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Yup, I had the same doubt, and got stuck there too..
But the ms had it pretty clearly explained, the range of f(x) is the domain of f^-1(x)
In f^-1(x) --- x > k-4
f(x) = (x-2)^2 -4 + k
Take it as if f^-1(x) is y and switch as always
x = (y-2)^2 - 4 + k
x+4-k = (y-2)^2
y-2 = ±√(x+4-k)
f^-1(x) = ±√(x+4-k) + 2
Replace x with anything less than k-4, like k-5
f^-1(x) = ±√(k-5+4-k) + 2
y - 2 = ±√(k-5+4-k)
The two k's cancel each other (FINALLY)
y-2 = ±√(5+4)
y-2 = ±√(9)
y - 2 = ±3
y = 5 or y = -1
The range of f^-1(x) is... f >= 2
So y = 5, so +.
f^-1(x) = +√(x+4-k) + 2
Why did ya take k-5 as x in f^-1 when x is supposed to be greater than k-4 in f^-1 ? That part I didn't understand.. everything else is clear cut.