PLEASE HELP ME OUT I've already filled out the first few blanks, but I really need your help on the rest of this problem
PLEASE SHOW YOUR STEPS
I WILL GIVE YOU THE BRAINLIEST IF ANSWERED PROPERLY
Thank youu!

Explanation:

Part D

For d, the very first thing you need to do is figure out which one of the steps you are going to use. You have 2 in b^2 + 2, so even if b = 0 the two still matters. It means that you use f(x) = -x + 3 because that's what you use when you have 2 or above.

The second thing you have to realize is that f(x) = -x + 3 has the meaning of what ever you see on the left in the place of x, you put on the right wherever there is an x.

In this case f(b^2 + 2) = -b^2 - 2 + 3 = 1 - b^2

I'm not sure enough to give you an answer for the domain and range, not this time of night.

9514 1404 393

  a) 9

  b) 1

  c) -1

  d) -b² +1

  domain: (-∞, ∞)

  range: (-∞, 1] ∪ [3, ∞)

Explanation:

When you evaluate a piecewise function, the first step is to determine what piece is applicable. Then, you fill in the argument value and do the arithmetic.

a) -6 < 2, so the first piece applies. f(-6) = |-6| +3 = 9

b) 2 = 2, so the second piece applies. f(2) = -2 +3 = 1

c) 4 > 2, so the second piece applies. f(4) = -4 +3 = -1

d) b² +2 ≥ 2, so the second piece applies. f(b² +2) = -(b² +2) +3 = -b² +1

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The function is defined for all values of x, so the domain is (-∞, ∞).

The minimum value of the first piece is +3. The maximum value of the second piece is f(2) = 1. So, values of y between 1 and 3 are not part of the range.

The range is (-∞, 1] ∪ [3, ∞).

Additional comment

The square of a real number can never be negative, so b²+2 cannot be less than 2.

a.10 children i think