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

\frac{4}{x} - 3

f (x) = \frac{7}{x + 8}

f (x) = x^{2} - 17

f (x) = x^{2} - 25

f (x) = \frac{7}{x + 3}

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