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Alecsey [184]
2 years ago
15

Given f(x)=√(3x-12)+2, what is the range in set notation? {yly € R, y ≥

Mathematics
1 answer:
Gnesinka [82]2 years ago
6 0

Answer:

{y€R | x\geq12}

Step-by-step explanation:

First by looking at the function, f(x)=sqrt(3x-12)+2, we have a square root so the domain would be restricted.

Note that, \sqrt{ 3x-12} cannot be less than 0 since the domain of square root functions are all positive numbers within its domain including 0.

So x\geq 12 for the domain of this function.

Therefore, the range would be {y€R | x\geq12}

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