Question

Suppose that the function h is defined, for all real numbers, as follows: h(x) = { 5x + 2 if x < -2, (x - 1)^2 - 2 if -2 ≤ x ≤ 5, 1 if x > 2. Find h(-2), h(-1), and h(3). h(-2) = 0, h(-1) = 8^(-1) = 0, h(3) = ?

Answer 1

To find h(-2), h(-1), and h(3), use the function definition for the appropriate interval. The results are h(-2) = 7, h(-1) = 2, and h(3) = 2.

Step-by-step explanation:

The goal is to find the values of the function h(x) at x = -2, x = -1, and x = 3. By looking at the definition of the function, we use the appropriate expressions for h(x) based on the intervals provided.

Firstly, h(-2) falls within the interval where it's defined as (x - 1)^2 - 2. So we compute h(-2) as follows:
(-2 - 1)^2 - 2 = (-3)^2 - 2 = 9 - 2 = 7. Therefore, h(-2) = 7.

Next, we calculate h(-1), which is also within the interval defined by (x - 1)^2 - 2. So h(-1) is:
(-1 - 1)^2 - 2 = (-2)^2 - 2 = 4 - 2 = 2. Thus, h(-1) = 2.

Finally, for h(3), it also falls in the same interval and thus is computed in the same manner:
(3 - 1)^2 - 2 = 2^2 - 2 = 4 - 2 = 2. Hence, h(3) = 2.