Precalculus Examples

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

Step 1

Identify the piece that describes the function at $x = - 2$ . In this case, $x = - 2$ falls within the interval $x \leq - 2$ , therefore use $3 x + 6$ to evaluate $f (- 2)$ .

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

Step 2

The function is equal to $3 x + 6$ at $x = - 2$ .

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

Step 3

Evaluate the function at $x = - 2$ .

$f (- 2) = 3 (- 2) + 6$

Step 4

Multiply $3$ by $- 2$ .

$f (- 2) = - 6 + 6$

Step 5

Add $- 6$ and $6$ .

$f (- 2) = 0$