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Chapter 0: Problem 27

Express the function in piecewise form without using absolute values. (a) \(f(x)=|x|+3 x+1\) (b) \(g(x)=|x|+|x-1|\)

Short Answer

Expert verified
(a) \(f(x) = \begin{cases} 4x+1, & x \geq 0 \\ 2x+1, & x < 0 \end{cases}\) (b) \(g(x) = \begin{cases} 2x-1, & x \geq 1 \\ 1, & 0 \leq x < 1 \\ 1-2x, & x < 0 \end{cases}\)

Step by step solution

01

Understanding Absolute Value

For any real number, the absolute value \(|a|\) is defined as \(a\) if \(a \geq 0\) and \(-a\) if \(a < 0\). We'll use this definition to remove the absolute value signs from the functions.
02

Rewrite \(f(x)\) for \(x \geq 0\)

For the function \(f(x) = |x| + 3x + 1\), when \(x \geq 0\), the absolute value of \(x\) is \(x\) itself. Therefore, \(f(x) = x + 3x + 1 = 4x + 1\).
03

Rewrite \(f(x)\) for \(x < 0\)

When \(x < 0\), the absolute value of \(x\) is \(-x\). Thus, the function becomes \(f(x) = -x + 3x + 1 = 2x + 1\).
04

Piecewise Form for \(f(x)\)

Putting the two scenarios together, the piecewise function for \(f(x)\) is: \[f(x) = \begin{cases} 4x + 1, & \text{if } x \geq 0 \ 2x + 1, & \text{if } x < 0 \end{cases}\]
05

Rewrite \(g(x)\) for \(x \geq 1\)

For \(g(x) = |x| + |x-1|\), when \(x \geq 1\), we have \(|x| = x\) and \(|x-1| = x-1\). Thus, \(g(x) = x + (x-1) = 2x - 1\).
06

Rewrite \(g(x)\) for \(0 \leq x < 1\)

When \(0 \leq x < 1\), the absolute value function \(|x| = x\) and \(|x-1| = 1-x\). Therefore, \(g(x) = x + (1-x) = 1\).
07

Rewrite \(g(x)\) for \(x < 0\)

For \(x < 0\), \(|x| = -x\) and \(|x-1| = 1-x\), so \(g(x) = -x + (1-x) = 1 - 2x\).
08

Piecewise Form for \(g(x)\)

Combining the different cases, the piecewise function for \(g(x)\) is: \[g(x) = \begin{cases} 2x - 1, & \text{if } x \geq 1 \ 1, & \text{if } 0 \leq x < 1 \ 1 - 2x, & \text{if } x < 0 \end{cases}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Absolute Value
The absolute value of a number is like its distance from zero on a number line, and it always results in a non-negative value. This is a fundamental concept in mathematics and is very useful in evaluating expressions where the sign of a number changes depending on its value.
For example, the absolute value of \( -5 \) is 5, because \( -5 \) is 5 units away from zero.
  • If the number is positive or zero, its absolute value is the number itself: \( |a| = a \) for \( a \geq 0 \).
  • If the number is negative, its absolute value is its opposite: \( |a| = -a \) for \( a < 0 \).
This rule helps us transform equations with absolute values into piecewise functions by considering different cases, depending on whether the expression within the absolute value is positive or negative.
Function Definition
A function can be thought of as a mathematical machine that produces an output from a given input according to a specific rule. For every input, there is exactly one output. This is an important property of functions.
The definition of a function requires the specification of:
  • The **domain**, which is the set of all possible inputs.
  • The **range**, which is the set of all possible outputs.
  • A **rule** that tells us which input is connected to which output.
In the context of our exercise, the functions are piecewise defined. This means they have different definitions depending on the value of the input, and this is particularly useful when dealing with absolute values that change behavior based on the sign of the input.
Function Transformation
Function transformation refers to the operations that alter the appearance or position of the graph of a function. These transformations include shifting, stretching, and reflecting functions. Predictably transforming a function鈥檚 graph is a valuable skill in both basic and advanced mathematics.
When working with piecewise functions that include absolute values,
  • We often translate or shift these functions horizontally or vertically based on changes to the equation.
  • Reflections occur when we substitute elements, such as changing a positive slope to a negative one.
  • Stretching or compressing graphs involves multiplying terms within the function, which alters the rate at which the function increases or decreases.
In our solutions, removing the absolute value requires formulating each part of the piecewise function, effectively shifting and reflecting components depending on the domain segment. This method provides clarity and visual understanding, helping with graphing and interpretation.

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Most popular questions from this chapter

A manufacturer of cardboard drink containers wants to construct a closed rectangular container that has a square base and will hold \(\frac{1}{10}\) liter \(\left(100 \mathrm{~cm}^{3}\right)\). Estimate the dimensions of the container that will require the least amount of material for its manufacture.

Find the fallacy in the following "proof" that \(\frac{1}{8}>\frac{1}{4}\). Multiply both sides of the inequality \(3>2\) by \(\log \frac{1}{2}\) to get $$ \begin{aligned} 3 \log \frac{1}{2} &>2 \log \frac{1}{2} \\ \log \left(\frac{1}{2}\right)^{3} &>\log \left(\frac{1}{2}\right)^{2} \\ \log \frac{1}{8} &>\log \frac{1}{4} \\ \frac{1}{8} &>\frac{1}{4} \end{aligned} $$

Find $$\frac{f(x+h)-f(x)}{h} \text { and } \frac{f(w)-f(x)}{w-x}$$ Simplify as much as possible. $$ f(x)=1 / x $$

True-False Determine whether the statement is true or false. Explain your answer. The domain of \(f+g\) is the intersection of the domains of \(f\) and \(g\).

Graph the functions on the same screen of a graphing utility. [Use the change of base formula (6), where needed.] \(\log _{2} x, \ln x, \log _{5} x, \log x\)
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