What is the inverse of 1/(x^2-6x+9) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of 1/(x^2-6x+9)

inverse

\frac{1}{x ^{2} - 6 x + 9}

\frac{3 x + x}{x}, \frac{3 x - x}{x}

Solution steps

y = \frac{1}{x ^{2} - 6 x + 9}

Replace

x

with

y

x = \frac{1}{y ^{2} - 6 y + 9}

Solve for

y, x = \frac{1}{y ^{2} - 6 y + 9}

\frac{3 x + x}{x}, \frac{3 x - x}{x}

f (x) = y^{2}

f (x) = - \frac{9}{10 x}

4 x + 5

10 e^{2 x}

\frac{x - 1}{2}

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