Question

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$g(x) = \

$$\begin{cases} x^2 - 5 &, \\ x \\in (-\\infty, -7) \\\\ 9x - 17 &, \\ x \\in -7, 2 \\\\ (x + 1)(x - 5) &, \\ x \\in (2, \\infty) \\end{cases}$$

$
$g(7) = \boxed{}$
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worked example: evaluating piecewise functions

Explanation:

Step1: Determine the domain for \( x = 7 \)

We need to find which piece of the piecewise function \( g(x) \) applies when \( x = 7 \). The domains are \( (-\infty, -7) \), \( [-7, 2] \), and \( (2, \infty) \). Since \( 7 > 2 \), \( x = 7 \) is in the domain \( (2, \infty) \).

Step2: Use the corresponding function

For \( x \in (2, \infty) \), the function is \( g(x) = (x + 1)(x - 5) \). Substitute \( x = 7 \) into this function:
\[

$$\begin{align*} g(7) &= (7 + 1)(7 - 5)\\ &= (8)(2)\\ &= 16 \end{align*}$$

\]

Answer:

\( 16 \)