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

The inverse of (sqrt(2x^2+1))/(3x-2) is (6x^2+sqrt(17x^2-2))/(9x^2-2),(6x^2-sqrt(17x^2-2))/(9x^2-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (sqrt(2x^2+1))/(3x-2)

inverse

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

\frac{6 x ^{2} + 17 x ^{2} - 2}{9 x ^{2} - 2}, \frac{6 x ^{2} - 17 x ^{2} - 2}{9 x ^{2} - 2}

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{6 x ^{2} + 17 x ^{2} - 2}{9 x ^{2} - 2}, \frac{6 x ^{2} - 17 x ^{2} - 2}{9 x ^{2} - 2}

3 x^{2} - 27

lo g_{10} (x - 2) - 4

f (x) = 13.5 - 5 x

f (x) = 7^{5^{x}}

g (x) = \frac{9 x}{8 x - 7}

What is the inverse of (sqrt(2x^2+1))/(3x-2) ?
The inverse of (sqrt(2x^2+1))/(3x-2) is (6x^2+sqrt(17x^2-2))/(9x^2-2),(6x^2-sqrt(17x^2-2))/(9x^2-2)