Question

Solve absolute value inequality. |2(x-1)+4|\leq8

Piecewise-Defined Functions
ANSWERED
asked 2021-02-25
Solve absolute value inequality.
\(\displaystyle{\left|{2}{\left({x}-{1}\right)}+{4}\right|}\leq{8}\)

Answers (1)

2021-02-26

Step 1
We have to solve the absolute value inequality:
\(\displaystyle{\left|{2}{\left({x}-{1}\right)}+{4}\right|}\leq{8}\)
We know for any modulus function,
\(|f(x)|=\begin{cases}f(x) & if\ f(x)\geq 0\\-f(x) & if\ f(x)<0\end{cases}\)
We also know that if \(\displaystyle{\left|{f{{\left({x}\right)}}}\right|}\leq\) a then it must satisfy that
\(\displaystyle-{a}\leq{f{{\left({x}\right)}}}\leq{a}\)
Applying above condition for given inequality,
\(\displaystyle-{a}\leq{f{{\left({x}\right)}}}\leq{a}\)
\(\displaystyle-{8}\leq{2}{\left({x}-{1}\right)}+{4}\leq{8}\)
Step 2
Simplifying further,
\(-8-4 \leq 2(x-1)+4-4 \leq 8-4\) (substracted 4)
\(\displaystyle-{12}\leq{2}{\left({x}-{1}\right)}\leq{4}\)
\(\displaystyle{\frac{{-{12}}}{{{2}}}}\leq{\frac{{{2}{\left({x}-{1}\right)}}}{{{2}}}}\leq{\frac{{{4}}}{{{2}}}}\) (divided all sides by 2)
\(\displaystyle-{6}\leq{\left({x}-{1}\right)}\leq{2}\)
\(\displaystyle-{6}+{1}\leq{x}-{1}+{1}\leq{2}+{1}\)
\(\displaystyle-{5}\leq{x}\leq{3}\)
In interval notation we can write it as \(\displaystyle{x}\in{\left[-{5},{3}\right]}\).
Hence, solution of the absolute value inequality is \(\displaystyle{x}\in{\left[-{5},{3}\right]}\).

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