)\,$ on graph? $(-a)^2b^3 - (-a)^4b \overset{?}{=} b$ $a^2b^3 - a^4b \overset{?}{=} b$

Yes!

$x$-axis:

Is $\,(a,-b)\,$ on graph? $a^2(-b)^3 - a^4(-b) \overset{?}{=} b$ $-a^2b^3 + a^4b \overset{?}{=} -b$ ($\times -1$) $a^2b^3 - a^4b \overset{?}{=} b$

Yes!

ORIGIN:

Is $\,(-a,-b)\,$ on graph? $(-a)^2(-b)^3 - (-a)^4(-b) \overset{?}{=} -b$ $-a^2b^3 + a^4b \overset{?}{=} -b$

Yes!

EVALUATING DIFFERENCE QUOTIENTS

An EXPRESSION of the form

$$ \frac{f(x+h)-f(x)}{(x+h)-x} = \frac{f(x+h)-f(x)}h $$

is called a DIFFERENCE QUOTIENT (a QUOTIENT of DIFFERENCES!)

Example: Let $\,f(x) = x^2 - 3x\,.$ FIND $\frac{f(x+h)-f(x)}h\,$ and SIMPLIFY; your final name should have NO $\,h\,$ in the denominator.

$$ \begin{align} &\frac{f(x+h) - f(x)}h\cr\cr &\quad = \frac{(x+h)^2 - 3(x+h) - (x^2-3x)}{h}\cr\cr &\quad = \frac{\cancel{x^2} + 2xh + h^2 - \cancel{3x} - 3h - \cancel{x^2} + \cancel{3x}}h\cr\cr &\quad = \frac{\cancel{h}(2x +h-3)}{\cancel{h}}\cr\cr &\quad = 2x + h - 3 \end{align} $$

WHAT DOES A DIFFERENCE QUOTIENT TELL US?

[labels on graphs] $\,y = f(x)\,,$ points $\,\bigl(x,f(x)\bigr)\,$ and $\,\bigl(x+h,f(x+h)\bigr)\,,$ $\,h\gt 0\,,$ $\,h\lt 0$

In both cases,

$$ \begin{align} &\frac{f(x+h)-f(x)}h\cr &\quad = \text{slope of line through}\cr &\qquad \bigl(x,f(x)\bigr) \ \text{and}\ \bigl(x+h,f(x+h)\bigr)\ ; \end{align} $$

as $\,h\,$ APPROACHES $\,0\,,$ this number gets close to the slope of the TANGENT LINE at $\,\bigl(x,f(x)\bigr)\,$

FINDING THE DOMAIN OF A FUNCTION

FROM A FORMULA:

NOT ALLOWED:

• DIVISION BY ZERO
• EVEN ROOTS OF NEGATIVE NUMBERS

Example: FIND domain of $\,g(x) = \frac{\sqrt{x-3}}{x-5}\,.$

$$ \begin{gather} x-3\ge 0\ \ \text{AND}\ \ x-5\ne 0\cr\cr x\ge 3\ \ \text{AND}\ \ x\ne 5\cr\cr \text{dom}(g) = [3,5)\cup (5,\infty) \end{gather} $$

FROM A GRAPH:

“Collapse” each point into its $x$-value:

$$ \text{dom}(f) = (-1,1)\cup (1,2] $$

[label on graph] $y = f(x)$

FINDING THE RANGE OF A FUNCTION

The RANGE of a function is its OUTPUT set.

... is easiest when you have the GRAPH of the function!

“Collapse’ each pt [point] into its $y$-value.

Example: Find the range of $\,f(x) = |x-5| + 3\,.$

Graph: $\,y = |x|\,,$ shift right $\,5\,,$ up $\,3\,$:

$$ \text{ran}(f) = [3,\infty) $$

WORKING WITH LINEAR FCTS [functions]...

Recall: $\,\text{slope} = \frac{\Delta y}{\Delta x} = \frac{y_2-y_1}{x_2-x_1}$

Spse [suppose] you have a KNOWN (old) pt [point] $\,(x_{\text{old}},y_{\text{old}})\,$ and want a NEW pt $\,(x_{\text{new}},y_{\text{new}})\,$:

$$ \begin{align} &m = \frac{y_{\text{new}} - y_{\text{old}}}{x_{\text{new}} - x_{\text{old}}}\cr &\quad\implies y_{\text{new}} - y_{\text{old}} = m(\overbrace{x_{\text{new}} - x_{\text{old}}}^{\Delta x})\cr\cr &\quad\implies y_{\text{new}} = y_{\text{old}} + m\Delta x \end{align} $$

EXAMPLE

Find $\,f(1.4)\,.$ Use nearby endpt [endpoint] $\,(1,-5)\,$:

[labels on graph] points $\,(1,-5)\,$ and $\,(3,10)\,,$ $\,y = f(x)\,,$ $\,m = \frac{15}2 = 7.5$

$$ y_{\text{new}} = \underbrace{-5}_{y_{\text{old}}} + \underbrace{7.5}_m \underset{ \substack{\uparrow \\ \text{$\Delta x\,$ in going}\\ \text{from old pt}\\ \text{to new pt}}}{(0.4)} $$

READING INFORMATION FROM A GRAPH

Recall: $\,y\,$ is a function of $\,x\,$ iff [if and only if] for every $\,x\,,$ there is EXACTLY one $\,y$

You must be able to read a WIDE VARIETY of info from the graph of a fct!!

[labels on graph] $\,m = 5\,,$ $\,y = f(x)\,,$ points $\,(-2,10)\,,\ \ (0,10)\,,\ \ (2,-10)\,$

$$ \begin{gather} f(0) = 10\cr f(\frac 12) = 5\cr f(1) = 0\cr f(-1.7) = 10 + 5(0.3) = 11.5\cr f(2.03) = -10\cr \text{dom}(f) = [-2,3)\cr \text{ran}(f) = [-10,20)\cr f \text{ increases on } [-2,0)\cr f \text{ decreases on } [0,2] \end{gather} $$

How can we tell this is the graph of a fct [function]? It passes the VERTICAL LINE TEST.

$$ \begin{gather} \{x\ |\ f(x) = 10\} = \{-2,0\}\cr\cr \{t\ |\ f(t) \gt 0 \} = [-2,1) \end{gather} $$

INCREASING/DECREASING FUNCTIONS

DEFINITIONS:

$f\,$ increases on an interval $\,I$	iff [if and only if]	whenever $\,a\lt b\,$ in $\,I\,,$ $\,f(a) \lt f(b)$
$f\,$ decreases on an interval $\,I$	iff	whenever $\,a\lt b\,$ in $\,I\,,$ $\,f(a) \gt f(b)$

“BASIC MODELS” YOU MUST KNOW!!

Given the EQUATION, produce the GRAPH. Given the GRAPH, provide the EQUATION.

[labels on graphs] $y = k\,,$ $\,y = x^2\,$ (and higher even powers), $\,y = x^3\,$ (and higher odd powers), $\,y = \frac 1x\,,$ $\,y = \sqrt x\,,$ $\,y = |x|$

PIECEWISE-DEFINED FUNCTIONS

Different rules apply for different inputs!

Given a formula, produce the graph. Given the graph, produce the formula.

$$ f(x) = \cases{ -x+3\,, & \text{if}\ {-}1\le x\lt 1\cr 5\,, & \text{if}\ 1\le x\lt 2\cr x^2\,, & \text{if}\ x\ge 2 } $$

NOTE:
$\text{dom}(f) = [-1,\infty)$
$\text{ran}(f) = (2,\infty)$

[graph] point $\,(-1,4)\,,$ hollow dot $\,(1,2)\,,$ point $\,(1,5)\,,$ hollow dot $\,(2,5)\,,$ point $\,(2,4)$

DIRECT VARIATION

The following are EQUIVALENT (completely INTERCHANGEABLE):

• $y = kx\,,$ for $\,k\ne 0$
• $y\,$ varies directly as $\,x$
• $y\,$ is directly proportional to $\,x$
• $y\,$ is proportional to $\,x$

(The constant $\,k\,$ is called the CONSTANT OF PROPORTIONALITY.)

INVERSE VARIATION

The Following Are EQUIVALENT (TFAE):

• $y = \frac kx\,,$ for $\,k\ne 0$
• $y\,$ is inversely proportional to $\,x$
• $y\,$ varies inversely as $\,x$

TFAE:

• $z = kxy\,,$ $\,k\ne 0$
• $z\,$ varies jointly as $\,x\,$ and $\,y$
• $z\,$ is jointly proportional to $\,x\,$ and $\,y$

TFAE:

• $z = k\frac xy\,,$ $\,k\ne 0$
• $z\,$ is proportional to $\,x\,$ and inversely proportional to $\,y$

PROPORTIONALITY PROBLEMS

Write an equation for:

“$A\,$ is proportional to the square of $\,t\,$ and inversely proportional to the cube of $\,x\,.$”

$$ A = k\frac{t^2}{x^3} $$

Now, if $\,A = 3\,$ when $\,t = 1\,$ and $\,x = 2\,,$ find the constant of proportionality.

$$ 3 = k\cdot \frac{1^2}{2^3}\,,\ \text{so}\ \ k = 24 $$

PROPORTIONALITY PRACTICE

Write a sentence that describes the relationship:

$$y = \frac kx$$

ANS [answer]: “$y\,$ varies inversely as $\,x$”

$$y = k\cdot\frac{\sqrt t}{v^2}$$

ANS: “$y\,$ varies directly as the square root of $\,t\,,$ and inversely as the square of $\,v$”

(no transcript needed for this card)

What is the GRAPH of $\,y = f(x)\,$?

ANSWER: It’s the set of ALL POINTS of the form $\,\bigl(x,\underbrace{f(x)}_{y}\bigr)\,.$

Graph of $\,y = f(x+3)\,$?
Set of pts [points]: $\,\bigl(x,f(x+3)\bigr)$

What does it mean when we say that $\,(a,b)\,$ is on the graph of $\,y = f(x)\,$?

ANS [answer]: When you substitute ‘$\,a\,$’ for ‘$\,x\,$’ and ‘$\,b\,$’ for ‘$\,y\,$’, the resulting equation is TRUE. That is, ‘$\,b = f(a)\,$’ is TRUE.

Using the “GEOMETRIC TRANSFORMATIONS” SHEET

Given ANY ENTRY in a row, be able to produce all other entries; also, UNDERSTAND why a certain change in the equation produces a certain change in the graph.

GIVEN:
GRAPH OF $\,y = f(x)\,$;
PTS of FORM $\,\bigl(x,f(x)\bigr)$

WANTED:
GRAPH OF $\,y = f(x) + 3\,$;
PTS of FORM $\,\bigl(x,f(x) + 3\bigr)$

[labels on graph] point $\,\bigl(x,f(x)\bigr)\,$ moves up to point $\,\bigl(x,f(x)+3\bigr)$

EACH PT [point] moves UP $\,3$

TRANSFORMATIONS INVOLVING ‘$\,y\,$’ are INTUITIVE!!

TRANSFORMATIONS INVOLVING ‘$\,x\,$’ are COUNTER-INTUITIVE ... WHY?

GIVEN: GRAPH OF $\,y = f(x)\,$;
PTS OF FORM $\,\bigl(x,f(x)\bigr)$

WANTED: GRAPH OF $\,y = f(x+3)\,$;
PTS OF FORM $\,\bigl(x,f(x+3)\bigr)$

[labels on graph] points $\,\bigl(x,f(x)\bigr)\,,$ $\,\bigl(x+3,f(x+3)\bigr)\,,$ and $\,\bigl(x,f(x+3)\bigr)$

Step 1: Locate $\,x+3$

Step 2: Find $\,f(x+3)$

Step 3: Plot $\,\bigl(x,f(x+3)\bigr)$

EACH PT [point] MOVES LEFT $\,3\,$!!

TRANSFORMATIONS INVOLVING ‘$\,x\,$’ ARE COUNTER-INTUITIVE!!

A TYPICAL PROBLEM

Build up $\,y = 3|{-}x+1|-5\,$ from a basic model.

EQUATION	ACTION	GRAPHICAL RESULT
$y = |x|$	BASIC MODEL
$y = 3|x|$	multiply previous $y$-values by $\,3$	vertical stretch, factor of $\,3$
$y = 3|x|-5$	subtract $\,5\,$ from previous $y$-values	DOWN $\,5$
$y = 3|x+1|-5$	replace every $\,x\,$ with $\,x+1$	LEFT $\,1$
$y = 3|{-}x+1|-5$	replace every $\,x\,$ with $\,-x$	reflect about $y$-axis

HORIZONTAL ELONGATION

GIVEN: GRAPH OF $\,y = f(x)\,$;
PTS [points] $\,\bigl(x,f(x)\bigr)$

WANTED: GRAPH OF $\,y = f(\frac x2)\,$;
PTS $\,\bigl(x,f(\frac x2)\bigr)$

[labels on graph] points $\,\bigl(\frac x2,f(\frac x2)\bigr)\,,$ $\,\bigl(x,f(\frac x2)\bigr)\,,$ $\,\bigl(x,f(x)\bigr)$

Replacing every $\,x\,$ by $\,\frac x2\,$ causes $\,(a,b)\,$ to move to $\,(2a,b)\,$; each $x$-value DOUBLES!

EX [example]
[labels on lower left graph] $\,1.5\,,$ $\,3\,,$ $\,y = f(x)$
[labels on lower right graph] $\,3\,,$ $\,6\,,$ $\,y = f(\frac x2)$

HORIZONTAL COMPRESSION

GIVEN: GRAPH OF $\,y = f(x)\,$;
PTS [points] $\,\bigl(x,f(x)\bigr)$

WANTED: GRAPH OF $\,y = f(2x)\,$;
PTS $\,\bigl(x,f(2x)\bigr)$

[labels on graph] points $\,\bigl(x,f(x)\bigr)\,,$ $\,\bigl(2x,f(2x)\bigr)\,,$ $\,\bigl(x,f(2x)\bigr)$

Replacing every $\,x\,$ by $\,2x\,$ causes $\,(a,b)\,$ to move to $\,(\frac a2,b)\,$; each $x$-value gets DIVIDED by $\,2\,$!

EX [example]
[labels on lower left graph] $\,1.5\,,$ $\,3\,,$ $\,y = f(x)$
[labels on lower right graph] $\,1.5\,,$ $\,y = f(2x)$

EVEN FCTS [functions]

DEFN [definition]. A fct [function] $\,f\,$ is EVEN     iff [if and only if]     $\,f(x) = f(-x)$

When INPUTS are OPPOSITES, OUTPUTS are the SAME.

EVEN FCTS have SYMMETRY ABOUT the $y$-AXIS.

NOTE: $\,y = x^2\,,$ $\,y = x^4\,,$ $\,y = x^6\,$ etc. are EVEN FCTS

EX [example] Is $\,f(x) = x^2 - x^4\,$ even?

$$ \begin{align} f(-x) &= (-x)^2 - (-x)^4\cr &= x^2 - x^4 = f(x)\,;\ \ \text{YES} \end{align} $$

ODD FCTS

DEFN. A fct $\,f\,$ is ODD   iff   $\,f(-x) = -f(x)$

When INPUTS are OPPOSITES, OUTPUTS are OPPOSITES.

ODD FCTS have SYMMETRY ABOUT the ORIGIN (fold twice; once along $\,x\,,$ once along $\,y\,$; graph coincides)

NOTE: $\,y = x^3\,,$ $\,y = x^5\,,$ $\,y = x^7\,$ etc are ODD FCTS

EX [example] Is $\,g(x) = x^3 - 1\,$ odd?

$$ \begin{align} g(-x) &= (-x)^3 - 1\cr &= -x^3 - 1\ne -g(x)\ \ \text{NO!!} \end{align} $$

EXTREME VALUES of FUNCTIONS

DEFN. An EXTREME VALUE of a fct [function] is the greatest or least value of the fct on some interval.

[labels on left graph] point $\,(2,10)\,,$ $\,y = f(x)\,,$ numbers going into the $\,f\,$ box, greatest # is $\,10$

We say: the MAXIMUM VALUE is $\,10\,$ (it occurs when $\,x = 2\,$)

[labels on right graph] point $\,(5,3)\,,$ $\,y = g(t)\,,$ numbers going into the $\,g\,$ box, least # is $\,3$

We say: the MINIMUM VALUE is $\,3\,$ (it occurs when $\,x = 5\,$)

FINDING EXTREME VALUES ON SOME CALCULATORS

(1) Graph the fct in a window where you can SEE the desired MAX/MIN

(2) 2nd CALC — MAX (or MIN)

(3) Move to left of desired MAX, press ENTER

(4) Move to right of desired MAX, press ENTER

(For steps (3) and (4), you're telling WHICH MAX YOU WANT!)

(5) Move close to desired MAX, press ENTER

(6) The $y$-value of the pt [point] is the MAXIMUM VALUE; the $x$-value is where it occurs.

[labels on graph] $\,y = x^3 - 8\,,$ window is from $\,-5\,$ to $\,5\,$ on $x$-axis, from $\,-20\,$ to $\,20\,$ on $y$-axis

MAX/MIN VALUES OF QUADRATIC FCTS [functions]

$$ y = a(x-h)^2 + k $$

VERTEX FORM of a QUADRATIC FCT

$$ \begin{gather} a \gt 0\cr \underbrace{a(x-h)^2}_{\ge 0} + k \overbrace{\ge}^{\text{always!}} k \end{gather} $$

[labels on graph] point $\,(h,k)\,,$ “holds water”

$(h,k)\,$ is a MINIMUM POINT

$$ \begin{gather} a \lt 0\cr\cr \underbrace{a(x-h)^2}_{\le 0} + k \le k \end{gather} $$

[labels on graph] point $\,(h,k)\,,$ “sheds water”

$(h,k)\,$ is a MAXIMUM POINT

VERTEX FORMULA for a QUADRATIC FCT

$$ \begin{align} f(x) &= \overbrace{ax^2 + bx + c}^{\text{standard form}}\cr\cr &= a \overbrace{\left(x^2 + \frac bax + \bigl(\frac b{2a}\bigr)^2\right)}^{\text{Complete the square!}} + c - \frac{b^2}{4a}\cr\cr &= a\bigl(x + \frac b{2a}\bigr)^2 + \text{stuff} \end{align} $$

The vertex of the parabola occurs at:

$$ \left( -\frac b{2a}\,,\,f\bigl(-\frac{b}{2a}\bigr) \right) $$

VERTEX FORMULA: MEMORIZE!!

INTERCEPTS of a QUADRATIC FCT [function]

$y$-INTERCEPT: Set $\,x = 0\,$; solve for $\,y$

$x$-INTERCEPTS: Set $\,y = 0\,$; solve for $\,x$

$$ \begin{gather} \text{(first column)}\cr y = ax^2+ bx + c\cr\cr \text{Use Quadratic Formula, if needed:}\cr\cr ax^2 + bx + c = 0\cr \text{iff}\cr x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gather} $$ $$ \begin{gather} \text{(second column)}\cr y = a(x-h)^2 + k\cr\cr 0 = a(x-h)^2 + k\cr\cr \text{EX [example]}\cr\cr 0 = 2(x-1)^2 - 5\cr 5 = 2(x-1)^2\cr \frac 52 = (x-1)^2\cr x-1 = \pm\sqrt{\frac 52}\cr x = 1\pm \sqrt{\frac 52} \end{gather} $$

PUTTING a QUADRATIC FCT in VERTEX FORM by COMPLETING THE SQUARE

$$ \begin{align} f(x) &= 5x^2 - 30x + 49\cr &= 5(x^2 - 6x\ \ \ \ \ \ \ ) + 49\cr &= 5(x^2 - 6x + 9) + 49 - 45\cr &\qquad \text{(Note: $(\frac{-6}2)^2 = 9\,$)}\cr &= 5(x-3)^2 + 4 \end{align} $$

Vertex @ [at] $\,(3,4)\,$; holds water

MAX/MIN PROBLEMS

Sample: Among all rectangles that have a perimeter of $\,20\,$ ft, find the dimensions of the one with the largest area.

FIND FUNCTION TO BE MAXIMIZED; express it in terms of ONLY ONE VARIABLE.

$$ \begin{align} A(x) &= xy\cr &= x(10-x)\ \ \text{(from below)}\cr &= -x^2 + 10x \end{align} $$ $$ \begin{gather} 2x + 2y = 20\cr 2y = 20-2x\cr y = 10-x \end{gather} $$

[graph] max point $\,(5,25)\,,$ $\,y = A(x)$

Using VERTEX FORMULA, max occus when $\,x = -\frac b{2a} = -\frac{10}{2(-1)} = 5\,$; $\,A(5) = 5(5) = 25\,.$

The dimensions are $\,5\text{ ft}\times 5\text{ ft}\,.$

MAX/MIN PROBLEMS

Sample: FIND $\,2\,$ positive #s [numbers] whose SUM is $\,100\,$ and the sum of whose squares is a MINIMUM.

Let $\,x,\,y\,$ be the two positive #s.

$$ \begin{gather} x + y = 100\cr y = 100-x \end{gather} $$ $$ \begin{align} \text{Sum of squares} &= x^2 + y^2\cr S(x) &= x^2 + (100-x)^2 \end{align} $$

[graph] From CALCULATOR: minimum point $\,(50,5000)$

The #s are $\,50\,$ and $\,50\,.$

COMBINING FUNCTIONS to get NEW Functions

SUM FCT [function] $\,f+g\,$ is defined by:

$$ \begin{gather} (f+g)(x) = f(x) + g(x)\cr\cr \text{dom}(f+g) = \{ x\ |\ x\in\text{dom}(f) \text{ and } x\in\text{dom}(g)\} \end{gather} $$

[image] $x\,$ goes into box labeled $\,f+g\,,$ (name of new fct), $\,f(x)\,$ comes out of $\,f\,$ box, $\,g(x)\,$ comes out of $\,g\,$ box, $\,f(x)\,$ and $\,g(x)\,$ are dropped into the ‘$\,+\,$’ box, $\,f(x) + g(x)\,$ comes out of bottom

Be able to write defn [definition] and domain for $\,f\pm g\,,$ $\,fg\,,$ $\,\frac fg\,$!

COMPOSITION OF FUNCTIONS
“stacking boxes”

The fct [function] $\,g\circ f\,$ is defined by:

$$ \begin{gather} (g\circ f)(x) = g\bigl(f(x)\bigr)\cr\cr \text{dom}(g\circ f) = \{x\ |\ x\in\text{dom}(f) \text{ and } f(x)\in\text{dom}(g)\} \end{gather} $$

[image] $\,x\,$ goes into $\,f\,$ box, $\,f(x)\,$ comes out and is dropped into $\,g\,$ box, and $\,g\bigl(f(x)\bigr)\,$ comes out bottom; name chosen: $\,g\circ f\,,$ read as: “$g\,$ circle $\,f\,$” or “$g\,$ composed with $\,f\,$”

NOTE: For the fct $\,g\circ f\,,$ $\,f\,$ acts FIRST, $\,g\,$ acts SECOND!

PRACTICE WITH FUNCTION COMPOSITION

Let $\,f(x) = x^2 - 1\,$ and $\,g(x) = 2 - 3x\,.$ FIND $\,(f\circ g)(x)\,$ and $\,(g\circ f)(x)\,.$

$$ \begin{align} (f\circ g)(x) &= f(g(x))\cr &= f(2-3x)\cr &= (2-3x)^2 - 1\cr &= 4 - 12x + 9x^2 - 1 \end{align} $$

Write a CMS [Complete Mathematical Sentence]!! Line up ‘$\,=\,$’ signs.

$$ \begin{align} (g\circ f)(x) &= g(f(x))\cr &= g(x^2-1)\cr &= 2 - 3(x^2-1)\cr &= 2 - 3x^2 + 3\cr &= -3x^2 + 5 \end{align} $$

Is fct [function] composition commutative? NO!!

WRITING A FCT [function] as a COMPOSITION

Let $\,h(x) = 5(x-4)^3 - 7\,.$ Find fcts [functions] $\,f\,$ and $\,g\,$ such that $\,h(x) = (f\circ g)(x)\,.$

SOLN: [Solution]

$$ \begin{align} x&\overset{-4}{\longrightarrow} x-4\cr &\overset{\text{cube}}{\longrightarrow} (x-4)^3\cr &\overset{\times 5}{\longrightarrow} 5(x-4)^3\cr &\overset{- 7}{\longrightarrow} 5(x-4)^3 - 7 \end{align} $$

[Let $\,g\,$ take the first two tasks] $\,g(x) = (x-4)^3$

[Let $\,f\,$ take the last two tasks] $\,f(x) = 5x-7$

Check:

$$ \begin{align} (f\circ g)(x) &= f(g(x))\cr &= f((x-4)^3)\cr &= 5(x-4)^3 - 7 \end{align} $$

Find $\,2\,$ different sets of fcts [functions] $\,f\,$ and $\,g\,$ that work!!

When can we use a FUNCTION BOX BACKWARDS?

[image] function box, named “SQUARE”, the number $\,4\,$ goes in the bottom, coming out the top: $\,2\,$?   $\,-2\,$?

[image] function box, named “CUBE” the number $\,8\,$ goes in the bottom, $\,2\,$ comes out the top

[graph] $\,y = x^2\,,$ showing the output $\,4\,$ going back to two inputs, $\,2\,$ and $\,-2$

[graph] Graph of $\,y = x^3\,,$ showing the output $\,8\,$ going back to the input $\,2\,$

ANS [answer]: When each OUTPUT has exactly one corresponding INPUT!

ONE-TO-ONE FUNCTIONS

DEFN [definition]   A function $\,f\,$ is one-to-one (1-1) iff [if and only if] whenever $\,x\ne y\,$ (INPUTS are different), $\,f(x)\ne f(y)\,$ (the corresponding OUTPUTS are different).

The graph of a FUNCTION passes a VERTICAL LINE TEST, because each INPUT has exactly one OUTPUT.

The graph of a 1-1 FCT [function] also passes a HORIZONTAL LINE TEST, because each OUTPUT has exactly one INPUT!

“UNDOING” a 1-1 FUNCTION

If $\,f\,$ is 1-1, then there exists a unique function $\,f^{-1}\,$ ($\,f\,$ inverse) that ‘UNDOES’ what $\,f\,$ did!

$$ \overset{(3)}{f^{-1}(y)} = x\quad \begin{matrix} \overset{f}{\longrightarrow} \cr \underset{\strut f^{-1}}{\longleftarrow} \end{matrix} \quad \overset{(1)}{\strut f(x)} = \overset{(2)}{\strut y} $$

That is:

$$ \begin{gather} \underbrace{f(x) = y}_{\text{$f\,$ takes $\,x\,$ to $\,y$}}\cr\cr \text{iff [if and only if]}\cr\cr \underbrace{f^{-1}(y) = x}_{\text{$f^{-1}\,$ takes $\,y\,$ to $\,x$}}\cr\cr \end{gather} $$

PROPERTIES of INVERSE FUNCTIONS

$$ f^{-1}\bigl(f(x)\bigr) = x\quad \begin{matrix} \overset{f}{\longrightarrow} \cr \underset{\strut f^{-1}}{\longleftarrow} \end{matrix} \quad f(x) $$

$f^{-1}\bigl(f(x)\bigr) = x\,$ for every $\,x\in\text{dom}(f)$

$$ f\bigl(f^{-1}(y)\bigr) = y\quad \begin{matrix} \overset{f^{-1}}{\longrightarrow} \cr \underset{\strut f}{\longleftarrow} \end{matrix} \quad f^{-1}(y) $$

$f\bigl(f^{-1}(y)\bigr) = y\,$ for every $\,y\in\text{ran}(f)$

NOTE:

$$ \begin{gather} \text{dom}(f) = \text{ran}(f^{-1})\cr \text{ran}(f) = \text{dom}(f^{-1}) \end{gather} $$

INPUT/OUTPUT roles for $\,f\,$ and $\,f^{-1}\,$ are REVERSED!

FINDING INVERSE FCTS [functions]
(when there's only one ‘$\,x\,$’ in the formula)

FIND the inverse:

$$ f(x) = 4 - \root 3\of{x+1} = -\root 3\of{x+1} + 4 $$

[interpretation of mapping diagram]

• Start with: $\,x$
• Add $\,1\,$ to get: $\,x+1$
• Take the cube root to get: $\,\root 3\of{x+1}$
• Multiply by $\,-1\,$ to get: $\,-\root 3\of{x+1}$
• Add $\,4\,$ to get: $\,4 - \root 3\of{x+1}$

Now, UNDO these operations:

• Start with: $\,4 - \root 3\of{x+1}$
• Subtract $\,4\,$ to get: $\,-\root 3\of{x+1}$
• Multiply by $\,-1\,$ to get: $\,\root 3\of{x+1}$
• Cube to get: $\,x+1$
• Subtract $\,1\,$ to get: $\,x$

$$ \begin{align} f^{-1}(x) &= [(x-4)(-1)]^3 - 1\cr &= (-x+4)^3 - 1\cr &= (4-x)^3 - 1 \end{align} $$

Check:

$$ \begin{align} f\bigl( f^{-1}(x) \bigr) &= f\bigl( (4-x)^3 - 1\bigr)\cr &= 4 - \root 3\of{(4-x)^3 - 1 + 1}\cr &= 4 - (4-x)\cr &= x \end{align} $$

FINDING INVERSE FCTS
(for more complicated formulas)

KEY IDEA: the INPUT/OUTPUT role for $\,f\,$ and $\,f^{-1}\,$ are reversed!

FIND the inverse: $\,f(x) = \frac{1-3x}{5 + 2x}$

1. SIMPLIFY NOTATION; let $\,y = f(x)\,$:     $\,y = \frac{1-3x}{5 + 2x}$
2. Switch roles of $\,x\,$ and $\,y\,$:     $\,x = \frac{1-3y}{5 + 2y}$
3. Solve for $\,y\,$; rename as $\,f^{-1}(x)\,$:

$$ \begin{gather} x(5+2y) = 1 - 3y\cr\cr 5x + 2xy = 1-3y\cr\cr 2xy + 3y = 1 - 5x\cr\cr y(2x+3) = 1-5x\cr\cr y = \frac{1-5x}{2x+3}\cr\cr \text{and check!} \end{gather} $$

GRAPH of $\,f^{-1}$

If $\,(a,b)\,$ is a point on the graph of $\,f\,,$ then $\,(b,a)\,$ is a point on the graph of $\,f^{-1}\,.$

Use the line $\,y = x\,$ to “FIND” the pt [point] $\,(b,a)\,.$

[graph] Start with point $\,(a,b)\,.$ Draw an arrow horizontally to the line $\,y = x\,$ to locate $\,b\,$ on the $x$-axis (step 1). Then, draw an arrow from $\,(a,b)\,$ vertically to the line $\,y = x\,$ to locate $\,a\,$ on the $y$-axis (step 2). Finally, plot the point $\,(b,a)\,$ (step 3).

NOTE: the shaded figure in a SQUARE! (Why?)

CONCLUSION: To get from a pt on the graph of $\,f\,,$ to a pt on the graph of $\,f^{-1}\,,$ FOLD ALONG THE LINE $\,y = x\,.$

GRAPHING a fct [function] and its INVERSE on the TI-83

Y1 = (ORIGINAL FCT)   (e.g., Y1 = x³ + 2) (use regular line-style)

Y2 = (its INVERSE)   (e.g., Y2 = (x-2)^(1/3) (use “THICK” line-style)

Y3 = x (use DOTTED line-style)

Set desired $x$-values; YMIN = 0, YMAX = 1 (set these) and use ZOOMSQUARE.

POLYNOMIALS—terminology

DEFN [definition] A POLYNOMIAL is a sum of terms, each of the form $\,ax^n\,,$ where $\,a\in\Bbb R\,,$ and $\,n\in\{0,1,2,\ldots\}$

DEGREE of poly’l [polynomial]: the HIGHEST exponent

LEADING COEFFICIENT: the coefficient of the highest power term

EX [example] $$ P(x) = 2 - 3x^7 + \frac 43x^2 - \sqrt3\,x^4 $$

$4$ terms; degree $= 7\,$; leading coeff [coefficient] $= -3$

STANDARD FORM: (highest power 1st, walk on down)

$$ P(x) = -3x^7 - \sqrt 3\,x^4 + \frac 43x^2 + 2 $$

ZEROS (ROOTS) of POLYNOMIALS

Let $\,P\,$ be a polynomial. The following are EQUIVALENT:

• $c\,$ is a zero (root) of $\,P$
• $P(c) = 0\,$ ($\,c\,$ is an INPUT whose OUTPUT is $\,0\,$)
• the graph of $\,P\,$ crosses the $x$-axis at $\,c\,$
• $x = c\,$ is a solution of the equation $\,P(x) = 0$

NOTE: These [first four] hold for ANY fct [function] $\,f$

• $x-c\,$ is a factor of $\,P(x)$
• $x-c\,$ goes into $\,P(x)\,$ evenly (remainder $= 0$)
• $P(x) = (x-c)(\text{stuff})$

These [last three] hold ONLY FOR POLYNOMIALS

EX [example]:

$2\,$ is a zero $\,\cdots\,$ $\,x-2\,$ is a factor

$-3\,$ is a zero $\,\cdots\,$ $\,x-(-3) = x+3\,$ is a factor

END BEHAVIOR of POLYNOMIALS

Key idea: For large values of $\,x\,,$ a poly’l [polynomial] “behaves like” its HIGHEST POWER TERM.

$n$	$a$	EX [example]	END BEHAVIOR
$n\,$ even $(2,4,6,8,\ldots)$	$a \gt 0$	$3x^2$	As $\,x\rightarrow\infty\,,$ $y\rightarrow\infty$ As $\,x\rightarrow -\infty\,,$ $y\rightarrow\infty$
	$a\lt 0$	$-3x^2$	As $\,x\rightarrow\infty\,,$ $y\rightarrow -\infty$ As $\,x\rightarrow -\infty\,,$ $y\rightarrow -\infty$
$n\,$ odd $(1,3,5,7,\ldots)$	$a \gt 0$	$3x^5$	As $\,x\rightarrow\infty\,,$ $y\rightarrow\infty$ As $\,x\rightarrow -\infty\,,$ $y\rightarrow -\infty$
	$a\lt 0$	$-3x^5$	As $\,x\rightarrow\infty\,,$ $y\rightarrow -\infty$ As $\,x\rightarrow -\infty\,,$ $y\rightarrow \infty$

“TURNING POINTS” of POLYNOMIALS; HOW MANY CAN THEY HAVE?

[graph] local MAX; local MIN

A “TURNING POINT” is a point where there is a local MAX [maximum] or a local MIN [minimum].

FACT: A poly’l of degree $\,n\,$ can have AT MOST (less than or equal to) $\,n-1\,$ turning points. (This is proved in a CALCULUS course.)

LONG DIVISION of POLYNOMIALS

EX [example]

$$ \frac{x^3 - 8x + 2}{x+3} \overbrace{=}^{\text{from below}} x^2 - 3x + 1 + \frac{-1}{x+3} $$

SPOT-CHECK: EQUAL when $\,x = 0\,$?

[pieces of division problem]

[numerator, with $0$'s for MISSING TERMS] $x^3 + 0x^2 - 8x + 2$

DIVISOR: $\,x+3$

QUOTIENT: $\,x^2 - 3x + 1$

[explanation] $(\frac{-3x^2}x = -3x)$

[explanation, after multiplying the $\,-3x\,$ by $\,x+3\,$] Put in $\,(\,)\,$; then SUBTRACT

REMAINDER: $\,-1\,$ (STOP when $\,\text{deg}(\text{REM}) \lt \text{deg}(\text{DIVISOR})$

The DIVISION ALGORITHM

If $\,P(x)\,$ and $\,D(x)\,$ are polynomials with $\,D(x)\ne 0\,,$ then there exist unique poly's [polynomials] $\,Q(x)\,$ and $\,R(x)\,$ such that

$$ \frac{P(x)}{D(x)} = \overbrace{Q(x)}^{\text{QUOTIENT}} + \frac{\overbrace{R(x)}^{\text{REMAINDER}}}{\underbrace{D(x)}_{\text{DIVISOR}}} $$

or, equivalently

$$P(x) = D(x)Q(x) + R(x)\,,$$

where either

• $R(x) = 0\,$ (when $\,D(x)\,$ goes into $\,P(x)\,$ evenly); or
• $\text{deg}\bigl(R(x)\bigr) \lt \text{deg}\bigl(D(x)\bigr)$

“TRAPPING” the ROOTS of a POLYNOMIAL; the CONFINEMENT THEOREM

COMPLEX NUMBERS

A COMPLEX # [number] is an expression of the form

$$a + bi$$

where $\,a\,$ (REAL part of the complex #) and $\,b\,$ (IMAGINARY part of the complex #) are real #s [numbers], and $\,\overbrace{i^2 = -1}^{i = \sqrt{-1}}\,.$

[graph] point $\,1 + 2i\,$ is over $\,1\,$ and up $\,2\,$ in the COMPLEX PLANE; point $\,-2-i\,$ is to the left $\,2\,$ and down $\,1$

A PURE IMAGINARY # has real part equal to $\,0\,$: like $\,7i$

A REAL # has imaginary part equal to $\,0\,$: like $\,5$

EX [example] $\,3-4i\,$: real part $\,= 3\,$ and imaginary part $\,= -4$

ARITHMETIC WITH COMPLEX #s [numbers]

ADDITION: $\,(a + bi) + (c + di) = (a+c) + (b+d)i$
Verbalize: To add $\,2\,$ complex #s, add the real parts and add the imaginary parts

EX [example] $\,(1-3i) + (4 + 5i) = 5 + 2i$

SUBTRACTION:

$$ \begin{align} (a+bi) - (c+di) &= a + bi - c - di\cr\cr &= (a-c) + (b-d)i \end{align} $$

MULTIPLICATION:

$$ \begin{align} (a+bi)(c+di) &= ac + adi + bci + \overbrace{bdi^2}^{= -bd}\cr\cr &= (ac - bd) + (ad + bc)i \end{align} $$

Don't memorize formula! Multiply out as needed.

DIVISION:

$$ \frac{a+bi}{c+di} = \frac{a+bi}{c+di}\cdot\frac{c-di}{c-di}\ \ \text{ and simplify!!} $$

NOTE: The COMPLEX CONJUGATE of $\,c + di\,$ is $\,c - di$

SQUARE ROOTS of NEGATIVE #s [numbers] & other ideas

(1)

$$ \begin{align} \sqrt{-36} &= \sqrt{(-1)(36)} = \sqrt{-1}\sqrt{36}\cr &= i6 = 6i \end{align} $$

(2) Solve: $\,x^2 = -36\,$; $\,x = \pm\sqrt{-36}\,$; $\,x = \pm 6i$

$$ \begin{gather} i\cr i^2 = -1\cr i^3 = -i\cr i^4 = 1 \end{gather} $$

(3)

$$ \begin{align} i^{\overbrace{1523}^{\substack{\text{get}\\ \text{another}\\ \text{name}}}} &= i^{\,4\cdot 380 + 3}\cr &= i^{\,4\cdot 380}\,i^3\cr &= (i^4)^{380}\, i^3\cr &= (1)^{380}\,i^3\cr &= i^3 \end{align} $$

[division problem] Divide $\,1523\,$ by $\,4\,$; it goes in $\,380\,$ times, with a remainder of $\,3$

(4)

$$ \begin{align} \sqrt{-5}\sqrt{-3} &= i\sqrt{5}\ i\sqrt{3}\cr &= i^2\sqrt{15}\cr &= -\sqrt{15} \end{align} $$

Careful!!

$$ \sqrt{-5}\,\sqrt{-3} \overbrace{\ne}^{\text{is NOT equal to!}} \sqrt{(-5)(-3)} $$

more on GRAPHING COMPLEX #s (the COMPLEX PLANE)

[graph] vertical axis is the IMAGINARY AXIS; horizontal axis is the REAL AXIS; labels $\,bi\,$ and $\,-bi\,$ on the imaginary axis; label $\,a\,$ on real axis; $\,z = a+bi\,$ and $\,\bar z = a - bi$

Let $\,z = a + bi\,.$ Then

$$ \overbrace{\bar z}^{\text{“$z$ bar”}} \overbrace{:=}^{\text{equal, by defn}} a - bi $$

is the COMPLEX CONJUGATE of $\,z\,.$

Also,

$$ \begin{align} &\overbrace{|z|}^{\substack{\text{called the}\\ \text{MODULUS of $\,z$}}}\cr\cr &\quad = {\small\text{the LENGTH of the vector}}\cr &\qquad \ \ {\small\text{(arrow) representing $\,z$}}\cr\cr &\quad = \sqrt{a^2 + b^2} \end{align} $$

the QUADRATIC FORMULA—REVISITED

Recall: $\,ax^2 + bx + c = 0\,$   ($\,a\ne 0\,$) has solutions:

$$ x = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

$b^2 - 4ac\,$ is called the DISCRIMINANT, because it helps us to discriminate between HOW MANY real solutions the equation has! (See below...)

ZEROs of QUADRATIC FUNCTIONS (the DISCRIMINANT)

[graphs] $\,y = ax^2 + bx + c$

First graph: $\,b^2 - 4ac\gt 0\,,$ $\,2\,$ different real # [number] solns [solutions]

Second graph: $\,b^2 - 4ac = 0\,,$ exactly $\,1\,$ real # soln

Third graph: $\,b^2 - 4ac \lt 0\,,$ NO real # solutions

NOTE: $\,ax^2 + bx + c\,$ is called IRREDUCIBLE if $\,b^2 - 4ac\lt 0$

MULTIPLICITY of a ZERO

Let $\,P(x)\,$ be a polynomial.

EXAMPLE:

$$ \begin{align} P(x) &= 7x^5(x-1)^3(2x-3)^4(x+5)\cr &= 7x^5(x-1)^3\cdot 2^4(x-\frac 32)^4(x+5)\ \ \text{(How?)} \end{align} $$

zero	multiplicity
$0$	$5$
$1$	$3$
$\frac 32$	$4$
$-5$	$1$   (a zero of multiplicity $\,1\,$ is called a SIMPLE ZERO)

MULTIPLICITY OF ZEROS: GRAPHICAL CONSEQUENCES

$P(x) =$	GRAPH NEAR $\,c$
$(x-c)(\text{other stuff})$	[graph] OR [graph]	Cuts straight through!
$\underbrace{(x-c)^2}_{\text{always }\gt\, 0}(\text{stuff})$	[graph] OR [graph]	EVEN POWERS: NO SIGN CHANGE
$\underbrace{(x-c)^3}_{\substack{x\,\gt\, c:\, +\cr x\,\lt\,c:\,-}}(\text{stuff})$	[graph] OR [graph]	ODD POWERS: SIGN CHANGE

$x$	$(x-3)^2$	$(x-3)^4$	$(x-3)^6$
$3.1$	$0.01$	$0.0001$	$0.000001$

GRAPH gets FLATTER near $\,c\,$ as MULTIPLICITY INCREASES!

the FUNDAMENTAL THEOREM OF ALGEBRA

NOTE: $\,\Bbb C = \,$ the set of COMPLEX #s [numbers]

Consequence: Every poly'l [polynomial] of degree $\,n\,$ with complex coeffs [coefficients] has EXACTLY $\,n\,$ zeros in $\,\Bbb C\,$ (counting multiplicities). (WHY is this a consequence?)

POLYNOMIALS with REAL # COEFFICIENTS

Spse [suppose] that $\,P(x)\,$ has REAL # coefficients.

• If $\,a + bi\,$ ($\,b\ne 0\,$) is a zero, then so is $\,a - bi\,$ (i.e., NON-REAL zeros must occur in complex conjugate pairs).
•   $$ \begin{align} &\bigl(x - (a+bi)\bigr)\bigl(x-(a-bi)\bigr)\cr\cr &\quad \overset{\text{Check!}}{=} \underbrace{x^2 - 2ax + (a^2 + b^2)}_{\text{irreducible quadratic}} \end{align} $$
• $P(x)\,$ can be factored into LINEAR FACTORS and IRREDUCIBLE QUADRATICS

SOLVING POLYNOMIAL EQUATIONS

degree $\,\le 2\,$: NO PROBLEM!!

EXAMPLE: SOLVE

$$x^5 - 2 = 2x^3 + 2x^2 + 3x$$

(Find ALL complex # [number] solutions, if possible.)

STEP 1: Write in form $\,P(x) = 0$

$$ \underbrace{x^5 - 2x^3 - 2x^2 - 3x - 2}_{P(x)} = 0 $$

STEP 2: GRAPH $\,P\,$: use Confinement Thm [theorem]; read off any real zeros

STEP 3: Use long division, as needed, to completely factor $\,P\,.$

(FINISH the problem on card 34a!)

Should get these solutions:

$$ \begin{align} &x^5 - 2x^3 - 2x^2 - 3x - 2\cr &\quad = (x+1)^2(x-2)(x^2 + 1)\cr &\quad = 0 \end{align} $$

SOLNS [solutions]: $\{\underbrace{-1}_{\text{MULT } 2},\, 2,\, \pm i\}$

RATIONAL FCTS [functions]

EX [example] $\displaystyle\,y = \frac{x-1}{x-2}$

[graph] The line $\,y = 1\,$ is the HORIZONTAL ASYMPTOTE; the line $\,x = 1\,$ is the VERTICAL ASYMPTOTE

DOMAIN: $\,\{x\ |\ \text{denominator} \ne 0\}$

$y$-INTERCEPT: set $\,x = 0$

$x$-INTERCEPT: set $\,\text{NUM [numerator]} = 0$
(WHY? The only way a FRACTION can equal zero, is if the NUMERATOR equals zero)

ASYMPTOTES

An ASYMPTOTE is a curve (usually a line) that another curve gets arbitrarily close to as $\,x\,$ approaches $\,+\infty\,,$ $\,-\infty\,,$ or some finite # [number].

[dashed vertical line $\,x = 2\,$] $\,x=2\,$ is a VERTICAL ASY [asymptote]:

$$ \begin{gather} \text{as} \ \underbrace{x\rightarrow 2^+}_{\text{“}\,x\,\text{approaches}\, 2\,\text{from the right”}}\,,\ y\rightarrow +\infty\cr \cr \text{as}\ x\rightarrow 2^-\,, y\rightarrow -\infty \end{gather} $$

[dashed horizontal line $\,y = 1\,$] $\,y = 1\,$ is a HORIZ ASY [asymptote]:

$$ \begin{gather} \text{as}\ x\rightarrow \infty\,,\ y\rightarrow 1 \cr\cr \text{as}\ x\rightarrow -\infty\,,\ y\rightarrow 1 \end{gather} $$

FINDING VERTICAL ASYMPTOTES

... they occur where:

$$ \begin{gather} \text{DENOMINATOR} = 0\,,\cr \text{and}\ \ \text{NUMERATOR} \ne 0 \end{gather} $$

EX [example]

$$f(x) = \frac{-3}{2x+1}$$

[graph] Graph $\,y = 2x+1\,,$ which is zero at $\,-\frac 12\,,$ negative to the left of $\,-\frac12\,,$ and positive to the right of $\,-\frac 12\,.$

$$ \begin{gather} 2x + 1 = 0\cr x = -\frac 12 \end{gather} $$

As $\,x\rightarrow -\frac 12^+\,,$ $\,f(x)\approx \frac{(-)}{(\text{small}\, +)} \rightarrow -\infty$

As $\,x\rightarrow -\frac 12^-\,,$ $\,f(x)\approx \frac{(-)}{(\text{small}\, -)} \rightarrow +\infty$

[graph of $\,y = \frac{-3}{2x+1}\,,$ showing vertical asymptote at $\,x = -\frac 12\,$]   Confirm w/ [with] graphing calculator!

“PUNCTURE POINTS” of GRAPHS

GRAPH:

$$ \begin{align} f(x) &= \frac{x+1}{x^2 - x - 2}\cr\cr &= \frac{(x+1)}{(x+1)(x-2)}\cr &\quad \text{(NUM & DEN are both $\,0\,$ when $\,x=-1\,$)}\cr\cr &= \frac{1}{x-2}\,,\ \ \text{for}\ x\ne -1\ \text{and}\ x\ne 2 \end{align} $$

[NUM = numerator; DEN = denominator]

Multiplying by $\,\frac{x+1}{x+1}\,$ “punctures” the graph at $\,x = -1\,$!

[graph of $\,y = \frac{x+1}{x^2 - x - 2}\,,$ showing puncture point at $\,x = -1\,$ and vertical asymptote at $\,x = 2\,$]

FINDING HORIZONTAL ASYMPTOTES

Let $\,R(x) = \frac{N(x)}{D(x)}\,.$

$\text{deg}\, N(x) \lt \text{deg}\, D(x)\,$;

EX [example]: $\,R(x) = \frac{3x-1}{5x^2 + 2x - 1}\,$;

denominator gets bigger faster than numerator;

as $\,x\rightarrow \pm\infty\,,$ $\,y\rightarrow 0\,$;

$\,y = 0\,$ is a HORIZONTAL ASYMPTOTE.

$\text{deg}\, N(x) = \text{deg}\, D(x)\,$;

EX:

$$ \begin{align} R(x) &= \frac{3x-1}{5x + 2}\cdot\frac{\frac1x}{\frac1x}\cr &= \frac{3 - \frac 1x}{5 + \frac 2x} \end{align} $$

NUM [numerator] and DENOM [denominator] “grow” at same rate;

as $\,x\rightarrow \pm\infty\,,$ $\,y\,$ looks like ratio of leading terms;

$\,y = \frac{\text{leading coeff [coefficient] of $\,N(x)$}} {\text{leading coeff of $\,D(x)$}}\,$ is HORIZ ASY [horizontal asymptote].

FINDING SLANT ASYMPTOTES

If degree of NUMERATOR is exactly $\,1\,$ more than degree of DENOMINATOR ($\,\text{deg}\,N(x) = 1 + \text{deg}\,D(x)\,$) then you have a SLANT ASYMPTOTE. Do a LONG DIVISION!

$$ \begin{align} R(x) &= \frac{x^2 - 4x - 5}{x-3} \cr &= x-1 + \frac{-8}{x-3} \end{align} $$

[long division, showing $\,x^2 - 4x - 5\,$ being divided by $\,x-3\,,$ getting a quotient of $\,x-1\,$ with a remainder of $\,-8\,$]

As $\,x\rightarrow \pm\infty\,,$ $\,\frac{-8}{x-3}\rightarrow 0\,.$

$\,y = x-1\,$ is a SLANT ASYMPTOTE.

EXPONENTIAL FUNCTION with base $\,a$

EX [example]: $\,2^x\,,$ $\,(\frac 12)^x\,,$ $\,\text{e}^x\,,$ $\,3.1^x\,$ ... all EXP [exponential] fcts [functions].

WHY $\,a\gt 0\,$?   $\,(-4)^{1/2}\,$ is not a real #

WHY $\,a\ne 1\,$?   $\,1^x \overset{\text{always}}{=} 1$

GRAPHS of EXPONENTIAL FCTS [functions]

$\,a\gt 1\,,$ $\,f(x) = 2^x$

$x$	$2^x$
$-1$	$\frac 12$
$0$	$1$
$1$	$2$
$2$	$4$
$3$	$8$

[graph] plotted points $\,(-1,\frac 12)\,,$ $\,(0,1)\,,$ $\,(1,2)\,,$ graph is labeled $\,y = 2^x\,,$ INCREASING

$0\lt a\lt 1$

$$ \begin{align} f(x) &= (\frac 12)^x\cr &= (2^{-1})^x\cr &= 2^{-x} \end{align} $$

(Replace $\,x\,$ by $\,-x\,$; reflect about $y$-axis)

[graph] $y = (\frac 12)^x\,,$ DECREASING

PROPERTIES OF EXPONENTIAL FCTS [functions]

$$f(x) = a^x$$

allowable bases: $\,a\gt 0\,, a\ne 1$

RANGE: $\,(0,\infty)\,$:   $\,a^x \overset{\text{Strict!!}}{\gt} 0\,$ for all $\,x$

DOMAIN: $\,(-\infty,\infty)\,$;   exponential fcts are defined everywhere

ONE-TO-ONE FUNCTIONS

For $\,a\gt 1\,,$ INC [increasing]: $\,x\lt y\,$ is equivalent to $\,a^x \lt a^y$

For $\,0 \lt a \lt 1\,,$ DEC [decreasing]: $\,x\lt y\,$ is equivalent to $\,a^x \gt a^y$

KEY PROPERTY OF EXPONENTIAL FCTS [functions]

Let $\,f(x) = a^x\,.$ Then:

$$ \begin{align} f(x+\Delta x) &= a^{x+\Delta x}\cr\cr &= a^x\,a^{\Delta x}\cr\cr &= \underbrace{a^{\Delta x}}_{\text{scaling factor}}\cdot f(x) \end{align} $$

so

$$ \frac{f(x+\Delta x)}{f(x)} = \text{CONSTANT!} $$ $$ \overbrace{(x,y)}^{\substack{\text{pt on}\cr \text{graph of}\cr \text{exp fct}}} \rightarrow \bigl( \overbrace{x+\Delta x}^{\text{$x\,$ changes by $\,\Delta x$}}\,,\, \underbrace{a^{\Delta x}\cdot y}_{\substack{ \text{$y$-value}\cr \text{gets multiplied}\cr \text{by a}\cr \text{scaling factor}}} \bigr) $$

$f(x) = 2^x$
$x$	$2^x$
$2$	$\frac 14$
$-1$	$\frac 12$
$0$	$1$
$1$	$2$
$2$	$4$
$3$	$8$
$a = 2$ $\Delta x = 1$ $a^{\Delta x} = 2^1 = 2$

[Note: In this table, note that as the $x$-values successively increase by $\,1\,$ ($\,\Delta x = 1\,$), the $y$-values get successively multiplied by the scaling constant $\,2\,.$]

LINEAR versus EXPONENTIAL FCTS [functions]

FIND equal changes in $\,x\,$; how does $\,y\,$ change?

Be able to RECOGNIZE lin/exp [linear/exponential] fcts [functions] from a table, and FIND A FORMULA!

$x$	$f(x)$
$-2$	$\frac 29$
$0$	$2$
$2$	$18$
$4$	$162$
EXPONENTIAL!

[Note the equal changes in $\,x\,$:]

• from $\,-2\,$ to $\,0\,$ is $\,\color{red}{2}$
• from $\,0\,$ to $\,2\,$ is $\,\color{red}{2}$
• from $\,2\,$ to $\,4\,$ is $\,\color{red}{2}$

[Also note that the $y$-values successively get multiplied by $\,9\,$:]

• $\frac 29\,$ times $\,\color{red}{9}\,$ gives $\,2$
• $2\,$ times $\,\color{red}{9}\,$ gives $\,18$
• $18\,$ times $\,\color{red}{9}\,$ gives $\,162$

$x$	$g(x)$
$-2$	$-5$
$-1$	$-1$
$0$	$3$
$1$	$7$
$2$	$11$
LINEAR!

[Note the equal changes in $\,x\,$: the $x$-values keep increasing by $\,1\,.$ Also note the equal changes in $\,y\,$: the $y$-values keep increasing by $\,4\,$, so that $\,\Delta y = 4\,.$]

[On side of cards:]

$$ g(x) = mx + b $$ $$ \begin{align} g(x + \Delta x) &= m(x + \Delta x) + b\cr &= mx + m\Delta x + b\cr &= (mx + b) + m\Delta x\cr &= g(x) + m\Delta x \end{align} $$

For linear fcts: equal changes in $\,x\,$ give rise to equal changes in $\,y\,.$

FINDING the FORMULA ...

$$ \begin{gather} \text{[for the exponential function]}\cr f(x) = K\cdot a^x\cr \text{(Find $\,K\,$ and $\,a\,$)}\cr\cr 18 = K\cdot a^2\cr 2 = K\cdot a^0\,\text{; so } K = 2\cr\cr 18 = 2\cdot a^2\cr 9 = a^2\cr a = 3\cr\cr f(x) = 2\cdot 3^x \text{ and check!!} \end{gather} $$
$$ \begin{gather} \text{[for the linear function]}\cr g(x) = mx + b\cr \text{(Find $\,m\,$ and $\,b\,$)}\cr\cr m = \frac{7-3}{1-0} = 4\cr\cr 3 = 4(0) + b\cr b = 3\cr\cr g(x) = 4x + 3\text{ and check!!} \end{gather} $$

the NATURAL EXPONENTIAL FUNCTION

... is the function $\,f(x) = {\text{e}}^x\,$ where the base $\,\text{e}\,$ is the IRRATIONAL # [number] defined by:

$$ \text{e} = \underbrace{\lim_{n\rightarrow\infty}}_{\substack{\text{the limit,}\cr \text{as $\,n\,$ goes}\cr \text{to infinity}} } \bigl(1+\frac 1n\bigr)^n \approx 2.72 $$

$n$	$\bigl(1 + \frac 1n\bigr)^n$
$10$	$2.5937$
$100$	$2.7048$
$1000$	$2.7169$
$10{,}000$	$2.7181$
$\downarrow$	$\downarrow$
$\infty$	$\text{e}$

We can get these #s [the numbers in the second column] AS CLOSE TO $\,\text{e}\,$ as desired, by making $\,n\,$ sufficiently large!

[graph] labeled $\,y = \text{e}^x\,,$ “THE” exponential fct [function]

SPECIAL PROPERTY of the NATURAL EXPONENTIAL FCT [function]

[graph] shows point $\,(x,\text{e}^x)\,$ on graph of $\,y = \text{e}^x\,,$ slope of tangent line $\,= \text{e}^x\,,$ $y$-value of point tells the slope of the tangent line!

Imagine walking along a curve, from left to right. AT EACH INSTANT, the slope of the tangent line tells how you're rising/falling.

[graph] shows a point on the graph of $\,y = \text{e}^x\,$ with $y$-value equal to $\,100\,$; at this point, the $y$-values are changing $\,100\,$ times faster than the $x$-values!

If $\,x\,$ changes by:	we expect $\,y\,$ to change by about:
$1$	$100$
$0.1$	$0.1(100) = 10$
$-0.01$	$-0.01(100) = -1$
$\Delta x$	$100\Delta x$

NOTE:

$$ \begin{gather} \text{SIMPLE INTEREST}\cr = (\text{PRINCIPAL})(\text{RATE})(\text{TIME}) \end{gather} $$

COMPOUND INTEREST

... involves ADDING IN INTEREST at regular intervals.

Let:

• $P = \,$ initial investment
• $r = \,$ annual interest rate ($\,2\% = 0.02\,$)
• $n = \,$ # [number] of times, per year, that interest is added in
  $\hphantom{n} = \,$ # of COMPOUNDING PERIODS per year

  NOTE:

  $$ n\ \frac{\text{cmping per}}{\text{yr}} = \frac{1 \text{ cmping per}}{\frac 1n \text{ yrs}} $$ [cmping per = compounding period]
• $t = \,$ # yrs [years]
• $A = \,$ AMT [amount] after $\,t\,$ yrs

After:	[Principal + Interest:]
$\color{red}{1}\,$ cmpding [compounding] period:	$$ \begin{align} &P + P(r)(\frac 1n)\cr &\quad = \color{red}{P(1 + \frac rn)^1} \end{align} $$
$\color{red}{2}\,$ [cmpding periods]:	$$ \begin{align} &P(1 + \frac rn)\cr &\quad + P(1+\frac rn)(r)(\frac 1n)\cr &\quad = \color{red}{P(1+\frac rn)^2} \end{align} $$
$\vdots$	$\vdots$
$\color{red}{n}\,$ ($\,1\,$ yr)	$\color{red}{P(1+\frac rn)^n}$
$\vdots$	$\vdots$
$\color{red}{2n}\,$ ($\,2\,$ yrs)	$\color{red}{P(1+\frac rn)^{n\cdot 2}}$
$$ A = P(1 + \frac rn)^{nt} $$

CONTINUOUS COMPOUNDING

$$ \begin{gather} {\small\text{What happens as we}}\cr \underbrace{\small\text{add in interest more & more frequently?}}_{\text{let } n\rightarrow\infty} \end{gather} $$

Let $\,\frac 1m = \frac rn\,,$ so $\,n = mr\,.$

$$ \begin{align} &P(1+\frac rn)^{nt} \cr &\quad = P(1 + \frac 1m)^{mrt}\cr &\quad = P\left[(1+\frac 1m)^m\right]^{rt}\cr &\quad \rightarrow P{\text{e}}^{rt} \end{align} $$

$$A = P{\text{e}}^{rt}$$

CONTINUOUS COMPOUNDING: interest is added in at the INSTANT it's earned!

Exponential Growth

... is modeled by:

$$ \underbrace{n(t)}_{\text{# at time $\,t$}} = \underbrace{n_0}_{\substack{\text{# at}\cr \text{time $\,0\,,$}\cr n(0)}}\,{\text{e}}^{rt} $$

($\,r = \,$ relative rate of growth)

Key idea, from Calculus: [graph shows point $\,(t,n_o{\text{e}}^{rt})\,$ with tangent line at this point; slope of tangent line is $\,r\cdot n_0{\text{e}}^{rt}\,$]

How fast is population increasing at time $\,t\,$?

ANSWER: $\,r\,$ times as fast as current population!!

The growth rate is proportional to the current size!

A Typical EXP [exponential] Growth Problem

The pop [population] of a city was $\,680{,}000\,$ in $\,1992\,$; relative growth rate is $\,12\%/\text{yr}\,.$ What is the population $\,t\,$ years after $\,1992\,$? In $\,2000\,$?

First Solution: Let $\,t = 0\,$ in $\,1992\,$; measure $\,n(t)\,$ in THOUSANDS (i.e., $\,680\,$ means $\,680\,$ THOUSAND).

$$ n(t) = 680{\text{e}}^{0.12t} $$

The year $\,2000\,$ corresponds to $\,t = 8\,$:

$$ \begin{align} n(8) &= 680{\text{e}}^{0.12(8)}\cr &\approx 1775{.}954 \end{align} $$

In $\,2000\,,$ the pop [population] will be about $\,1{,}775{,}954\,$ people.

Alternative solution:

$$ n(t) = 680{,}000{\text{e}}^{0.12t} $$ $$ \begin{align} n(8) &= 680{,}000{\text{e}}^{0.12t}\cr &\approx 1{,}775{,}954 \end{align} $$

LOGARITHMIC FUNCTIONS

Exponential fcts [functions] are $\,1{-}1\,$ [one-to-one], hence have inverses!!

$$ \overbrace{y = {\text{a}}^x}^{\text{exponential form}} \ \text{iff}\ \overbrace{\log_a y = x}^{\text{logarithmic form}} $$

[Note: ‘iff’ means ‘if and only if’. On the logarithmic form, arrow starts with $\,a\,,$ moves to $\,x\,,$ then to $\,y\,,$ giving $\,a^x = y\,.$]

[Sketch shows $\,x\,$ being dropped into the ‘EXP fct [exponential function], base $\,a\,$’ box, with $\,a^x = y\,$ as the output. Then, $\,y\,$ is dropped into the ‘$\,\text{log}_a\,$’ box, ‘LOG fct [logarithmic function], base $\,a\,$’ giving the output $\,x\,.$]

What is the # [number] $\,\text{log}_a y\,$?

It is the EXPONENT that ‘$\,a\,$’ must be raised to, to get $\,y\,$!

TWO IMPORTANT VIEWPTS [viewpoints]

1. Logarithmic fcts [functions] ‘UNDO’ exponential fcts!
2. A logarithm IS an exponent!

PROPERTIES of LOGARITHMIC FCTS [functions]

[first graph, $\,a\gt 1\,,$ showing the graph of $\,y = a^x\,$ and its inverse $\,y = \text{log}_a x\,,$ mirrored about the line $\,y = x\,.$]

[second graph, $\,0\lt a\lt 1\,,$ showing the graph of $\,y = a^x\,$ and its inverse $\,y = \text{log}_a x\,,$ mirrored about the line $\,y = x\,.$]

• LOG fcts [logarithmic functions] are $\,1{-}1\,$ [one-to-one]
• $\text{log}_a 1 = 0$
• DOMAIN: $\,(0,\infty)$
• RANGE: $\,(-\infty,\infty)$
• VERTICAL ASY [asymptote]: $\,x = 0$
• $\text{log}_a a = 1$
• SPECIAL NAMES:
  $\text{log}_{10}x\,$   COMMON LOG
  $\text{log}_e x = \ln x\,$   NATURAL LOG
• INVERSE PROPERTIES:
  $\text{log}_a a^x = x$
  $a^{\text{log}_a x} = x$

LAWS of LOGARITHMS

Let $\,a\gt 0\,,$ $\,a\ne 1\,,$ $\,A\gt 0\,,$ $\,B\gt 0\,,$ $\,c\in\Bbb R\,.$ Then,

$$ \log_a(AB) = \log_a A + \log_a B $$

(the log of a PRODUCT is the SUM of the logs)

$$ \log_a(\frac AB) = \log_a A - \log_a B $$

(the log of a QUOTIENT is the DIFFERENCE of the logs)

$$ \log_a(A^c) = c\,\log_a A $$

(You can bring exponents down front)

Sample Proof

Spse [suppose] $\,\log_a A = u\,$ and $\,\log_a B = v\,,$ so $\,a^u = A\,$ and $\,a^v = B\,.$ Then,

$$ \begin{align} \log_a AB &= \log_a a^u a^v\ \text{(substitution)}\cr &= \log_a a^{u+v}\ \text{(law of exponents)}\cr &= u + v\ \text{(inverse property)}\cr &= \log_a A + \log_a B\ \text{(subst [substitution])} \end{align} $$

CHANGE of BASE FORMULA

Let $\,a\,$ and $\,b\,$ be positive real #s [numbers] that are not equal to $\,1\,,$ and let $\,x \gt 0\,.$ Then:

$$ \log_b x = \frac{\log_a x}{\log_a b} $$

You can switch from any base to any other base—the price you pay: you must $\,\div\,$ [divide] by the log, to the new base, of the old base.

Consequence: any log curve is just a VERTICAL SCALING of any other log curve!

DERIVATION

The following equations are EQUIVALENT:

$$ \begin{gather} y = \log_b x\cr \text{(LHS [left-hand-side]}\cr \text{of Change-of-base formula)}\cr\cr b^y = x\cr \text{(equiv [equivalent]}\cr \text{EXP [exponential] form)}\cr\cr \log_a b^y = \log_a x\cr \text{(log fcts are 1-1)}\cr\cr y \log_a b = \log_a x\cr \text{(property of logs)}\cr\cr y = \frac{\log_a x}{\log_a b}\cr \text{(divide by $\,\log_a b\,$)} \end{gather} $$

SOLVING EXPONENTIAL EQUATIONS

(an exponential equation has at least one VARIABLE in an EXPONENT)

• ISOLATE an exponential expression (i.e., get something of the form $\,a^{(\text{stuff})}\,$ all by itself on one side of equation)
• ‘UNDO’ the exponential fct [function] with a log! (usually use $\,\log\,$ or $\,\ln\,$)
• SOLVE for the variable
• CHECK in original equation

EXAMPLES

SOLVE:

$$ \begin{gather} 8 \cdot 5^{2x-1} = 20\cr\cr 5^{2x-1} = \frac{20}{8} = \frac{5}{2}\cr\cr \ln 5^{2x-1} = \ln \frac{5}{2}\cr\cr (2x-1) \ln 5 = \ln 2.5\cr\cr 2x \ln 5 - \ln 5 = \ln 2.5\cr\cr 2x \ln 5 = \ln 2.5 + \ln 5\cr\cr x = \frac{\ln 2.5 + \ln 5}{2 \ln 5} \end{gather} $$

(NO APPROXIMATION until final step!)

SOLVE:

$$\text{e}^{2x} - \text{e}^{x} - 6 = 0$$

(‘fake’ quadratic!)

$$(\text{e}^x)^2 - (\text{e}^x) - 6 = 0$$

Let $\,u =\text{e}^x\,$:

$$ \begin{gather} u^2 - u - 6 = 0\cr\cr (u-3)(u+2) = 0\cr\cr (\text{e}^x-3)(\text{e}^x+2) = 0\cr\cr \text{e}^x = 3 \quad \text{or} \quad \overbrace{\text{e}^x = -2}^{\text{exp fcts can't be negative}}\cr\cr x = \ln 3 \end{gather} $$

[Don't forget to check your solutions in the original equations!]