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Wize University Calculus 1 Textbook > Pre-Calculus (Review)

Absolute Value Functions

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The Absolute Value Function

The absolute value function is defined as follows:

∣x∣={xif x≥0−xif x<0\boxed{|x| = \begin{cases} x\hspace{1.06cm} \text{if } x\geq0\\ -x\hspace{0.8cm}\text{if } x<0\\ \end{cases}}



More generally, if we would like to consider the composition of the absolute value function with another function:

∣f(x)∣={f(x)for x such that f(x)≥0−f(x)for x such that f(x)<0\boxed{|f(x)| = \begin{cases} f(x)\hspace{1.06cm} \text{for x such that } f(x)\geq0\\ -f(x)\hspace{0.8cm} \text{for x such that } f(x)<0\\ \end{cases} }

In these cases, we must determine the intervals on which our function is positive and negative respectively.




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Example: Absolute Value Functions

Write out a piece-wise definition for the function

f(x)=∣3x2−1∣f(x) = |3x^2-1|

First we should consider when 3x2−13x^2-1 is positive.
3x2−1≥03x2≥1x2≥13x≤−13 or x≥13\begin{aligned} & 3x^2-1 \geq 0 \\ & 3x^2 \geq 1 \\ & x^2 \geq \frac{1}{3} \\ & x \leq \frac{-1 }{\sqrt{3}} \text{ or } x \geq \frac{1 }{\sqrt{3}} \end{aligned}
Next, we can consider when 3x2−13x^2-1 is negative.
3x2−1<03x2<1x2<13−13<x<13\begin{aligned} & 3x^2-1 < 0 \\ & 3x^2 < 1 \\ & x^2 < \frac{1}{3} \\ & \frac{-1 }{\sqrt{3}} < x < \frac{1 }{\sqrt{3}} \end{aligned}
Finally, we can determine the piecewise definition of f(x)f(x).
f(x)=∣3x2−1∣={3x2−1 if x≤−13 or x≥13−3x2+1 if −13<x<13f(x) = |3x^2-1| = \begin{cases} 3x^2-1 \hspace{1.0 cm}\text{ if } x \leq \frac{-1 }{\sqrt{3}} \text{ or } x \geq \frac{1 }{\sqrt{3}} \\ -3x^2 + 1 \hspace{0.74 cm}\text{ if } \frac{-1 }{\sqrt{3}} < x < \frac{1 }{\sqrt{3}} \end{cases}
Extra Practice
Absolute Value Functions: Function Properties
If f(x)=∣x−2∣, g(x)=3−x+1f\left(x\right)=\left|x-2\right|,\ g\left(x\right)=3-\sqrt{x+1}, which of the following is true about h(x)=g(f(x))h\left(x\right)=g\left(f\left(x\right)\right)?
Solve the following inequality ∣2x−3∣+∣x−4∣≥3\left|2x-3\right|+\left|x-4\right|\ge3
Absolute Value Functions: Inequalities with Absolute Values
Solve the following inequality: ∣x+1∣≥x+42\displaystyle\left|x+1\right|\ge\frac{x+4}{2}
Absolute Value Functions: Inequality with Absolute Values
Find all xx that satisfy the inequality ∣x2−3x+1∣<1\left|x^2−3x+1\right|<1
Absolute Value Functions: Function Properties
If f(x)=∣x−2∣, g(x)=3−x+1f\left(x\right)=\left|x-2\right|,\ g\left(x\right)=3-\sqrt{x+1}, which of the following is true about h(x)=g(f(x))h\left(x\right)=g\left(f\left(x\right)\right)?
Absolute Value Functions
Solve the inequality for xx, ∣3x−4∣+∣x−2∣≥5|3x-4|+|x-2| \geq 5.
Solve the following inequality ∣2x−3∣+∣x−4∣≥3\left|2x-3\right|+\left|x-4\right|\ge3
Absolute Value Functions: Inequalities with Absolute Values
Solve the following inequality: ∣x+1∣≥x+42\displaystyle\left|x+1\right|\ge\frac{x+4}{2}
Absolute Value Functions: Inequality with Absolute Values
Find all xx that satisfy the inequality ∣x2−3x+1∣<1\left|x^2−3x+1\right|<1