Precalculus Examples

$f (x) = {\begin{cases} - \frac{1}{2} x - 1 & x < - 2 \\ x + 2 & - 2 \leq x \leq 3 \\ - 2 x + 7 & x > 3 \end{cases}$ Find $f (- 2)$

Step 1

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

$f (x) = {\begin{cases} - \frac{1}{2} x - 1 & x < - 2 \\ x + 2 & - 2 \leq x \leq 3 \\ - 2 x + 7 & x > 3 \end{cases}$

Step 2

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

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

Step 3

Evaluate the function at $x = - 2$ .

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

Step 4

Add $- 2$ and $2$ .

$f (- 2) = 0$