What is the inverse of 3/(sqrt(1+x^2)) ?

The inverse of 3/(sqrt(1+x^2)) is (sqrt(9-x^2))/x ,-(sqrt(9-x^2))/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of 3/(sqrt(1+x^2))

inverse

\frac{3}{1 + x ^{2}}

\frac{9 - x ^{2}}{x}, - \frac{9 - x ^{2}}{x}

Solution steps

y = \frac{3}{1 + x ^{2}}

Replace

x

with

y

x = \frac{3}{1 + y ^{2}}

Solve for

y, x = \frac{3}{1 + y ^{2}}

\frac{9 - x ^{2}}{x}, - \frac{9 - x ^{2}}{x}

What is the inverse of 3/(sqrt(1+x^2)) ?
The inverse of 3/(sqrt(1+x^2)) is (sqrt(9-x^2))/x ,-(sqrt(9-x^2))/x