Find the Inverse f(x)=(x+15)/(x-5)

$f (x) = \frac{x + 15}{x - 5}$

Step 1

Write $f (x) = \frac{x + 15}{x - 5}$ as an equation.

$y = \frac{x + 15}{x - 5}$

Step 2

Interchange the variables.

$x = \frac{y + 15}{y - 5}$

Step 3

Solve for $y$ .

Multiply the equation by $y - 5$ .

$(y - 5) (x) = (\frac{y + 15}{y - 5}) (y - 5)$

Simplify the left side.

Apply the distributive property.

$y x - 5 x = (\frac{y + 15}{y - 5}) (y - 5)$

Simplify the right side.

Cancel the common factor of $y - 5$ .

Cancel the common factor.

$y x - 5 x = \frac{y + 15}{y - 5} (y - 5)$

Rewrite the expression.

$y x - 5 x = y + 15$

Solve for $y$ .

Subtract $y$ from both sides of the equation.

$y x - 5 x - y = 15$

Add $5 x$ to both sides of the equation.

$y x - y = 15 + 5 x$

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

Factor $y$ out of $y x$ .

$y (x) - y = 15 + 5 x$

Factor $y$ out of $- y$ .

$y (x) + y \cdot - 1 = 15 + 5 x$

Factor $y$ out of $y (x) + y \cdot - 1$ .

$y (x - 1) = 15 + 5 x$

Divide each term in $y (x - 1) = 15 + 5 x$ by $x - 1$ and simplify.

Divide each term in $y (x - 1) = 15 + 5 x$ by $x - 1$ .

$\frac{y (x - 1)}{x - 1} = \frac{15}{x - 1} + \frac{5 x}{x - 1}$

Simplify the left side.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$\frac{y (x - 1)}{x - 1} = \frac{15}{x - 1} + \frac{5 x}{x - 1}$

Divide $y$ by $1$ .

$y = \frac{15}{x - 1} + \frac{5 x}{x - 1}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{15 + 5 x}{x - 1}$

Factor $5$ out of $15 + 5 x$ .

Factor $5$ out of $15$ .

$y = \frac{5 \cdot 3 + 5 x}{x - 1}$

Factor $5$ out of $5 \cdot 3 + 5 x$ .

$y = \frac{5 (3 + x)}{x - 1}$

Step 4

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

$f^{- 1} (x) = \frac{5 (3 + x)}{x - 1}$

Step 5

Verify if $f^{- 1} (x) = \frac{5 (3 + x)}{x - 1}$ is the inverse of $f (x) = \frac{x + 15}{x - 5}$ .

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 $f^{- 1} (\frac{x + 15}{x - 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (3 + \frac{x + 15}{x - 5})}{(\frac{x + 15}{x - 5}) - 1}$

Remove parentheses.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (3 + \frac{x + 15}{x - 5})}{(\frac{x + 15}{x - 5}) - 1}$

Simplify the numerator.

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{3 (x - 5)}{x - 5} + \frac{x + 15}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{3 (x - 5) + x + 15}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Reorder terms.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{x + 3 (x - 5) + 15}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Rewrite $\frac{x + 3 (x - 5) + 15}{x - 5}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{x + 3 x + 3 \cdot - 5 + 15}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Multiply $3$ by $- 5$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{x + 3 x - 15 + 15}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Add $x$ and $3 x$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x - 15 + 15}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Add $- 15$ and $15$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x + 0}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Add $4 x$ and $0$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{x + 15}{x - 5} - 1}$

Simplify the denominator.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x - 5}{x - 5}$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{x + 15}{x - 5} - 1 \cdot \frac{x - 5}{x - 5}}$

Combine $- 1$ and $\frac{x - 5}{x - 5}$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{x + 15}{x - 5} + \frac{- (x - 5)}{x - 5}}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{x + 15 - (x - 5)}{x - 5}}$

Rewrite $\frac{x + 15 - (x - 5)}{x - 5}$ in a factored form.

Apply the distributive property.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{x + 15 - x + 5}{x - 5}}$

Multiply $- 1$ by $- 5$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{x + 15 - x + 5}{x - 5}}$

Subtract $x$ from $x$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{0 + 15 + 5}{x - 5}}$

Add $0$ and $15$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{15 + 5}{x - 5}}$

Add $15$ and $5$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{5 (\frac{4 x}{x - 5})}{\frac{20}{x - 5}}$

Combine $5$ and $\frac{4 x}{x - 5}$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{\frac{5 (4 x)}{x - 5}}{\frac{20}{x - 5}}$

Multiply $5$ by $4$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{\frac{20 x}{x - 5}}{\frac{20}{x - 5}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{20 x}{x - 5} \cdot \frac{x - 5}{20}$

Cancel the common factor of $20$ .

Factor $20$ out of $20 x$ .

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{20 (x)}{x - 5} \cdot \frac{x - 5}{20}$

Cancel the common factor.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{20 x}{x - 5} \cdot \frac{x - 5}{20}$

Rewrite the expression.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{x}{x - 5} \cdot (x - 5)$

Cancel the common factor of $x - 5$ .

Cancel the common factor.

$f^{- 1} (\frac{x + 15}{x - 5}) = \frac{x}{x - 5} \cdot (x - 5)$

Rewrite the expression.

$f^{- 1} (\frac{x + 15}{x - 5}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{5 (3 + x)}{x - 1})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{(\frac{5 (3 + x)}{x - 1}) + 15}{(\frac{5 (3 + x)}{x - 1}) - 5}$

Cancel the common factor of $(\frac{5 (3 + x)}{x - 1}) + 15$ and $(\frac{5 (3 + x)}{x - 1}) - 5$ .

Factor $5$ out of $\frac{5 (3 + x)}{x - 1}$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1}) + 15}{(\frac{5 (3 + x)}{x - 1}) - 5}$

Factor $5$ out of $15$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1}) + 5 (3)}{(\frac{5 (3 + x)}{x - 1}) - 5}$

Factor $5$ out of $5 \frac{3 + x}{x - 1} + 5 (3)$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1} + 3)}{(\frac{5 (3 + x)}{x - 1}) - 5}$

Cancel the common factors.

Factor $5$ out of $\frac{5 (3 + x)}{x - 1}$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1} + 3)}{5 (\frac{3 + x}{x - 1}) - 5}$

Factor $5$ out of $- 5$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1} + 3)}{5 (\frac{3 + x}{x - 1}) + 5 (- 1)}$

Factor $5$ out of $5 \frac{3 + x}{x - 1} + 5 (- 1)$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1} + 3)}{5 (\frac{3 + x}{x - 1} - 1)}$

Cancel the common factor.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{5 (\frac{3 + x}{x - 1} + 3)}{5 (\frac{3 + x}{x - 1} - 1)}$

Rewrite the expression.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{\frac{3 + x}{x - 1} + 3}{\frac{3 + x}{x - 1} - 1}$

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

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

$f (\frac{5 (3 + x)}{x - 1}) = \frac{x - 1}{x - 1} \cdot \frac{\frac{3 + x}{x - 1} + 3}{\frac{3 + x}{x - 1} - 1}$

Combine.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{(x - 1) (\frac{3 + x}{x - 1} + 3)}{(x - 1) (\frac{3 + x}{x - 1} - 1)}$

Apply the distributive property.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{(x - 1) (\frac{3 + x}{x - 1}) + (x - 1) \cdot 3}{(x - 1) (\frac{3 + x}{x - 1}) + (x - 1) \cdot - 1}$

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{(x - 1) (\frac{3 + x}{x - 1}) + (x - 1) \cdot 3}{(x - 1) (\frac{3 + x}{x - 1}) + (x - 1) \cdot - 1}$

Rewrite the expression.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{3 + x + (x - 1) \cdot 3}{(x - 1) (\frac{3 + x}{x - 1}) + (x - 1) \cdot - 1}$

Cancel the common factor of $x - 1$ .

Cancel the common factor.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{3 + x + (x - 1) \cdot 3}{(x - 1) (\frac{3 + x}{x - 1}) + (x - 1) \cdot - 1}$

Rewrite the expression.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{3 + x + (x - 1) \cdot 3}{3 + x + (x - 1) \cdot - 1}$

Simplify the numerator.

Apply the distributive property.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{3 + x + x \cdot 3 - 1 \cdot 3}{3 + x + (x - 1) \cdot - 1}$

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

$f (\frac{5 (3 + x)}{x - 1}) = \frac{3 + x + 3 \cdot x - 1 \cdot 3}{3 + x + (x - 1) \cdot - 1}$

Multiply $- 1$ by $3$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{3 + x + 3 \cdot x - 3}{3 + x + (x - 1) \cdot - 1}$

Subtract $3$ from $3$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{x + 3 x + 0}{3 + x + (x - 1) \cdot - 1}$

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

$f (\frac{5 (3 + x)}{x - 1}) = \frac{x + 3 x}{3 + x + (x - 1) \cdot - 1}$

Add $x$ and $3 x$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{3 + x + (x - 1) \cdot - 1}$

Simplify the denominator.

Apply the distributive property.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{3 + x + x \cdot - 1 - 1 \cdot - 1}$

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

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{3 + x - 1 \cdot x - 1 \cdot - 1}$

Multiply $- 1$ by $- 1$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{3 + x - 1 \cdot x + 1}$

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

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{3 + x - x + 1}$

Add $3$ and $1$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{x - x + 4}$

Subtract $x$ from $x$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{0 + 4}$

Add $0$ and $4$ .

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{5 (3 + x)}{x - 1}) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f (\frac{5 (3 + x)}{x - 1}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 (3 + x)}{x - 1}$ is the inverse of $f (x) = \frac{x + 15}{x - 5}$ .

$f^{- 1} (x) = \frac{5 (3 + x)}{x - 1}$

Find the Inverse f(x)=(x+15)/(x-5)

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