Find the Inverse f(x)=-1/(x-1)+3

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

Step 1

Write $f (x) = - \frac{1}{x - 1} + 3$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $- \frac{1}{y - 1} + 3 = x$ .

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

Subtract $3$ from both sides of the equation.

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

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.

$y - 1, 1, 1$

Remove parentheses.

$y - 1, 1, 1$

The LCM of one and any expression is the expression.

$y - 1$

Multiply each term in $- \frac{1}{y - 1} = x - 3$ by $y - 1$ to eliminate the fractions.

Multiply each term in $- \frac{1}{y - 1} = x - 3$ by $y - 1$ .

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

Simplify the left side.

Cancel the common factor of $y - 1$ .

Move the leading negative in $- \frac{1}{y - 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$- 1 = x (y - 1) - 3 (y - 1)$

Simplify the right side.

Simplify each term.

Apply the distributive property.

$- 1 = x y + x \cdot - 1 - 3 (y - 1)$

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

$- 1 = x y - 1 \cdot x - 3 (y - 1)$

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

$- 1 = x y - x - 3 (y - 1)$

Apply the distributive property.

$- 1 = x y - x - 3 y - 3 \cdot - 1$

Multiply $- 3$ by $- 1$ .

$- 1 = x y - x - 3 y + 3$

Solve the equation.

Rewrite the equation as $x y - x - 3 y + 3 = - 1$ .

$x y - x - 3 y + 3 = - 1$

Move all terms not containing $y$ to the right side of the equation.

Add $x$ to both sides of the equation.

$x y - 3 y + 3 = - 1 + x$

Subtract $3$ from both sides of the equation.

$x y - 3 y = - 1 + x - 3$

Subtract $3$ from $- 1$ .

$x y - 3 y = x - 4$

Factor $y$ out of $x y - 3 y$ .

Factor $y$ out of $x y$ .

$y x - 3 y = x - 4$

Factor $y$ out of $- 3 y$ .

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

Factor $y$ out of $y x + y \cdot - 3$ .

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

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

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

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

Simplify the left side.

Cancel the common factor of $x - 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Combine the numerators over the common denominator.

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

Step 4

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

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

Step 5

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

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

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

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

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

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

Combine.

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

Apply the distributive property.

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

Simplify by cancelling.

Cancel the common factor of $x - 1$ .

Move the leading negative in $- \frac{1}{x - 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 1$ .

Move the leading negative in $- \frac{1}{x - 1}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the numerator.

Reorder $x - 1$ and $- 4$ .

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

Subtract $4 \cdot (x - 1)$ from $3 \cdot (x - 1)$ .

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

Factor $- 1$ out of $- 1 - (x - 1)$ .

Reorder $- 1$ and $- (x - 1)$ .

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

Rewrite $- 1$ as $- 1 (1)$ .

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

Factor $- 1$ out of $- (x - 1) - 1 (1)$ .

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $- 1$ by $3$ .

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

Apply the distributive property.

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

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

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

Multiply $- 1$ by $- 3$ .

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

Subtract $3$ from $- 1$ .

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

Subtract $3 x$ from $3 x$ .

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

Subtract $4$ from $0$ .

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

Add $- 4$ and $3$ .

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

Reduce the expression by cancelling the common factors.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Simplify the denominator.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{x - 3}{x - 3}$ .

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

Combine $- 1$ and $\frac{x - 3}{x - 3}$ .

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

Combine the numerators over the common denominator.

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

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

Apply the distributive property.

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

Multiply $- 1$ by $- 3$ .

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

Subtract $x$ from $x$ .

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

Subtract $4$ from $0$ .

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

Add $- 4$ and $3$ .

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

Move the negative in front of the fraction.

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

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x - 3$ by $1$ .

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

Apply the distributive property.

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

Multiply $- 1$ by $- 3$ .

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

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $3$ .

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

Combine the opposite terms in $x - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse f(x)=-1/(x-1)+3

Related Questions

Support LCMGCF.com by disabling your adblocker.