Can you find the inverse of this please...?

Question

Can anyone find the inverse function of this:

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

Thanks!

Answer 1

Rewrite equation as y = x² + 2x - 1

Switch x and y

x = y² + 2y - 1

Now solve for y

x = (y² + 2y) - 1 = (y² + 2y + 1) - 1 - 1

Here I completed the square within the brackets by adding 1, but I must now balance it out by subtracting 1 outside bracket.

x = ( y² + 2y + 1) - 2

x = (y + 1)² - 2

(y + 1)² = x + 2

y + 1 = √ (x+2)

y = √ (x+2) - 1

Answer: f⁻¹(x) = √ (x+2) - 1

To verify that this is true, we can check that f(f⁻¹(x)) = x

f(f⁻¹(x)) = f(√ (x+2) - 1)

= (√ (x+2) - 1)² + 2(√ (x+2) - 1) - 1

= x + 2 - 2(√ (x+2)) + 1 + 2(√ (x+2)) - 2 -1

= x + 2 + 1 - 2 - 1

= x

Answer 2

Rewrite the equation as x= f(y) = y^2+2y-1

Solve for new y which is the inverse y.

Since you have a -1 it is not easy....you have to complete the square.

Rearranging we have

y^2+2y= x+1

Completing the square by taking 1/2 midle term then squaring it we have

y^2+2y+1=x+1+1

y^2+2y+1=x+2

Factoring left side

(y+1)^2=x+2

Taking square root of each side

y+1=sqrt(x+2)

y=sqrt(x+2)-1 answer

Answer 3

Replace x with y, then solve for y.

y = x² + 2x - 1

x = y² + 2y - 1

To solve for y, I will complete the square.

x + 1 = y² + 2y

x + 1 + 1 = y² + 2y + 1

x + 2 = √(y + 1)²

√(x + 2) = y + 1

√(x + 2) - 1 = y

y = √(x + 2) -1