The function f(x)is defined as

f(x)={-2x+3: if -2 less than or equal to < 1

f(x)={5: if 1 less than or equal to x less than or equal to 2

f(x)={x^2+2: if x > 2

a. find f(0), f(2) and f(5)
b. determine the domain of f(x)

Please show work

GRAPH IT !!!

when x is </= 1 it is a straight line sloping down from (0,3) to (1,1)

so
f(0) = 3

f is constant 5 from 1 to 2 but leaps up to 6 at x = 2 and then is a parabola headed up
t exactly x = 2, f(2) = 5 but then jumps up

at x = 5, f(5) = 5^2+2 = 27

f is defined for all real x

To find the values of f(0), f(2), and f(5), we need to apply the given function rules.

a. Finding f(0):
The given function rules state:
For -2 ≤ x < 1, f(x) = -2x + 3
For 1 ≤ x ≤ 2, f(x) = 5
For x > 2, f(x) = x^2 + 2

For f(0), we need to use the rule where -2 ≤ x < 1. So substituting x = 0 into the first rule:
f(0) = -2(0) + 3
f(0) = 0 + 3
f(0) = 3

Therefore, f(0) = 3.

b. Finding f(2):
Again, we use the given function rules:
For -2 ≤ x < 1, f(x) = -2x + 3
For 1 ≤ x ≤ 2, f(x) = 5
For x > 2, f(x) = x^2 + 2

For f(2), we use the rule where 1 ≤ x ≤ 2. So substituting x = 2 into the second rule:
f(2) = 5

Therefore, f(2) = 5.

c. Finding f(5):
Finally, using the last rule for x > 2: f(x) = x^2 + 2, we can substitute x = 5:
f(5) = 5^2 + 2
f(5) = 25 + 2
f(5) = 27

Therefore, f(5) = 27.

Now let's determine the domain of f(x).

The domain of a function is the set of all possible x-values for which the function is defined. In this case, we can see from the given function rules that f(x) is defined for three different ranges of x:

1. For -2 ≤ x < 1, the function is defined as f(x) = -2x + 3.
2. For 1 ≤ x ≤ 2, the function is defined as f(x) = 5.
3. For x > 2, the function is defined as f(x) = x^2 + 2.

So, the domain of f(x) is the union of these three intervals:
Domain = (-2, 1) ∪ [1, 2] ∪ (2, ∞)