# What is the inverse of f(x) = x + sqrt(x)?

No matter what I end up with x^2 = y^2 + y and have no idea how to get it to just one y.

Relevance

f(x) = x + √x

1.

To find the inverse, first change to x, y notation.

y = x + √x

Then interchange the positions of x and y and solve for y.

x = y + √y

The solution for the new y is the inverse function.

x^2 = y^2 + y + 2y^(3/2)

x =

• Anonymous
6 years ago

ƒ(x) = x + √x

y - x = √x

y² - 2xy + x² = x

x² + (-2y - 1)x + y² = 0

x = 2y + 1 ± √[4y² + 4y + 1 - 4y²] ÷ 2

x = 2y + 1 ± √[4y + 1] ÷ 2

Using the original equation, if x = 1, y = 2. Subbing in y = 2 into the inverse gets 4 if we use the positive root or 1 if we use the negative. Therefore, the negative root is the right one.

ƒ⁻¹(x) = x - 1/2 √[4x + 1] + 1/2

• DON'T know how you got x² = y² + y...you appear to use [ a + b]² = a² + b² !!!

solve y = x + √x for x....[ y - x ]² = x = y² - 2xy + x² ---> x² + x [ - 2y -1 ] + y² = 0...

quadratic equation yields x = { (2y + 1) -√ [ (2y + 1)² - 4 y² ] } / 2...

if you demand the dependent variable be y then interchange x & y..

note : no ± since when y = 0 so must x = 0