What is the inverse of f(x)=x^2+2x,x>= 0 ?

The inverse of f(x)=x^2+2x,x>= 0 is -1+sqrt(x+1),-sqrt(x+1)-1

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inverse of f(x)=x^2+2x,x>= 0

inverse

f (x) = x^{2} + 2 x, x \geq 0

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

Solution steps

y = x^{2} + 2 x

Replace

x

with

y

x = y^{2} + 2 y

Solve for

y, x = y^{2} + 2 y

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

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

f (x) = lo g_{2} (4 x - 10)

f (x) = 2 (x - 3)^{2} - 4

f (x) = sinh (5 x)

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

What is the inverse of f(x)=x^2+2x,x>= 0 ?
The inverse of f(x)=x^2+2x,x>= 0 is -1+sqrt(x+1),-sqrt(x+1)-1