What is the inverse of (x-2)^2+1,x>= 2 ?

The inverse of (x-2)^2+1,x>= 2 is sqrt(x-1)+2,-sqrt(x-1)+2

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inverse of (x-2)^2+1,x>= 2

inverse

(x - 2)^{2} + 1, x \geq 2

x - 1 + 2, - x - 1 + 2

Solution steps

y = (x - 2)^{2} + 1

Replace

x

with

y

x = (y - 2)^{2} + 1

Solve for

y, x = (y - 2)^{2} + 1

x - 1 + 2, - x - 1 + 2

88.4 x^{- 1.11}

f (x) = cos (x + 1) - 12

g (x) = - \frac{4}{3} x + 2

4 x^{2}

f (x) = x + \frac{x ^{2}}{2}

What is the inverse of (x-2)^2+1,x>= 2 ?
The inverse of (x-2)^2+1,x>= 2 is sqrt(x-1)+2,-sqrt(x-1)+2