What is the inverse of (x^2+3x-1)/(7x-4) ?

The inverse of (x^2+3x-1)/(7x-4) is (-3+7x+sqrt(49x^2-58x+13))/2 ,(-3+7x-sqrt(49x^2-58x+13))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of (x^2+3x-1)/(7x-4)

inverse

\frac{x ^{2} + 3 x - 1}{7 x - 4}

\frac{- 3 + 7 x + 49 x ^{2} - 58 x + 13}{2}, \frac{- 3 + 7 x - 49 x ^{2} - 58 x + 13}{2}

Solution steps

y = \frac{x ^{2} + 3 x - 1}{7 x - 4}

Replace

x

with

y

x = \frac{y ^{2} + 3 y - 1}{7 y - 4}

Solve for

y, x = \frac{y ^{2} + 3 y - 1}{7 y - 4}

\frac{- 3 + 7 x + 49 x ^{2} - 58 x + 13}{2}, \frac{- 3 + 7 x - 49 x ^{2} - 58 x + 13}{2}

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

f (x) = (x + 4)^{2}, x \geq 0

x^{2} + 6 x + 9

\frac{x ^{2} + 3 x - 10}{x ^{2} - 4}

What is the inverse of (x^2+3x-1)/(7x-4) ?
The inverse of (x^2+3x-1)/(7x-4) is (-3+7x+sqrt(49x^2-58x+13))/2 ,(-3+7x-sqrt(49x^2-58x+13))/2