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

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

inverse

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

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

f (x) = 2^{x} - 64

f (x) = - t^{2} + 2 t + 10

f (x) = \frac{200000}{100 + 900 e ^{- x}}

+ x^{2} - 7 x

f (x) = ln (x) + 1, x > 1

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