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

The inverse of (t^2+1)/(t-1) is (t+sqrt(t^2-4(1+t)))/2 ,(t-sqrt(t^2-4(1+t)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (t^2+1)/(t-1)

inverse

\frac{t ^{2} + 1}{t - 1}

\frac{t + t ^{2} - 4 ( 1 + t )}{2}, \frac{t - t ^{2} - 4 ( 1 + t )}{2}

Solution steps

y = \frac{t ^{2} + 1}{t - 1}

Replace

t

with

y

t = \frac{y ^{2} + 1}{y - 1}

Solve for

y, t = \frac{y ^{2} + 1}{y - 1}

\frac{t + t ^{2} - 4 ( 1 + t )}{2}, \frac{t - t ^{2} - 4 ( 1 + t )}{2}

\frac{1}{3 x + 4}

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

f (x) = \frac{x}{5} - 9

3 x^{2} + 12 x - 7

y = \frac{x + 28}{x - 7}

What is the inverse of (t^2+1)/(t-1) ?
The inverse of (t^2+1)/(t-1) is (t+sqrt(t^2-4(1+t)))/2 ,(t-sqrt(t^2-4(1+t)))/2