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

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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 4 + x - 2

f (x) = x + 2 + 5

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

f (x) = 2 (x + 3)^{2} - 5, x \leq - 3

f (x) = \frac{3 x + 1}{2}

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