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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

y = y = x^{2} - 6

\frac{1 - 2 x}{1 - 3 x}

2 cos (2 π (x + \frac{2 π}{3})) - 5

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

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

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