What is the inverse of f(x)=(x+3)^2-1,x>=-3 ?

The inverse of f(x)=(x+3)^2-1,x>=-3 is sqrt(x+1)-3,-sqrt(x+1)-3

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inverse of f(x)=(x+3)^2-1,x>=-3

inverse

f (x) = (x + 3)^{2} - 1, x \geq - 3

x + 1 - 3, - x + 1 - 3

Solution steps

y = (x + 3)^{2} - 1

Replace

x

with

y

x = (y + 3)^{2} - 1

Solve for

y, x = (y + 3)^{2} - 1

x + 1 - 3, - x + 1 - 3

f (x) = 12 + 2 x - 2

f (x) = 8 x, x > 4

f (x) = \frac{2}{x - 1} - 4

\frac{1}{n}

\frac{5 x + 3}{2 x - 7}

What is the inverse of f(x)=(x+3)^2-1,x>=-3 ?
The inverse of f(x)=(x+3)^2-1,x>=-3 is sqrt(x+1)-3,-sqrt(x+1)-3