2.3E Exercises
- Page ID
- 152902
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Translate \( (-8, 2] \cup [3, 5) \) into English.
- Translate \( [3, \infty) \) into English.
- Translate "the set of all real numbers except 7" into math.
- Translate "the set of all real numbers strictly between 1 and 4" into math.
- Translate \( \{ x \: | \: x > 0 \} \) into English.
- Translate \( \{ x \: | \: x \neq 0 \} \) into English.
- Draw a number line representing the set \( (-\infty, 5] \).
- What set does the number line below represent?
- Answer
-
- "The set of real numbers strictly greater than \(-8\) but less than or equal to \(2\), together with the set of real numbers greater than or equal to \(3\) but strictly less than \(5\)."
- "The set of all real numbers greater than or equal to \(3\)."
- \( \{ x \: | \: x \neq 7 \} \), or \( x \in \mathbb{R}, x \neq 7 \), etc.
- \( \{ x \: | \: 1 < x < 4 \}\), or in interval notation, \( (1, 4) \).
- "The set of positive real numbers."
- "All real numbers \(x\) except for \(x = 0\)."
- \( \{ x \: | \: -4 < x < 3 \} \) or in interval notation, \( (-4, 3) \).
What is the domain of the expression?
- \( \dfrac{1}{3-x} \)
- \( \dfrac{2}{(x-1)(x-2)} \)
- \( \dfrac{3}{x^2 - 2x - 3} \)
- \( \dfrac{1}{x^2 + 5x + 6} \)
- \( 1+ x+ x^2 + x^3 \)
- \( \sqrt{x+4} \)
- \( \dfrac{1}{\sqrt{x} }\)
- \( (x-7)(x+2) \)
- \( \sqrt[3]{x+8} \)
- Answer
-
- \( \{x \: | \: x \neq 3 \} \)
- \( \{ x \: | \: x \neq 1, x \neq 2 \} \)
- \( \{ x \: | \: x \neq -1, 3 \} \)
- \( \{ x \: | \: x \neq -2, -3 \} \)
- No legality issues, domain is all real numbers.
- \( \{ x \: | \: x \geq -4 \} \)
- \( \{ x \: | \: x > 0 \} \)
- No legality issues, domain is all real numbers.
- Careful! No legality issues because we are allowed to put negative numbers into odd roots like cube roots! All real numbers.
Simplify.
1. \( \dfrac{ 8y}{y^2 - 16} - \dfrac{4}{y-4} \)
2. \( \dfrac{3}{x-3} + \dfrac{2}{x-2} \)
3. \( \dfrac{x^2 - 6x}{x^2 - 1} - \dfrac{3x+2}{1-x^2} \)
4. \( \dfrac{5}{2x - 10} - \dfrac{7}{4x-12} \)
- Answer
-
1. \( \dfrac{4}{y+4} \)
2. \( \dfrac{5x-12}{(x-3)(x-2)} \)
3. \( \dfrac{x-2}{x+1} \)
4. \( \dfrac{3x+5}{4(x-5)(x-3)} \)
Simplify.
1. \( \dfrac{ 3x}{x^2 -5x + 6} \cdot \dfrac{ 9x -27}{x^3} \)
2. \( \dfrac{2x+1}{x^2 -4} \cdot \dfrac{ x^3 - 4x}{2x^2 -5x - 3} \)
3. \( \dfrac{t^2 + 2t - 3}{t^2 - 2t - 3} \cdot \dfrac{3-t}{3+t} \)
4. \( \dfrac{\frac{x^3}{x+1}}{\frac{x}{x^2 + 2x + 1}} \)
- Answer
-
1. \( \dfrac{27}{x^2(x-2)} \)
2. \( \dfrac{x}{x-3} \)
3. \( -\dfrac{t-1}{t+1} \) or \( \dfrac{1-t}{t+1} \)
4. \( x^2(x+1)\)
In the compound fractions, get common denominators and add or subtract as needed in numerator and denominator until you have a single fraction of rational expressions. Then compute and simplify if necessary.
1. \( \dfrac{1 + \frac{1}{x}}{\frac{1}{x} - 1} \)
2. \( \dfrac{ 1+\frac{1}{x-2}}{ 1 - \frac{1}{x+3}} \)
- Answer
-
1. \( \dfrac{x+1}{1-x} \)
2. \( \dfrac{(x-1)(x+3)}{(x-2)(x+2)} \)