What is the inverse of f(x)=(x-9)^2,x>= 9 ?

The inverse of f(x)=(x-9)^2,x>= 9 is sqrt(x)+9,-sqrt(x)+9

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inverse of f(x)=(x-9)^2,x>= 9

inverse

f (x) = (x - 9)^{2}, x \geq 9

x + 9, - x + 9

Solution steps

y = (x - 9)^{2}

Replace

x

with

y

x = (y - 9)^{2}

Solve for

y, x = (y - 9)^{2}

x + 9, - x + 9

f (x) = 1 + x

arcsin (\frac{1}{2})

f (x) = \frac{( x - 1 )}{x + 1}

f (x) = 4 (x + 17)

f (x) = \frac{3}{x ^{2} - 1}

What is the inverse of f(x)=(x-9)^2,x>= 9 ?
The inverse of f(x)=(x-9)^2,x>= 9 is sqrt(x)+9,-sqrt(x)+9