# Calculus/Precalculus/Solutions
## Convert to interval notation

$\{x:-4<x<2\}\,$
$\mathbf {(-4,2)}$
$\mathbf {(-4,2)}$
$\{x:-{\frac {7}{3}}\leq x\leq -{\frac {1}{3}}\}$
$\mathbf {[-{\frac {7}{3}},-{\frac {1}{3}}]}$
$\mathbf {[-{\frac {7}{3}},-{\frac {1}{3}}]}$
$\{x:-\pi \leq x<\pi \}$
$\mathbf {[-\pi ,\pi )}$
$\mathbf {[-\pi ,\pi )}$
$\{x:x\leq 17/9\}$
$\mathbf {(-\infty ,{\frac {17}{9}}]}$
$\mathbf {(-\infty ,{\frac {17}{9}}]}$
$\{x:5\leq x+1\leq 6\}$
$4\leq x\leq 5$
$\mathbf {[4,5]}$
$4\leq x\leq 5$
$\mathbf {[4,5]}$
$\{x:x-1/4<1\}\,$
$x<1{\frac {1}{4}}={\frac {5}{4}}$
$\mathbf {(-\infty ,{\frac {5}{4}})}$
$x<1{\frac {1}{4}}={\frac {5}{4}}$
$\mathbf {(-\infty ,{\frac {5}{4}})}$
$\{x:3>3x\}\,$
$1>x$
$x<1$
$\mathbf {(-\infty ,1)}$
$1>x$
$x<1$
$\mathbf {(-\infty ,1)}$
$\{x:0\leq 2x+1<3\}$
$-1\leq 2x\leq 2$
$-{\frac {1}{2}}\leq x<1$
$\mathbf {[-{\frac {1}{2}},1)}$
$-1\leq 2x\leq 2$
$-{\frac {1}{2}}\leq x<1$
$\mathbf {[-{\frac {1}{2}},1)}$
$\{x:5<x{\mbox{ and }}x<6\}\,$
$5<x<6$
$\mathbf {(5,6)}$
$5<x<6$
$\mathbf {(5,6)}$
$\{x:5<x{\mbox{ or }}x<6\}\,$
A number in the set can be on either the red or blue line, so the entire number line is included.
$\mathbf {(-\infty ,\infty )}$
A number in the set can be on either the red or blue line, so the entire number line is included.
$\mathbf {(-\infty ,\infty )}$

## State the following intervals using set notation

$[3,4]\,$
$\mathbf {\{x:3\leq x\leq 4\}}$
$\mathbf {\{x:3\leq x\leq 4\}}$
$[3,4)\,$
$\mathbf {\{x:3\leq x<4\}}$
$\mathbf {\{x:3\leq x<4\}}$
$(3,\infty )$
$\mathbf {\{x:x>3\}}$
$\mathbf {\{x:x>3\}}$
$(-{\frac {1}{3}},{\frac {1}{3}})\,$
$\mathbf {\{x:-{\frac {1}{3}}<x<{\frac {1}{3}}\}}$
$\mathbf {\{x:-{\frac {1}{3}}<x<{\frac {1}{3}}\}}$
$(-\pi ,{\frac {15}{16}})\,$
$\mathbf {\{x:-\pi <x<{\frac {15}{16}}\}}$
$\mathbf {\{x:-\pi <x<{\frac {15}{16}}\}}$
$(-\infty ,\infty )$
$\mathbf {\{x:x\in \Re \}}$
$\mathbf {\{x:x\in \Re \}}$

## Which one of the following is a true statement?

$|x+y|=|x|+|y|\,$
$x=-5,y=5$
$|x+y|=|-5+5|=|0|=0$, and $|x|+|y|=|-5|+|5|=5+5=10$ Thus, $|x+y|\neq |x|+|y|$
$x=-5,y=5$
$|x+y|=|-5+5|=|0|=0$, and $|x|+|y|=|-5|+|5|=5+5=10$ Thus, $|x+y|\neq |x|+|y|$
$|x+y|\geq |x|+|y|$
$|x+y|\ngeq |x|+|y|$
$|x+y|\ngeq |x|+|y|$
$|x+y|\leq |x|+|y|$

## Evaluate the following expressions

$8^{1/3}\,$
$(2^{3})^{1/3}=2^{1}=\mathbf {2}$
$(2^{3})^{1/3}=2^{1}=\mathbf {2}$
$(-8)^{1/3}\,$
$(-2^{3})^{1/3}=-2^{1}=\mathbf {-2}$
$(-2^{3})^{1/3}=-2^{1}=\mathbf {-2}$
${\bigg{\frac {1}{8}}{\bigg )}^{1/3}\,$
$({\frac {1}{2^{3}}})^{1/3}=(2^{-3})^{1/3}=2^{-1}=\mathbf {\frac {1}{2}}$
$({\frac {1}{2^{3}}})^{1/3}=(2^{-3})^{1/3}=2^{-1}=\mathbf {\frac {1}{2}}$
$(8^{2/3})(8^{3/2})(8^{0})\,$
$8^{{\frac {2}{3}}+{\frac {3}{2}}+0}=8^{{\frac {4}{6}}+{\frac {9}{6}}}=8^{\frac {13}{6}}=(2^{3})^{\frac {13}{6}}=\mathbf {2^{13/2}}$
$8^{{\frac {2}{3}}+{\frac {3}{2}}+0}=8^{{\frac {4}{6}}+{\frac {9}{6}}}=8^{\frac {13}{6}}=(2^{3})^{\frac {13}{6}}=\mathbf {2^{13/2}}$
${\bigg{\bigg{\frac {1}{8}}{\bigg )}^{1/3}{\bigg )}^{7}$
$(({\frac {1}{2^{3}}})^{1/3})^{7}=((2^{-3})^{1/3})^{7}=(2^{-1})^{7}=2^{-7}={\frac {1}{2^{7}}}=\mathbf {\frac {1}{128}}$
$(({\frac {1}{2^{3}}})^{1/3})^{7}=((2^{-3})^{1/3})^{7}=(2^{-1})^{7}=2^{-7}={\frac {1}{2^{7}}}=\mathbf {\frac {1}{128}}$
${\sqrt[{3}]{\frac {27}{8}}}$
$({\frac {27}{8}})^{1/3}=({\frac {3^{3}}{2^{3}}})^{1/3}={\frac {3^{1}}{2^{1}}}=\mathbf {\frac {3}{2}}$
$({\frac {27}{8}})^{1/3}=({\frac {3^{3}}{2^{3}}})^{1/3}={\frac {3^{1}}{2^{1}}}=\mathbf {\frac {3}{2}}$
${\frac {4^{5}\cdot 4^{-2}}{4^{3}}}$
$4^{5-2-3}=4^{0}=\mathbf {1}$
$4^{5-2-3}=4^{0}=\mathbf {1}$
${\bigg{\sqrt {27}}{\bigg )}^{2/3}$
$((3^{3})^{1/2})^{2/3}=(3^{\frac {3}{2}})^{\frac {2}{3}}=3^{1}=\mathbf {3}$
$((3^{3})^{1/2})^{2/3}=(3^{\frac {3}{2}})^{\frac {2}{3}}=3^{1}=\mathbf {3}$
${\frac {\sqrt {27}}{\sqrt[{3}]{9}}}$
${\frac {(3^{3})^{1/2}}{(3^{2})^{1/3}}}={\frac {3^{\frac {3}{2}}}{3^{\frac {2}{3}}}}=3^{{\frac {3}{2}}-{\frac {2}{3}}}=3^{{\frac {9}{6}}-{\frac {4}{6}}}=\mathbf {3^{5/6}}$
${\frac {(3^{3})^{1/2}}{(3^{2})^{1/3}}}={\frac {3^{\frac {3}{2}}}{3^{\frac {2}{3}}}}=3^{{\frac {3}{2}}-{\frac {2}{3}}}=3^{{\frac {9}{6}}-{\frac {4}{6}}}=\mathbf {3^{5/6}}$

## Simplify the following

$x^{3}+3x^{3}\,$
$\mathbf {4x^{3}}$
$\mathbf {4x^{3}}$
${\frac {x^{3}+3x^{3}}{x^{2}}}$
$\mathbf {4x}$
$\mathbf {4x}$
$(x^{3}+3x^{3})^{3}\,$
$\mathbf {64x^{9}}$
$\mathbf {64x^{9}}$
${\frac {x^{15}+x^{3}}{x}}$
$\mathbf {x^{14}+x^{2}}$
$\mathbf {x^{14}+x^{2}}$
$(2x^{2})(3x^{-2})\,$
$\mathbf {6}$
$\mathbf {6}$
${\frac {x^{2}y^{-3}}{x^{3}y^{2}}}$
$\mathbf {\frac {1}{xy^{5}}}$
$\mathbf {\frac {1}{xy^{5}}}$
${\sqrt {x^{2}y^{4}}}$
$\mathbf {xy^{2}}$
$\mathbf {xy^{2}}$
${\bigg{\frac {8x^{6}}{y^{4}}}{\bigg )}^{1/3}$
$\mathbf {\frac {2x^{2}}{y^{\frac {4}{3}}}}$
$\mathbf {\frac {2x^{2}}{y^{\frac {4}{3}}}}$

## Find the roots of the following polynomials

$x^{2}-1\,$
$x=\pm 1$
$x=\pm 1$
$x^{2}+2x+1\,$
$x=-1$
$x=-1$
$x^{2}+7x+12\,$
$x=-3,x=-4$
$x=-3,x=-4$
$3x^{2}-5x-2\,$
$x=2,x=-{\frac {1}{3}}$
$x=2,x=-{\frac {1}{3}}$
$x^{2}+5/6x+1/6\,$
$x=-{\frac {1}{3}},x=-{\frac {1}{2}}$
$x=-{\frac {1}{3}},x=-{\frac {1}{2}}$
$4x^{3}+4x^{2}+x\,$
$x=0,x=-{\frac {1}{2}}$
$x=0,x=-{\frac {1}{2}}$
$x^{4}-1\,$
$x=\pm i,x=\pm 1$
$x=\pm i,x=\pm 1$
$x^{3}+2x^{2}-4x-8\,$
$x=\pm 2$
$x=\pm 2$

## Factor the following expressions

$4a^{2}-ab-3b^{2}\,$
$(4a+3b)(a-b)$
$(4a+3b)(a-b)$
$(c+d)^{2}-4\,$
$(c+d+2)(c+d-2)$
$(c+d+2)(c+d-2)$
$4x^{2}-9y^{2}\,$
$(2x+3y)(2x-3y)$
$(2x+3y)(2x-3y)$

## Simplify the following

${\frac {x^{2}-1}{x+1}}\,$
$x-1,x\neq -1$
$x-1,x\neq -1$
${\frac {3x^{2}+4x+1}{x+1}}\,$
$3x+1,x\neq -1$
$3x+1,x\neq -1$
${\frac {4x^{2}-9}{4x^{2}+12x+9}}\,$
${\frac {2x-3}{2x+3}}$
${\frac {2x-3}{2x+3}}$
${\frac {x^{2}+y^{2}+2xy}{x(x+y)}}\,$
${\frac {x+y}{x}},x\neq -y$
${\frac {x+y}{x}},x\neq -y$

## Functions

52. Let $f(x)=x^{2}$.
$f(0)$
$f(2)$
$f(0)=0$
$f(2)=4$
$f(0)=0$
$f(2)=4$
$f$
$(-\infty ,\infty )$
$[0,\infty )$
$(-\infty ,\infty )$
$[0,\infty )$
$f$
$f$
$f(-1)=f(1)=1$
$f$
$f(-1)=f(1)=1$
53. Let $f(x)=x+2$, $g(x)=1/x$.
    a. Give formulae for
$f+g$
$(f+g)(x)=x+2+1/x=(x^{2}+2x+1)/x$
$(f+g)(x)=x+2+1/x=(x^{2}+2x+1)/x$
$f-g$
$(f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x$
$(f-g)(x)=x+2-1/x=(x^{2}+2x-1)/x$
$g-f$
$(g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x$
$(g-f)(x)=1/x-x-2=(1-x^{2}-2x)/x$
$f\times g$
$(f\times g)(x)=(x+2)/x$
$(f\times g)(x)=(x+2)/x$
$f/g$
$(f/g)(x)=x(x+2)$
$x\neq 0$
$f/g$
$g$
$(f/g)(x)=x(x+2)$
$x\neq 0$
$f/g$
$g$
$g/f$
$(g/f)(x)=1/[x(x+2)]$
$1/[x(x+2)]$
$(g/f)(x)=1/[x(x+2)]$
$1/[x(x+2)]$
$f\circ g$
$(f\circ g)(x)=1/x+2=(2x+1)/x$
$(f\circ g)(x)=1/x+2=(2x+1)/x$
$g\circ f$
$(g\circ f)(x)=1/(x+2)$
$(g\circ f)(x)=1/(x+2)$
$f(g(2))$
$g(f(2))$
$f(g(2))=5/2$
$g(f(2))=1/4$
$f(g(2))=5/2$
$g(f(2))=1/4$
$f$
$g$
$f^{-1}(x)=x-2$
$g^{-1}(x)=1/x$
$g$
$f^{-1}(x)=x-2$
$g^{-1}(x)=1/x$
$g$
55. Consider the following function
    $f(x)={\begin{cases}-{\frac {1}{9}}&{\mbox{if }}x<-1\\2&{\mbox{if }}-1\leq x\leq 0\\x+3&{\mbox{if }}x>0.\end{cases}}$
${(-\infty ,\infty )}$
${(-\infty ,\infty )}$
${(-1/9,\infty )}$
${(-1/9,\infty )}$
$f$
${x>0}$
${x>0}$
56. Consider the following function
    $f(x)={\begin{cases}x^{2}&{\mbox{if }}x>0\\-1&{\mbox{if }}x\leq 0.\end{cases}}$
${(-\infty ,\infty )}$
${(-\infty ,\infty )}$
${(-1,\infty )}$
${(-1,\infty )}$
$f$
${x>0}$
${x>0}$
57. Consider the following function
    $f(x)={\frac {\sqrt {2x-3}}{x-10}}$
${(3/2,10)\cup (10,\infty )}$
${(3/2,10)\cup (10,\infty )}$
${(-\infty ,\infty )}$
${(-\infty ,\infty )}$
$f$
${(3/2,10)and(x>10)}$
${(3/2,10)and(x>10)}$
58. Consider the following function
    $f(x)={\frac {x-7}{x^{2}-49}}$
${(-\infty ,-7)\cup (-7,\infty )}$
${(-\infty ,-7)\cup (-7,\infty )}$
${(-\infty ,\infty )}$
${(-\infty ,\infty )}$
$f$
${(-\infty ,-7)and(-7,\infty )}$
${(-\infty ,-7)and(-7,\infty )}$

## Graphing

$3x-y=4$
$3x-y=4$
$3x-2y=0$
$3x-2y=0$
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