4.11: Piecewise-Definition Functions
Piecewise-Defined Functions are functions that are defined using different equations for different parts of the domain.
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:
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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 |
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 |
Evaluate the following piecewise-defined functions for the given values of x, and graph the functions:.
-
\(f(x)=\left\{\begin{array}{cc}
x & x<0\\
2 x+1 &x\geq 0
\end{array}\right.\) - \(g(x) = \left\{\begin{array}{cc} 4 − x& x < 2\\ 2x − 2 &x \geq 2 \end{array} \right.\)
- \(h(x) = \left\{\begin{array}{cc} −x − 1 & x < −1 \\ 0& −1 \leq x \leq 1 \\ x + 1 & x > 1 \end{array} \right.\)
- \(g(x) = \left\{\begin{array}{cc} 6 & −8 \leq x < −4 \\ 3 &−4 \leq x \leq 5 \end{array}\right.\)
- \(f(x) = \left\{\begin{array}{cc} −x + 1 & −1 \leq x < 1 \\ \sqrt{x − 1 } &1 \leq x \leq 5\end{array}\right.\)