Misael Matthews

2022-06-25

Let $f\left(x\right)=7+|2x-1|$ . How do you find all x for which $f\left(x\right)<16$ ?

Xzavier Shelton

Expert

Step 1
Given: $f\left(x\right)=7+|2x-1|$ and $f\left(x\right)<16$
We can write the inequality:
$7+|2x-1|<16$
Subtract 7 from both sides:
$|2x-1|<9$
Because of the piecewise definition of the absolute value function,
$|A|=\left\{\begin{array}{l}A;A\ge 0\\ -A;A<0\end{array}$
we can separate the inequality into two inequalities:
$-\left(2x-1\right)<9$ and $2x-1<9$
Multiply both sides of the first inequality by -1:
$2x-1>-9$ and $2x-1<9$
Add 1 to both sides of both inequalities:
$2x>-8$ and $2x<10$
Divide both sides of both inequalities by 2:
$x>-4$ and $x<5$
This can be written as:
$-4
To check, I will verify that the end points equal 16:
$7+\mid 2\left(-4\right)-1\right)|=7+|-9\mid =16$
$7+|2\left(5\right)-1|=7+|9|=16$

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