Chapter 3

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

function

function

function

function

function

function

function

function

function

not a function

$f(-3)=-11;$
$f(2)=-1;$
$f(-a)=-2a-5;$
$-f(a)=-2a+5;$
$f(a+h)=2a+2h-5$

$f(-3)=\sqrt{5}+5;$
$f(2)=5;$
$f(-a)=\sqrt{2+a}+5;$
$-f(a)=-\sqrt{2-a}-5;$
$f(a+h)=\sqrt{2-a-h}+5$

$f(-3)=2;$ $f(2)=1-3=-2;$
$f(-a)=|-a-1|-|-a+1|;$
$-f(a)=-|a-1|+|a+1|;$
$f(a+h)=|a+h-1|-|a+h+1|$

$\frac{g(x)-g(a)}{x-a}=x+a+2,x\ne a$

a. $f(-2)=14;$ b. $x=3$

a. $f(5)=10;$ b. $x=-1$ or $\text{<mi>x</mi><mo>=</mo><mn>4</mn>}$

not a function

function

function

function

function

function

not a function so it is also not a one-to-one function

one-to- one function

function, but not one-to-one

function

function

not a function

$f(x)=1,x=2$

$\begin{array}{ccccc}f(-2)=14;& f(-1)=11;& f(0)=8;& f(1)=5;& f(2)=2\end{array}$

$\begin{array}{ccccc}f(-2)=4;& f(-1)=4.414;& f(0)=4.732;& f(1)=5;& f(2)=5.236\end{array}$

$\begin{array}{ccccc}f(-2)=\frac{1}{9};& f(-1)=\frac{1}{3};& f(0)=1;& f(1)=3;& f(2)=9\end{array}$

20

$[0,\text{100}]$

$[-0.001,\text{0}\text{.001}]$

$[-1,000,000,\text{1,000,000}]$

$[0,\text{10}]$

$[\mathrm{-0.1},\text{0.1}]$

$[-100,\text{100}]$

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

There is no restriction on $x$ for $f(x)=\sqrt[3]{x}$ because you can take the cube root of any real number. So the domain is all real numbers, $(-\infty ,\infty ).$ When dealing with the set of real numbers, you cannot take the square root of negative numbers. So $x$ -values are restricted for $f(x)=\sqrt[]{x}$ to nonnegative numbers and the domain is $[0,\infty ).$

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the $x$ -axis and $y$ -axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate $-\infty$ or $\text{<mi>∞</mi><mo>.</mo>}$ Combine the graphs to find the graph of the piecewise function.

$(-\infty ,\infty )$

$(-\infty ,3]$

$(-\infty ,\infty )$

$(-\infty ,\infty )$

$(-\infty ,-\frac{1}{2})\cup (-\frac{1}{2},\infty )$

$(-\infty ,-11)\cup (-11,2)\cup (2,\infty )$

$(-\infty ,-3)\cup (-3,5)\cup (5,\infty )$

$(-\infty ,5)$

$[6,\infty )$

$\left(-\infty ,-9\right)\cup \left(-9,9\right)\cup \left(9,\infty \right)$

domain: $(2,8],$ range $[6,8)$

domain: $[-4,\text{4],}$ range: $[0,\text{2]}$

domain: $[-5,\text{<mn>3</mn><mo stretchy="false">)</mo><mo>,</mo>}$ range: $\left[0,2\right]$

domain: $(-\infty ,1],$ range: $[0,\infty )$

domain: $\left[-6,-\frac{1}{6}\right]\cup \left[\frac{1}{6},6\right];$ range: $\left[-6,-\frac{1}{6}\right]\cup \left[\frac{1}{6},6\right]$

domain: $[-3,\text{<mi>∞</mi><mo stretchy="false">)</mo><mo>;</mo>}$ range: $[0,\infty )$

domain: $(-\infty ,\infty )$

domain: $(-\infty ,\infty )$

domain: $(-\infty ,\infty )$

domain: $(-\infty ,\infty )$

$\begin{array}{cccc}f(-3)=1;& f(-2)=0;& f(-1)=0;& f(0)=0\end{array}$

$\begin{array}{cccc}f(-1)=-4;& f(0)=6;& f(2)=20;& f(4)=34\end{array}$

$\begin{array}{cccc}f(-1)=-5;& f(0)=3;& f(2)=3;& f(4)=16\end{array}$

domain: $(-\infty ,1)\cup (1,\infty )$

$[0,\text{<mn>8</mn><mo stretchy="false">]</mo>}$

Many answers. One function is $f(x)=\frac{1}{\sqrt{x-2}}.$

Yes, the average rate of change of all linear functions is constant.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

$4\left(b+1\right)$

3

$4x+2h$

$\frac{-1}{13\left(13+h\right)}$

$3h^2+9h+9$

$4x+2h-3$

$\frac{4}{3}$

increasing on $\left(-\infty ,-2.5\right)\cup \left(1,\infty \right),$ decreasing on $(-2.5,\text{<mn>1</mn><mo stretchy="false">)</mo>}$

increasing on $\left(-\infty ,1\right)\cup \left(3,4\right),$ decreasing on $\left(1,3\right)\cup \left(4,\infty \right)$

local maximum: $(-3,\text{<mn>60</mn><mo stretchy="false">)</mo><mo>,</mo>}$ local minimum: $(3,\text{<mo>−</mo><mn>60</mn><mo stretchy="false">)</mo>}$

absolute maximum at approximately $(7,\text{<mn>150</mn><mo stretchy="false">)</mo><mo>,</mo>}$ absolute minimum at approximately $(\mathrm{-7.5},\text{<mn>−220</mn><mo stretchy="false">)</mo>}$

-4

27

–0.167

Local minimum at $(3,-22),$ decreasing on $(-\infty ,\text{<mn>3</mn><mo stretchy="false">)</mo><mo>,</mo>}$ increasing on $(3,\text{<mi>∞</mi><mo stretchy="false">)</mo>}$

Local minimum at $(-2,-2),$ decreasing on $(-3,-2),$ increasing on $(-2,\text{<mi>∞</mi><mo stretchy="false">)</mo>}$

Local maximum at $(-0.5,\text{<mn>6</mn><mo stretchy="false">)</mo><mo>,</mo>}$ local minima at $(-3.25,-47)$ and $(2.1,-32),$ decreasing on $(-\infty ,-3.25)$ and $(-0.5,\text{<mn>2.1</mn><mo stretchy="false">)</mo><mo>,</mo>}$ increasing on $(-3.25,\text{<mo>−</mo><mn>0.5</mn><mo stretchy="false">)</mo>}$ and $(2.1,\text{<mi>∞</mi><mo stretchy="false">)</mo>}$

A

$b=5$

2.7 gallons per minute

approximately –0.6 milligrams per day

Find the numbers that make the function in the denominator $g$ equal to zero, and check for any other domain restrictions on $f$ and $g,$ such as an even-indexed root or zeros in the denominator.

Yes. Sample answer: Let $f(x)=x+1\text{and }g(x)=x-1.$ Then $f(g(x))=f(x-1)=(x-1)+1=x$ and $g(f(x))=g(x+1)=(x+1)-1=x.$ So $f\circ g=g\circ f.$

$(f+g)\left(x\right)=2x+6,$ domain: $(-\infty ,\infty )$

$(f-g)\left(x\right)=2x^2+2x-6,$ domain: $(-\infty ,\infty )$

$(fg)\left(x\right)=-x^4-2x^3+6x^2+12x,$ domain: $(-\infty ,\infty )$

$\left(\frac{f}{g}\right)\left(x\right)=\frac{x^2+2x}{6-x^2},$ domain: $(-\infty ,-\sqrt{6})\cup (-\sqrt{6},\sqrt{6})\cup (\sqrt{6},\infty )$

$(f+g)\left(x\right)=\frac{4x^3+8x^2+1}{2x},$ domain: $(-\infty ,0)\cup (0,\infty )$

$(f-g)\left(x\right)=\frac{4x^3+8x^2-1}{2x},$ domain: $(-\infty ,0)\cup (0,\infty )$

$(fg)\left(x\right)=x+2,$ domain: $(-\infty ,0)\cup (0,\infty )$

$\left(\frac{f}{g}\right)\left(x\right)=4x^3+8x^2,$ domain: $(-\infty ,0)\cup (0,\infty )$

$(f+g)(x)=3x^2+\sqrt{x-5},$ domain: $[5,\infty )$

$(f-g)(x)=3x^2-\sqrt{x-5},$ domain: $[5,\infty )$

$(fg)(x)=3x^2\sqrt{x-5},$ domain: $[5,\infty )$

$\left(\frac{f}{g}\right)(x)=\frac{3x^2}{\sqrt{x-5}},$ domain: $(5,\infty )$

$f(g(x))=\sqrt{x^2+3}+2,g(f(x))=x+4\sqrt{x}+7$

$f(g(x))=\sqrt[3]{\frac{x+1}{x^3}}=\frac{\sqrt[3]{x+1}}{x},g(f(x))=\frac{\sqrt[3]{x}+1}{x}$