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

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

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

inverse

g (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 0^{(x + 2)} - 3

e^{x - 3}

\frac{x - 4}{x + 3}

x^{2} + 6 x

f (x) = 2 2 x

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