Please help I have no clue on this.
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f(x) is the right half of a parabola, with vertex (1, -1), opening up.
y = f(x) = (1/2)(x - 1)^2 - 1
x = (1/2)(y - 1)^2 - 1
x + 1 = (1/2)(y - 1)^2
2(x + 1) = (y - 1)^2
± √(2(x + 1)) = y - 1
y = 1 ± 2√(2(x + 1))
f^-1 (x) = 1 + 2√(2(x + 1)).
The domain of f is {x: x ≥ 1}, and the range is {y: y ≥ -1}. Its graph is the right half of the parabola y = (1/2)(x - 1)^2 - 1.
The domain of f^-1 is {x: x ≥ -1} and the range is {y: y ≥ 1}. Its graph if the upper half of the parabola x = (1/2)(y - 1)^2 - 1.
y = f(x) = (1/2)(x - 1)^2 - 1
x = (1/2)(y - 1)^2 - 1
x + 1 = (1/2)(y - 1)^2
2(x + 1) = (y - 1)^2
± √(2(x + 1)) = y - 1
y = 1 ± 2√(2(x + 1))
f^-1 (x) = 1 + 2√(2(x + 1)).
The domain of f is {x: x ≥ 1}, and the range is {y: y ≥ -1}. Its graph is the right half of the parabola y = (1/2)(x - 1)^2 - 1.
The domain of f^-1 is {x: x ≥ -1} and the range is {y: y ≥ 1}. Its graph if the upper half of the parabola x = (1/2)(y - 1)^2 - 1.