What is the inverse of f(x)=sqrt(x^2+2x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=sqrt(x^2+2x)

inverse

f (x) = x^{2} + 2 x

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

f (x) = \frac{1}{2} x + \frac{3}{2}

f (x) = - 8 x + 4

\frac{2 x}{x + 5}

f (x) = \frac{x + 18}{x - 6}

(1, 7.5), (3, 16.875)

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