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Given that the range of y = f(1-x) - 1 is 2 ≤ y ≤ 3, find the range of (a) f(x) (b) f(x+1) + 1

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Final answer:

The range of y = f(1-x) - 1 is given as 2 ≤ y ≤ 3. The range of f(x) is calculated as 3 ≤ f(x) ≤ 4, and the range of f(x+1) + 1 is 4 ≤ f(x+1) + 1 ≤ 5.

Step-by-step explanation:

The question asks us to determine the range of two functions based on the given range of y = f(1-x) - 1. Since the range of y is given as 2 ≤ y ≤ 3, we can use that to find the ranges of f(x) and f(x+1) + 1.

To find the range of f(x), we observe that y = f(1-x) - 1 can be written as y + 1 = f(1-x). Substituting t for 1-x, we get f(t) = y + 1. Since y ranges from 2 to 3, after adding 1, we get 3 ≤ f(t) ≤ 4. Therefore, the range of f(x) is 3 ≤ f(x) ≤ 4.

To find the range of f(x+1) + 1, we substitute t for x+1 and apply the previously calculated range of f(t), knowing that adding 1 to this will shift the entire range by 1 unit upwards. Thus, we get 4 ≤ f(t) + 1 ≤ 5, so the range of f(x+1) + 1 is 4 ≤ f(x+1) + 1 ≤ 5.

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