Find the Inverse f(x)=1/(x-2)

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

Write $f (x) = \frac{1}{x - 2}$ as an equation.

$y = \frac{1}{x - 2}$

Interchange the variables.

$x = \frac{1}{y - 2}$

Solve for $y$ .

Rewrite the equation as $\frac{1}{y - 2} = x$ .

$\frac{1}{y - 2} = x$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y - 2, 1$

Remove parentheses.

$y - 2, 1$

The LCM of one and any expression is the expression.

$y - 2$

Multiply each term in $\frac{1}{y - 2} = x$ by $y - 2$ to eliminate the fractions.

Multiply each term in $\frac{1}{y - 2} = x$ by $y - 2$ .

$\frac{1}{y - 2} (y - 2) = x (y - 2)$

Simplify the left side.

Cancel the common factor of $y - 2$ .

Cancel the common factor.

$\frac{1}{y - 2} (y - 2) = x (y - 2)$

Rewrite the expression.

$1 = x (y - 2)$

Simplify the right side.

Apply the distributive property.

$1 = x y + x \cdot - 2$

Move $- 2$ to the left of $x$ .

$1 = x y - 2 x$

Solve the equation.

Rewrite the equation as $x y - 2 x = 1$ .

$x y - 2 x = 1$

Add $2 x$ to both sides of the equation.

$x y = 1 + 2 x$

Divide each term in $x y = 1 + 2 x$ by $x$ and simplify.

Divide each term in $x y = 1 + 2 x$ by $x$ .

$\frac{x y}{x} = \frac{1}{x} + \frac{2 x}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x y}{x} = \frac{1}{x} + \frac{2 x}{x}$

Divide $y$ by $1$ .

$y = \frac{1}{x} + \frac{2 x}{x}$

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

$y = \frac{1}{x} + \frac{2 x}{x}$

Divide $2$ by $1$ .

$y = \frac{1}{x} + 2$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

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

Verify if $f^{- 1} (x) = \frac{1}{x} + 2$ is the inverse of $f (x) = \frac{1}{x - 2}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\frac{1}{x - 2})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x - 2$ by $1$ .

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

Combine the opposite terms in $x - 2 + 2$ .

Add $- 2$ and $2$ .

$f^{- 1} (\frac{1}{x - 2}) = x + 0$

Add $x$ and $0$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{1}{x} + 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify the denominator.

Subtract $2$ from $2$ .

$f (\frac{1}{x} + 2) = \frac{1}{\frac{1}{x} + 0}$

Add $\frac{1}{x}$ and $0$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{1}{x} + 2$ is the inverse of $f (x) = \frac{1}{x - 2}$ .

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

Find the Inverse f(x)=1/(x-2)

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