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

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

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

inverse

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

x + 2, - x + 2

Solution steps

y = (x - 2)^{2}

Replace

x

with

y

x = (y - 2)^{2}

Solve for

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

x + 2, - x + 2

\frac{x}{x ^{2} + 16}

f (x) = - (\frac{1}{x - 9})

f (x) = 6 x^{6} + 4 x^{4} - 3 x^{2} + 2

f (x) = - 6 x - 11

f (x) = \frac{x + 6}{x - 2}

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