Question

Suppose that the function $h$ is defined on the interval $(-2, 2]$ as follows. \begin{equation} h(x) = \begin{cases} -1 & \text{if } -2 < x \le -1 \\ 0 & \text{if } -1 < x \le 0 \\ 1 & \text{if } 0 < x \le 1 \\ 2 & \text{if } 1 < x \le 2 \end{cases} \end{equation} Find $h(-1)$, $h(0.5)$, and $h(1)$.

          Suppose that the function $h$ is defined on the interval $(-2, 2]$ as follows.
\begin{equation*}
h(x) = \begin{cases} -1 & \text{if } -2 < x \le -1 \\ 0 & \text{if } -1 < x \le 0 \\ 1 & \text{if } 0 < x \le 1 \\ 2 & \text{if } 1 < x \le 2 \end{cases}
\end{equation*}
Find $h(-1)$, $h(0.5)$, and $h(1)$.

Suppose that the function h is defined on the interval (-2, 2] as follows.

h(x) = -1 if -2 < x ≤ -1
0 if -1 < x ≤ 0
1 if 0 < x ≤ 1
2 if 1 < x ≤ 2

Find h(-1), h(0.5), and h(1).

Added by Julie G.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Vishal Parmar

05/09/2022

Step 1

5)), we need to find the value of h(0.5) first. Show more…

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Suppose that the function h is defined on the interval [-2,2] as follows: h(x) = 1 if 2 < x ≤ 1 0 if 1 < x ≤ 0 h(x) = 1 if 0 < x ≤ 1 if 1 ≤ x ≤ 2 Find h^(-1)(h(0.5)) and h(1). h^(-1)(h(0.5)) = h^(-1)(1) = h(1) = 1