Find the Domain and Range f^-1(x)=(x-3)/(1-x)

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

Set the denominator in $\frac{x - 3}{1 - x}$ equal to $0$ to find where the expression is undefined.

$1 - x = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, 1) \cup (1, \infty)$

Set-Builder Notation:

${x | x \neq 1}$

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - 1) \cup (- 1, \infty)$

Set-Builder Notation:

${y | y \neq - 1}$

Determine the domain and range.

Domain: $(- \infty, 1) \cup (1, \infty), {x | x \neq 1}$

Range: $(- \infty, - 1) \cup (- 1, \infty), {y | y \neq - 1}$