Find the Inverse f(x)=(ax+b)/(cx+d)

$f (x) = \frac{a x + b}{c x + d}$

Step 1

Write $f (x) = \frac{a x + b}{c x + d}$ as an equation.

$y = \frac{a x + b}{c x + d}$

Step 2

Interchange the variables.

$a = \frac{y x + b}{c x + d}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{y x + b}{c x + d} = a$ .

$\frac{y x + b}{c x + d} = a$

Multiply both sides by $c x + d$ .

$\frac{y x + b}{c x + d} (c x + d) = a (c x + d)$

Simplify.

Simplify the left side.

Simplify $\frac{y x + b}{c x + d} (c x + d)$ .

Cancel the common factor of $c x + d$ .

Cancel the common factor.

$\frac{y x + b}{c x + d} (c x + d) = a (c x + d)$

Rewrite the expression.

$y x + b = a (c x + d)$

Reorder $y$ and $x$ .

$x y + b = a (c x + d)$

Simplify the right side.

Apply the distributive property.

$x y + b = a c x + a d$

Solve for $y$ .

Subtract $b$ from both sides of the equation.

$x y = a c x + a d - b$

Divide each term in $x y = a c x + a d - b$ by $x$ and simplify.

Divide each term in $x y = a c x + a d - b$ by $x$ .

$\frac{x y}{x} = \frac{a c x}{x} + \frac{a d}{x} + \frac{- b}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x y}{x} = \frac{a c x}{x} + \frac{a d}{x} + \frac{- b}{x}$

Divide $y$ by $1$ .

$y = \frac{a c x}{x} + \frac{a d}{x} + \frac{- b}{x}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $x$ .

Cancel the common factor.

$y = \frac{a c x}{x} + \frac{a d}{x} + \frac{- b}{x}$

Divide $a c$ by $1$ .

$y = a c + \frac{a d}{x} + \frac{- b}{x}$

Move the negative in front of the fraction.

$y = a c + \frac{a d}{x} - \frac{b}{x}$

Step 4

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

$f^{- 1} (a) = a c + \frac{a d}{x} - \frac{b}{x}$

Step 5

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

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

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

Set up the composite result function.

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

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

$f^{- 1} (\frac{a x + b}{c x + d}) = (\frac{a x + b}{c x + d}) \cdot c + (\frac{a x + b}{c x + d}) \cdot \frac{d}{x} - \frac{b}{x}$

Simplify each term.

Combine $\frac{a x + b}{c x + d}$ and $c$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) c}{c x + d} + (\frac{a x + b}{c x + d}) \cdot \frac{d}{x} - \frac{b}{x}$

Multiply $\frac{a x + b}{c x + d}$ by $\frac{d}{x}$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) c}{c x + d} + \frac{(a x + b) d}{(c x + d) x} - \frac{b}{x}$

To write $\frac{(a x + b) c}{c x + d}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) c}{c x + d} \cdot \frac{x}{x} + \frac{(a x + b) d}{(c x + d) x} - \frac{b}{x}$

Simplify terms.

Multiply $\frac{(a x + b) c}{c x + d}$ by $\frac{x}{x}$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) c x}{(c x + d) x} + \frac{(a x + b) d}{(c x + d) x} - \frac{b}{x}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) c x + (a x + b) d}{(c x + d) x} - \frac{b}{x}$

Simplify each term.

Factor $a x + b$ out of $(a x + b) c x + (a x + b) d$ .

Factor $a x + b$ out of $(a x + b) c x$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) (c x) + (a x + b) d}{(c x + d) x} - \frac{b}{x}$

Factor $a x + b$ out of $(a x + b) d$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) (c x) + (a x + b) (d)}{(c x + d) x} - \frac{b}{x}$

Factor $a x + b$ out of $(a x + b) (c x) + (a x + b) (d)$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) (c x + d)}{(c x + d) x} - \frac{b}{x}$

Cancel the common factor of $c x + d$ .

Cancel the common factor.

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{(a x + b) (c x + d)}{(c x + d) x} - \frac{b}{x}$

Rewrite the expression.

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{a x + b}{x} - \frac{b}{x}$

Simplify terms.

Combine the numerators over the common denominator.

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{a x + b - b}{x}$

Combine the opposite terms in $a x + b - b$ .

Subtract $b$ from $b$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{a x + 0}{x}$

Add $a x$ and $0$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{a x}{x}$

Cancel the common factor of $x$ .

Cancel the common factor.

$f^{- 1} (\frac{a x + b}{c x + d}) = \frac{a x}{x}$

Divide $a$ by $1$ .

$f^{- 1} (\frac{a x + b}{c x + d}) = a$

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

Set up the composite result function.

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

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

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{(a c + \frac{a d}{x} - \frac{b}{x}) \cdot x + b}{c x + d}$

Simplify the numerator.

Apply the distributive property.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + \frac{a d}{x} \cdot x - \frac{b}{x} \cdot x + b}{c x + d}$

Simplify.

Cancel the common factor of $x$ .

Cancel the common factor.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + \frac{a d}{x} \cdot x - \frac{b}{x} \cdot x + b}{c x + d}$

Rewrite the expression.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + a d - \frac{b}{x} \cdot x + b}{c x + d}$

Cancel the common factor of $x$ .

Move the leading negative in $- \frac{b}{x}$ into the numerator.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + a d + \frac{- b}{x} \cdot x + b}{c x + d}$

Cancel the common factor.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + a d + \frac{- b}{x} \cdot x + b}{c x + d}$

Rewrite the expression.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + a d - b + b}{c x + d}$

Add $- b$ and $b$ .

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + a d + 0}{c x + d}$

Add $a c x + a d$ and $0$ .

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a c x + a d}{c x + d}$

Factor $a$ out of $a c x + a d$ .

Factor $a$ out of $a c x$ .

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a (c x) + a d}{c x + d}$

Factor $a$ out of $a d$ .

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a (c x) + a (d)}{c x + d}$

Factor $a$ out of $a (c x) + a (d)$ .

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a (c x + d)}{c x + d}$

Cancel the common factor of $c x + d$ .

Cancel the common factor.

$f (a c + \frac{a d}{x} - \frac{b}{x}) = \frac{a (c x + d)}{c x + d}$

Divide $a$ by $1$ .

$f (a c + \frac{a d}{x} - \frac{b}{x}) = a$

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

$f^{- 1} (a) = a c + \frac{a d}{x} - \frac{b}{x}$

Find the Inverse f(x)=(ax+b)/(cx+d)

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