Find the Inverse (4x+3)/(x-1)

$\frac{4 x + 3}{x - 1}$

Step 1

Interchange the variables.

$x = \frac{4 y + 3}{y - 1}$

Step 2

Solve for $y$ .

Multiply the equation by $y - 1$ .

$(y - 1) (x) = (\frac{4 y + 3}{y - 1}) (y - 1)$

Simplify the left side.

Simplify $(y - 1) (x)$ .

Apply the distributive property.

$y x - 1 x = (\frac{4 y + 3}{y - 1}) (y - 1)$

Rewrite $- 1 x$ as $- x$ .

$y x - x = (\frac{4 y + 3}{y - 1}) (y - 1)$

Simplify the right side.

Cancel the common factor of $y - 1$ .

Cancel the common factor.

$y x - x = \frac{4 y + 3}{y - 1} (y - 1)$

Rewrite the expression.

$y x - x = 4 y + 3$

Solve for $y$ .

Subtract $4 y$ from both sides of the equation.

$y x - x - 4 y = 3$

Add $x$ to both sides of the equation.

$y x - 4 y = 3 + x$

Factor $y$ out of $y x - 4 y$ .

Factor $y$ out of $y x$ .

$y (x) - 4 y = 3 + x$

Factor $y$ out of $- 4 y$ .

$y (x) + y \cdot - 4 = 3 + x$

Factor $y$ out of $y (x) + y \cdot - 4$ .

$y (x - 4) = 3 + x$

Divide each term in $y (x - 4) = 3 + x$ by $x - 4$ and simplify.

Divide each term in $y (x - 4) = 3 + x$ by $x - 4$ .

$\frac{y (x - 4)}{x - 4} = \frac{3}{x - 4} + \frac{x}{x - 4}$

Simplify the left side.

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$\frac{y (x - 4)}{x - 4} = \frac{3}{x - 4} + \frac{x}{x - 4}$

Divide $y$ by $1$ .

$y = \frac{3}{x - 4} + \frac{x}{x - 4}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{3 + x}{x - 4}$

Step 3

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

$f^{- 1} (x) = \frac{3 + x}{x - 4}$

Step 4

Verify if $f^{- 1} (x) = \frac{3 + x}{x - 4}$ is the inverse of $f (x) = \frac{4 x + 3}{x - 1}$ .

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 $f^{- 1} (\frac{4 x + 3}{x - 1})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{3 + \frac{4 x + 3}{x - 1}}{(\frac{4 x + 3}{x - 1}) - 4}$

Remove parentheses.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{3 + \frac{4 x + 3}{x - 1}}{(\frac{4 x + 3}{x - 1}) - 4}$

Multiply the numerator and denominator of the complex fraction by $x - 1$ .

Multiply $\frac{3 + \frac{4 x + 3}{x - 1}}{\frac{4 x + 3}{x - 1} - 4}$ by $\frac{x - 1}{x - 1}$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{x - 1}{x - 1} \cdot \frac{3 + \frac{4 x + 3}{x - 1}}{\frac{4 x + 3}{x - 1} - 4}$

Combine.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{(x - 1) (3 + \frac{4 x + 3}{x - 1})}{(x - 1) (\frac{4 x + 3}{x - 1} - 4)}$

Apply the distributive property.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{(x - 1) \cdot 3 + (x - 1) (\frac{4 x + 3}{x - 1})}{(x - 1) (\frac{4 x + 3}{x - 1}) + (x - 1) \cdot - 4}$

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{(x - 1) \cdot 3 + (x - 1) (\frac{4 x + 3}{x - 1})}{(x - 1) (\frac{4 x + 3}{x - 1}) + (x - 1) \cdot - 4}$

Rewrite the expression.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{(x - 1) \cdot 3 + 4 x + 3}{(x - 1) (\frac{4 x + 3}{x - 1}) + (x - 1) \cdot - 4}$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{(x - 1) \cdot 3 + 4 x + 3}{(x - 1) (\frac{4 x + 3}{x - 1}) + (x - 1) \cdot - 4}$

Rewrite the expression.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{(x - 1) \cdot 3 + 4 x + 3}{4 x + 3 + (x - 1) \cdot - 4}$

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{x \cdot 3 - 1 \cdot 3 + 4 x + 3}{4 x + 3 + (x - 1) \cdot - 4}$

Move $3$ to the left of $x$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{3 \cdot x - 1 \cdot 3 + 4 x + 3}{4 x + 3 + (x - 1) \cdot - 4}$

Multiply $- 1$ by $3$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{3 \cdot x - 3 + 4 x + 3}{4 x + 3 + (x - 1) \cdot - 4}$

Add $3 x$ and $4 x$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x - 3 + 3}{4 x + 3 + (x - 1) \cdot - 4}$

Add $- 3$ and $3$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x + 0}{4 x + 3 + (x - 1) \cdot - 4}$

Add $7 x$ and $0$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{4 x + 3 + (x - 1) \cdot - 4}$

Simplify the denominator.

Apply the distributive property.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{4 x + 3 + x \cdot - 4 - 1 \cdot - 4}$

Move $- 4$ to the left of $x$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{4 x + 3 - 4 \cdot x - 1 \cdot - 4}$

Multiply $- 1$ by $- 4$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{4 x + 3 - 4 \cdot x + 4}$

Subtract $4 x$ from $4 x$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{0 + 3 + 4}$

Add $0$ and $3$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{3 + 4}$

Add $3$ and $4$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\frac{4 x + 3}{x - 1}) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{4 x + 3}{x - 1}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{3 + x}{x - 4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{3 + x}{x - 4}) = \frac{4 (\frac{3 + x}{x - 4}) + 3}{(\frac{3 + x}{x - 4}) - 1}$

Multiply the numerator and denominator of the complex fraction by $x - 4$ .

Multiply $\frac{4 \frac{3 + x}{x - 4} + 3}{\frac{3 + x}{x - 4} - 1}$ by $\frac{x - 4}{x - 4}$ .

$f (\frac{3 + x}{x - 4}) = \frac{x - 4}{x - 4} \cdot \frac{4 (\frac{3 + x}{x - 4}) + 3}{\frac{3 + x}{x - 4} - 1}$

Combine.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4}) + 3)}{(x - 4) (\frac{3 + x}{x - 4} - 1)}$

Apply the distributive property.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4})) + (x - 4) \cdot 3}{(x - 4) (\frac{3 + x}{x - 4}) + (x - 4) \cdot - 1}$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4})) + (x - 4) \cdot 3}{(x - 4) (\frac{3 + x}{x - 4}) + (x - 4) \cdot - 1}$

Rewrite the expression.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4})) + (x - 4) \cdot 3}{3 + x + (x - 4) \cdot - 1}$

Simplify the numerator.

Factor $x - 4$ out of $(x - 4) \cdot 4 \frac{3 + x}{x - 4} + (x - 4) \cdot 3$ .

Factor $x - 4$ out of $(x - 4) \cdot 4 \frac{3 + x}{x - 4}$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4})) + (x - 4) \cdot 3}{3 + x + (x - 4) \cdot - 1}$

Factor $x - 4$ out of $(x - 4) \cdot 3$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4})) + (x - 4) (3)}{3 + x + (x - 4) \cdot - 1}$

Factor $x - 4$ out of $(x - 4) (4 \frac{3 + x}{x - 4}) + (x - 4) (3)$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (4 (\frac{3 + x}{x - 4}) + 3)}{3 + x + (x - 4) \cdot - 1}$

Combine $4$ and $\frac{3 + x}{x - 4}$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{4 (3 + x)}{x - 4} + 3)}{3 + x + (x - 4) \cdot - 1}$

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 4}{x - 4}$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{4 (3 + x)}{x - 4} + \frac{3 (x - 4)}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Combine the numerators over the common denominator.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{4 (3 + x) + 3 (x - 4)}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Rewrite $\frac{4 (3 + x) + 3 (x - 4)}{x - 4}$ in a factored form.

Apply the distributive property.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{4 \cdot 3 + 4 x + 3 (x - 4)}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Multiply $4$ by $3$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{12 + 4 x + 3 (x - 4)}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Apply the distributive property.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{12 + 4 x + 3 x + 3 \cdot - 4}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Multiply $3$ by $- 4$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{12 + 4 x + 3 x - 12}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Subtract $12$ from $12$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{4 x + 3 x + 0}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Add $4 x + 3 x$ and $0$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{4 x + 3 x}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Add $4 x$ and $3 x$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{3 + x + (x - 4) \cdot - 1}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{3 + x + x \cdot - 1 - 4 \cdot - 1}$

Move $- 1$ to the left of $x$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{3 + x - 1 \cdot x - 4 \cdot - 1}$

Multiply $- 4$ by $- 1$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{3 + x - 1 \cdot x + 4}$

Rewrite $- 1 x$ as $- x$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{3 + x - x + 4}$

Add $3$ and $4$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{x - x + 7}$

Subtract $x$ from $x$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{0 + 7}$

Add $0$ and $7$ .

$f (\frac{3 + x}{x - 4}) = \frac{(x - 4) (\frac{7 x}{x - 4})}{7}$

Reduce the expression by cancelling the common factors.

Factor $\frac{7 x}{x - 4}$ out of $\frac{(x - 4) \frac{7 x}{x - 4}}{7}$ .

$f (\frac{3 + x}{x - 4}) = \frac{7 x}{x - 4} \cdot \frac{x - 4}{7}$

Cancel the common factor of $7$ .

Factor $7$ out of $7 x$ .

$f (\frac{3 + x}{x - 4}) = \frac{7 (x)}{x - 4} \cdot \frac{x - 4}{7}$

Cancel the common factor.

$f (\frac{3 + x}{x - 4}) = \frac{7 x}{x - 4} \cdot \frac{x - 4}{7}$

Rewrite the expression.

$f (\frac{3 + x}{x - 4}) = \frac{x}{x - 4} \cdot (x - 4)$

Cancel the common factor of $x - 4$ .

Cancel the common factor.

$f (\frac{3 + x}{x - 4}) = \frac{x}{x - 4} \cdot (x - 4)$

Rewrite the expression.

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{3 + x}{x - 4}$ is the inverse of $f (x) = \frac{4 x + 3}{x - 1}$ .

$f^{- 1} (x) = \frac{3 + x}{x - 4}$

Find the Inverse (4x+3)/(x-1)

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