What is the inverse of f(x)=(x-1)^2,x<= 1 ?

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

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inverse of f(x)=(x-1)^2,x<= 1

inverse

f (x) = (x - 1)^{2}, x \leq 1

x + 1, - x + 1

Solution steps

y = (x - 1)^{2}

Replace

x

with

y

x = (y - 1)^{2}

Solve for

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

x + 1, - x + 1

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

f (x) = - \frac{5}{3} x + 5

x^{3} + 2 x^{2} + 9 x + 18

f (x) = - 3 cos (x)

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

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