What is the inverse of f(x)=(x-19)^2,[19,infinity ] ?

The inverse of f(x)=(x-19)^2,[19,infinity ] is sqrt(x)+19,-sqrt(x)+19

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inverse of f(x)=(x-19)^2,[19,infinity ]

inverse

f (x) = (x - 19)^{2}, [19, \infty]

x + 19, - x + 19

Solution steps

y = (x - 19)^{2}

Replace

x

with

y

x = (y - 19)^{2}

Solve for

y, x = (y - 19)^{2}

x + 19, - x + 19

f (x) = \frac{2}{- x + 4}

9 sin (x)

f (x) = 2 x^{- \frac{1}{2}}

\frac{1.05}{1 + 100 e ^{- 0.285 t}}

f (x) = \frac{( x - 2 )}{x + 5}

What is the inverse of f(x)=(x-19)^2,[19,infinity ] ?
The inverse of f(x)=(x-19)^2,[19,infinity ] is sqrt(x)+19,-sqrt(x)+19