3.11: Piecewise-Definition Functions

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Victoria Dominguez, Cristian Martinez, and Sanaa Saykali
Citrus College via ASCCC Open Educational Resources Initiative

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Definition: Piecewise-Defined Functions

Piecewise-Defined Functions are functions that are defined using different equations for different parts of the domain.

Example 4.11.1

Evaluate the following piecewise-defined function for the given values of $x$, and graph the function:

$f(x) = \left\{\begin{array}{cc}−2x + 1 & −1 \leq x < 0 \\ x^2 + 2 &0 \leq x \leq 2\end{array} \right.$

Solution

To graph this function, make a table of solutions:

Table of Solutions for $f(x) = −2x + 1 $ Domain $−1 \leq x < 0$
$x$	$f(x)$
-1	3
0	1 (open circle here, 0 not in the domain)

Table of Solutions for $f(x) = x^2 + 2$ Domain $0 \leq x \leq 2$
$x$	$f(x)$
0	2
1	3
2	6

$clipboard_e94fd5197718a7373772af1280306cf06.png$ Figure 4.11.1

Example 4.11.2

Evaluate the following piecewise-defined function for the given values of $x$, and graph the function:

$f(x) = \left\{\begin{array}{cc} −x + 1 &x \leq −1 \\ 2 & −1 < x \leq 1 \\ −x + 3 &x > 1 \end{array}\right.$

Solution

To graph this function, once again make a table of solutions:

Table of Solutions for $f(x) = −x + 1$ Domain $x \leq −1$
$x$	$f(x)$
-3	4
-2	3
-1	2 (closed circle here, -1 is in the domain)

Table of Solutions for $f(x) = 2$ Domain $−1 < x \leq 1$
$x$	$f(x)$
-1	2 (open circle filled in by the previous function, -1 not in the domain)
0	2
1	2 (closed circle here, 1 is in the domain)

Table of Solutions for $f(x) = −x + 3$ Domain $x > 1$
$x$	$f(x)$
1	2 (open circle filled in by the previous function, 1 not in the domain)
2	1
3	0

$clipboard_e795cd2fce50083772c8741bdcad72855.png$ Figure 4.11.2

Exercise 4.11.1

Evaluate the following piecewise-defined functions for the given values of x, and graph the functions:.

$f(x)=\left\{\begin{array}{cc}
x & x<0\\
2 x+1 &x\geq 0
\end{array}\right.$
$g(x) = \left\{\begin{array}{cc} 4 − x& x < 2\\ 2x − 2 &x \geq 2 \end{array} \right.$
$h(x) = \left\{\begin{array}{cc} −x − 1 & x < −1 \\ 0& −1 \leq x \leq 1 \\ x + 1 & x > 1 \end{array} \right.$
$g(x) = \left\{\begin{array}{cc} 6 & −8 \leq x < −4 \\ 3 &−4 \leq x \leq 5 \end{array}\right.$
$f(x) = \left\{\begin{array}{cc} −x + 1 & −1 \leq x < 1 \\ \sqrt{x − 1 } &1 \leq x \leq 5\end{array}\right.$

Table of Solutions for \(f(x) = −2x + 1 \) Domain \(−1 \leq x < 0\)
\(x\)	\(f(x)\)
-1	3
0	1 (open circle here, 0 not in the domain)

Table of Solutions for \(f(x) = −x + 3\) Domain \(x > 1\)
\(x\)	\(f(x)\)
1	2 (open circle filled in by the previous function, 1 not in the domain)
2	1
3	0

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Definition: Piecewise-Defined Functions

Example 4.11.1

Example 4.11.2

Exercise 4.11.1

Table of Solutions for \(f(x) = x^2 + 2\) Domain \(0 \leq x \leq 2\)
\(x\)	\(f(x)\)
0	2
1	3
2	6

Table of Solutions for \(f(x) = −x + 1\) Domain \(x \leq −1\)
\(x\)	\(f(x)\)
-3	4
-2	3
-1	2 (closed circle here, -1 is in the domain)

Table of Solutions for \(f(x) = 2\) Domain \(−1 < x \leq 1\)
\(x\)	\(f(x)\)
-1	2 (open circle filled in by the previous function, -1 not in the domain)
0	2
1	2 (closed circle here, 1 is in the domain)