What is the inverse of 1/(sqrt(x^2+7)) ?

The inverse of 1/(sqrt(x^2+7)) is (sqrt(1-7x^2))/x ,-(sqrt(1-7x^2))/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of 1/(sqrt(x^2+7))

inverse

\frac{1}{x ^{2} + 7}

\frac{1 - 7 x ^{2}}{x}, - \frac{1 - 7 x ^{2}}{x}

Solution steps

y = \frac{1}{x ^{2} + 7}

Replace

x

with

y

x = \frac{1}{y ^{2} + 7}

Solve for

y, x = \frac{1}{y ^{2} + 7}

\frac{1 - 7 x ^{2}}{x}, - \frac{1 - 7 x ^{2}}{x}

\frac{x + 1}{x - 1}

f (x) = \frac{x}{144}

y = 5 tan (5 x - π)

(- 4, 5), (- 1, 8)

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

What is the inverse of 1/(sqrt(x^2+7)) ?
The inverse of 1/(sqrt(x^2+7)) is (sqrt(1-7x^2))/x ,-(sqrt(1-7x^2))/x