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Easy: Values of √(ax + b)
A square root is only defined when the number inside is not negative. For f(x) = √(ax + b), solve ax + b ≥ 0. If a is positive, x ≥ −b/a. If a is negative, dividing by a flips the sign, so x ≤ −b/a.
Evaluate each function, or find its domain.
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A real square root of a negative number doesn't exist, so every x you can use must make ax + b ≥ 0. That set of x-values is the domain.
Solving −2x + 6 ≥ 0 gives x ≤ 3, not x ≥ 3, because dividing both sides by −2 flips the inequality.
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