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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)= 3/(x^2+1)

inverse

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

\frac{3 - x}{x}, - \frac{3 - x}{x}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{3 - x}{x}, - \frac{3 - x}{x}

f (x) = 6 + 3 + x

F (s) = \frac{3 - s}{s ^{2} - 4 s + 7}

f (x) = \frac{10 x - 9}{7 x - 1}

f (x) = (- 15 x - 13)^{2}

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