Question

evaluating piecewise functions algebraically.
$f(x) = \

$$\begin{cases} -\\dfrac{3}{2}x - 1, & x < -2 \\\\ x + 1, & -2 \\leq x < 1 \\\\ 3, & x > 1 \\end{cases}$$

$
evaluate: $2f(-4)$
evaluate: $3f(0)$
evaluate: $f(3)$
evaluate: $f(1)$

Explanation:

Step1: Evaluate $f(-4)$

Since $-4 < -2$, use $f(x)=-\frac{3}{2}x - 1$:
$f(-4)=-\frac{3}{2}(-4) - 1 = 6 - 1 = 5$

Step2: Calculate $2f(-4)$

$2f(-4)=2\times5=10$

Step3: Evaluate $f(0)$

Since $-2 \leq 0 < 1$, use $f(x)=x + 1$:
$f(0)=0 + 1 = 1$

Step4: Calculate $3f(0)$

$3f(0)=3\times1=3$

Step5: Evaluate $f(3)$

Since $3 > 1$, use $f(x)=3$:
$f(3)=3$

Step6: Evaluate $f(1)$

The function has no definition for $x=1$ (the pieces cover $x < -2$, $-2 \leq x < 1$, $x > 1$; $x=1$ is not included in any interval).

Answer:

$2f(-4)=10$
$3f(0)=3$
$f(3)=3$
$f(1)$ is undefined (not in any domain interval)