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Writing Piecewise Functions Solutions

Piecewise functions are functions that are constructed from two or more expressions over defined intervals of the domain.  
These functions allow you to model real-life situations more realistically because real-life scenarios can rarely be represented by just one function. 
The inequality symbols for the domain intervals determine whether the function will have open points, closed points, or arrows (for continuous functions).

$f (x) = \{\begin{cases} - 2 & if - 5 < x \leq - 2 \\ x^{2} - 4 & if - 2 < x < 2 \\ - 1.5 x + 6 & if x \geq 2.5 \end{cases}$

Note

Remember to use the VLT to check if a graph is a function. If an open and closed point are aligned vertically, then the piecewise graph is a function.

To determine if an ordered pair is a solution to a piecewise function: 
- First check what part of the domain inequality interval the x-value is in. 
- Then substitute the x-value in the corresponding expression to find f(x) .

Example 1

Write the expressions for the piecewise function given the graph and the intervals for the domain.

$f (x) = \{\begin{cases} 2 x + 5 & if x < - 1 \\ |x| & if - 1 \leq x < 3 \\ 0.5 x - 2.5 & if x \geq 3 \end{cases}$

Determine f (x) when $x = \{- 0.5, 2.99, 3\}$
$- 1 \leq - 0.5 < 3$

$f (- 0.5) = |- 0.5| f (- 0.5) = 0.5 - 1 \leq 2.99 < 3 f (2.99) = |2.99| f (2.99) = 2.99 3 \geq 3 f (3) = 0.5 (3) - 2.5 f (3) = - 1$

Example 2

Write the intervals for the domain given the graph and the expressions for the piecewise function.

$p (x) = \{\begin{cases} - |x + 3| + 2 \\ - 1 \\ - {(x - 3)}^{2} + 3 \end{cases} \begin{array}{l} if - 5 < x \leq - 1 \\ if - 1 < x \leq 1 \\ if x \geq 2 \end{array}$

Note

Pay careful attention to the open and closed points and the VLT to ensure that the domain inequalities result in a function.

Example 3

Write the piecewise function using the given graph.

$y = \{\begin{cases} x + 4 & if - 4 \leq x < 0 \\ \frac{5}{3} x - 2 & if 0 \leq x < 3 \\ 3 & if x > 3 \end{cases}$