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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)= x/(sqrt(x^2+1))

inverse

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

- \frac{x ^{2}}{x ^{2} - 1}, - - \frac{x ^{2}}{x ^{2} - 1}

Solution steps

y = \frac{x}{x ^{2} + 1}

Replace

x

with

y

x = \frac{y}{y ^{2} + 1}

Solve for

y, x = \frac{y}{y ^{2} + 1}

- \frac{x ^{2}}{x ^{2} - 1}, - - \frac{x ^{2}}{x ^{2} - 1}

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

f (x) = 8 x - 3

x^{2} - 3 x - 10

f (x) = 2 \cdot 3^{x} + 9^{x} + 5

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

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