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 \geq - 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

3 (\frac{1}{4})^{x - 2} + 4

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

4 t^{2} + 1

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

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

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