Precalculus Examples

$f (x) = {\begin{cases} {(x + 6)}^{2} + 3 & x < - 5 \\ 0 & x = - 5 \\ - 2 x - 8 & x > - 5 \end{cases}$ Find $f (- 7)$

Step 1

Identify the piece that describes the function at $x = - 7$ . In this case, $x = - 7$ falls within the interval $x < - 5$ , therefore use ${(x + 6)}^{2} + 3$ to evaluate $f (- 7)$ .

$f (x) = {\begin{cases} {(x + 6)}^{2} + 3 & x < - 5 \\ 0 & x = - 5 \\ - 2 x - 8 & x > - 5 \end{cases}$

Step 2

The function is equal to ${(x + 6)}^{2} + 3$ at $x = - 7$ .

$f (x) = {(x + 6)}^{2} + 3 (]_{x = - 7}^{})$

Step 3

Evaluate the function at $x = - 7$ .

$f (- 7) = {((- 7) + 6)}^{2} + 3$

Step 4

Simplify each term.

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Step 4.1

Add $- 7$ and $6$ .

$f (- 7) = {(- 1)}^{2} + 3$

Step 4.2

Raise $- 1$ to the power of $2$ .

$f (- 7) = 1 + 3$

Step 5

Add $1$ and $3$ .

$f (- 7) = 4$