What is the inverse of f(x)= 1/(x^2+3x+2) ?

The inverse of f(x)= 1/(x^2+3x+2) is (-3x+sqrt(x^2+4x))/(2x),(-3x-sqrt(x^2+4x))/(2x)

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/(x^2+3x+2)

inverse

f (x) = \frac{1}{x ^{2} + 3 x + 2}

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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

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

f (x) = lo g_{10} (3)

y (q) = 1000 \cdot (\frac{q}{450})

y = 3 + lo g_{2} (x - 1)

f (x) = \frac{- 3}{60}

What is the inverse of f(x)= 1/(x^2+3x+2) ?
The inverse of f(x)= 1/(x^2+3x+2) is (-3x+sqrt(x^2+4x))/(2x),(-3x-sqrt(x^2+4x))/(2x)