What is the inverse of f(x)=(x-18)^2,[18,infinity ] ?

The inverse of f(x)=(x-18)^2,[18,infinity ] is sqrt(x)+18,-sqrt(x)+18

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inverse of f(x)=(x-18)^2,[18,infinity ]

inverse

f (x) = (x - 18)^{2}, [18, \infty]

x + 18, - x + 18

Solution steps

y = (x - 18)^{2}

Replace

x

with

y

x = (y - 18)^{2}

Solve for

y, x = (y - 18)^{2}

x + 18, - x + 18

- x^{2} + 121 + 2

f (x) = 9 sin (x) - 5

f (x) = x^{2} - 14, x \geq 0

y = - (x - 5)^{2}

f (x) = cos (8 x)

What is the inverse of f(x)=(x-18)^2,[18,infinity ] ?
The inverse of f(x)=(x-18)^2,[18,infinity ] is sqrt(x)+18,-sqrt(x)+18