Precalculus Examples

$f (x) = {\begin{cases} {(x + 3)}^{2} + 5 & x < - 2 \\ 6 & - 2 \leq x \leq 2 \\ 2 x + 2 & x > 2 \end{cases}$ Find $f (- 1)$

Step 1

Identify the piece that describes the function at $x = - 1$ . In this case, $x = - 1$ falls within the interval $- 2 \leq x \leq 2$ , therefore use $6$ to evaluate $f (- 1)$ .

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

Step 2

The function is equal to $6$ at $x = - 1$ .

$f (x) = 6 (]_{x = - 1}^{})$

Step 3

Evaluate the function at $x = - 1$ .

$f (- 1) = 6$