Question

suppose that the function ( h ) is defined, for all real numbers, as follows. h(x)=\begin{cases} \frac{1}{4}x - 1 & \text{if } x < -2 \\ -(x - 1)^2 + 2 & \text{if } -2 leq x leq 2 \\ 2 & \text{if } x > 2 end{cases} find ( h(-5) ), ( h(-1) ), and ( h(2) ). ( h(-5) = ) ( h(-1) = ) ( h(2) = )

Explanation:

Step1: Evaluate $h(-5)$ (use first rule)

Since $-5 < -2$, use $h(x)=\frac{1}{4}x - 1$.
$h(-5)=\frac{1}{4}(-5) - 1 = -\frac{5}{4} - 1 = -\frac{9}{4}$

Step2: Evaluate $h(-1)$ (use second rule)

Since $-2 \leq -1 \leq 2$, use $h(x)=-(x-1)^2 + 2$.
$h(-1)=-(-1-1)^2 + 2 = -(-2)^2 + 2 = -4 + 2 = -2$

Step3: Evaluate $h(2)$ (use second rule)

Since $-2 \leq 2 \leq 2$, use $h(x)=-(x-1)^2 + 2$.
$h(2)=-(2-1)^2 + 2 = -(1)^2 + 2 = -1 + 2 = 1$

Answer:

$h(-5)=-\frac{9}{4}$
$h(-1)=-2$
$h(2)=1$