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

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

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

inverse

f (x) = x^{2} + 2 x - 1, x > 0

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

g (x) = 4 x - 6

f (x) = 2.5 x + 85

y = \frac{( x ^{2} - 2 x )}{( x ^{2} - 4 )}

x^{3} - 2 x^{2} + 2

f (x) = \frac{6 x - 9}{11 x + 5}

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