2.2: Solving and Graphing Inequalities, and Writing Answers in Interval Notation
- Page ID
- 32429
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)To solve and graph inequalities:
- Solve the inequality using the Properties of Inequalities from the previous section.
- Graph the solution set on a number line.
- Write the solution set in interval notation.
Solve the inequality, graph the solution set on a number line and show the solution set in interval notation:
- \(−1 ≤ 2x − 5 < 7\)
- \(x^2 + 7x + 10 < 0\)
- \(−6 < x − 2 < 4\)
Solution
- \(\begin{array} &&−1 ≤ 2x − 5 < 7 &\text{Example problem} \\ &−1 + 5 ≤ 2x − 5 + 5 < 7 + 5 &\text{The goal is to isolate the variable \(x\), so start by adding \(5\) to all three regions in the inequality.} \\ &4 ≤ 2x < 12 &\text{Simplify.} \\ &\dfrac{4}{2} ≤ 2x^2 < \dfrac{4}{2} &\text{Divide all by \(2\) to isolate the variable \(x\).} \\ &2 ≤ x < 6 &\text{Final answer written in inequality/solution set form.} \\ &[2, 6) &\text{Final answer written in interval notation (see section on Interval Notation for more details)} \end{array} \)
- \(\begin{array} &&x^2 + 7x + 10 < 0 &\text{Example problem} \\ &(x + 5)(x + 2) < 0 &\text{Factor the polynomial.} \\ &(x + 5)(x + 2) < 0 &\text{The product must be less than \(0\), which means that if \((x + 5) > 0\), then \((x + 2) < 0\). Similarly, if \((x + 5) < 0\), then \((x + 2) > 0\).} \\ &(x + 5) > 0 ∪ (x + 2) < 0 &\text{Find the intersection of each of these inequalities.} \\ &x > −5 ∪ x < −2 &\text{Find the intersection of each of these inequalities.} \end{array}\)
\(\begin{array} &&\;\;\;−5 < x < −2 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&\text{Final answer written in inequality/solution set form.} \\ &\;\;\;(−5, −2) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&\text{Final answer written in interval notation (see section on Interval Notation for more details).} \end{array}\)
- \(\begin{array}&&−6 < x − 2 ≤ 4 &\text{Example problem} \\ &−6 + 2 < x − 2 + 2 ≤ 4 + 2 &\text{The goal is to isolate the variable \(x\), so start by adding \(2\) to all three regions in the inequality.} \\ &−4 < x ≤ 6 &\text{Final answer written in inequality/solution set form.} \\ &(−4, 6] &\text{Final answer written in interval notation (see section on Interval Notation for more details).} \end{array}\)
Solve the inequalities, graph the solution sets on a number line and show the solution sets in interval notation:
- \(0 ≤ x + 1 ≤ 4\)
- \(0 < 2(x − 1) ≤ 4\)
- \(6 < 2(x − 1) < 12\)
- \(x^2 − 6x − 16 < 0\)
- \(2x^2 − x − 15 > 0\)