Given the function

A.
To calculate f(-1)
Let us look at where f(-1) is defirned under the function f(x).
When f(x) = 5x + 9, x < 0. We can see that -1 is less than zero. Therefore, f(-1) is defined under f(x) = 5x + 9
Thus:
f(x) = 5x + 9
Substitute x = -1 into f(x)
f(-1) = 5(-1) + 9
f(-1) = -5 + 9
f(-1) = 4
B.
To calculate f(0)
f(0) is defined when f(x) = 5x + 18 (because we are told that x is greater than or equal to xero here).

C. To calculate f(2):
f(2) is also defined when f(x) = 5x + 18 ( because x is greater than or equal to zero here and 2 is greater than 0)
Thus:

Hence,
f(-1) = 4
f(0) = 18
f(2) = 28