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

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

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

inverse

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

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

g (x) = \frac{x - 8}{3}

f (x) = - x^{2} + 4 x + 5

49 + 2.3 (x - 60)

f (x) = \frac{2 x - 1}{5 x - 1}

g (x) = \frac{0}{2 ^{x + 1}}

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