Question

Suppose that the function f is defined, for all real numbers, as follows.

[f(x) = begin{cases} -2, & text{if } x < -2 x+1, & text{if } -2 leq x leq 1 -x+2, & text{if } x > 1 end{cases}]
Find (f(-3)), (f(-2)), and (f(0)).
a) ((-2, -1, 1))
b) ((-2, 0, 2))
c) (-1, 1, 2))
d) ((-2, 1, 0))

Answer 1

The function values for f(-3), f(-2), and f(0) by applying the rule that corresponds to each x value result in (-2, -1, and 1), respectively, with f(-3) falling under the first rule, and f(-2) and f(0) falling under the second rule.

Step-by-step explanation:

The function f(x) is defined by different rules depending on the value of x. To find the values of f(-3), f(-2), and f(0), we must apply the appropriate rule for each value.

• For f(-3), since -3 is less than -2, we use the rule f(x) = -2. Therefore, f(-3) = -2.
• For f(-2), the value of x falls exactly on the boundary of the second rule, so we use f(x) = x + 1. Thus, f(-2) = -2 + 1 = -1.
• For f(0), since 0 is between -2 and 1,inclusive, we still use the rule f(x) = x + 1. So, f(0) = 0 + 1 = 1.

Therefore, the values of the function at f(-3), f(-2), and f(0) are (-2, -1, 1), respectively, matching answer option a.