What is the inverse of h(x)=(5x-1)/(2x+1) ?

The inverse of h(x)=(5x-1)/(2x+1) is (-1-x)/(2x-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of h(x)=(5x-1)/(2x+1)

inverse

h (x) = \frac{5 x - 1}{2 x + 1}

\frac{- 1 - x}{2 x - 5}

Solution steps

y = \frac{5 x - 1}{2 x + 1}

Replace

x

with

y

x = \frac{5 y - 1}{2 y + 1}

Solve for

y, x = \frac{5 y - 1}{2 y + 1}

\frac{- 1 - x}{2 x - 5}

f (x) = - (x + 1) 2 - 1

f (x) = \frac{4 x - 3}{2 - x}

f (x) = 2 sin (3 x - 4) + 5

f (x) = \frac{9 x}{2 x - 1}

lo g_{10} (X - 1) - lo g_{10} (X)

What is the inverse of h(x)=(5x-1)/(2x+1) ?
The inverse of h(x)=(5x-1)/(2x+1) is (-1-x)/(2x-5)