Suppose that the function $h$ is defined, for all real numbers, as follows. \[ h(x)=\left\{\begin{array}{ll} \frac{1}{4} x+2 & \text { if } x<-1 \\ -(x+1)^{2}+3 & \text { if }-1 \leq x \leq 2 \\ -2 & \text { if } x>2 \end{array}\right. \] Find $h(-4), h(-1)$, and $h(1)$.
Solution
Step 1 ：The function h(x) is defined piecewise, meaning it behaves differently depending on the value of x. To find h(-4), h(-1), and h(1), we need to determine which part of the function definition applies for each of these values and then substitute the value into the corresponding expression.
Step 2 ：For h(-4), since -4 < -1, we use the first part of the function definition: h(x) = \(\frac{1}{4}x + 2\).
Step 3 ：For h(-1), since -1 is between -1 and 2 (inclusive), we use the second part of the function definition: h(x) = \(-(x+1)^2 + 3\).
Step 4 ：For h(1), since 1 is also between -1 and 2 (inclusive), we again use the second part of the function definition: h(x) = \(-(x+1)^2 + 3\).
Step 5 ：The values of the function h at -4, -1, and 1 are 1.0, 3, and -1 respectively. So, h(-4) = 1.0, h(-1) = 3, and h(1) = -1.
Step 6 ：Final Answer: \(\boxed{1.0, 3, -1}\)
