Precalculus Examples

$f (x) = \frac{2 x}{4 - 3 x}$ $f^{- 1} (x)$

Step 1

Write $f (x) = \frac{2 x}{4 - 3 x}$ as an equation.

$y = \frac{2 x}{4 - 3 x}$

Step 2

Interchange the variables.

$x = \frac{2 y}{4 - 3 y}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{2 y}{4 - 3 y} = x$ .

$\frac{2 y}{4 - 3 y} = x$

Step 3.2

Find the LCD of the terms in the equation.

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Step 3.2.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$4 - 3 y, 1$

Step 3.2.2

Remove parentheses.

$4 - 3 y, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$4 - 3 y$

Step 3.3

Multiply each term in $\frac{2 y}{4 - 3 y} = x$ by $4 - 3 y$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{2 y}{4 - 3 y} = x$ by $4 - 3 y$ .

$\frac{2 y}{4 - 3 y} (4 - 3 y) = x (4 - 3 y)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $4 - 3 y$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 y}{4 - 3 y} (4 - 3 y) = x (4 - 3 y)$

Step 3.3.2.1.2

Rewrite the expression.

$2 y = x (4 - 3 y)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$2 y = x \cdot 4 + x (- 3 y)$

Step 3.3.3.2

Reorder.

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Step 3.3.3.2.1

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

$2 y = 4 \cdot x + x (- 3 y)$

Step 3.3.3.2.2

Rewrite using the commutative property of multiplication.

$2 y = 4 x - 3 x y$

Step 3.4

Solve the equation.

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Step 3.4.1

Add $3 x y$ to both sides of the equation.

$2 y + 3 x y = 4 x$

Step 3.4.2

Factor $y$ out of $2 y + 3 x y$ .

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Step 3.4.2.1

Factor $y$ out of $2 y$ .

$y \cdot 2 + 3 x y = 4 x$

Step 3.4.2.2

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

$y \cdot 2 + y (3 x) = 4 x$

Step 3.4.2.3

Factor $y$ out of $y \cdot 2 + y (3 x)$ .

$y (2 + 3 x) = 4 x$

Step 3.4.3

Divide each term in $y (2 + 3 x) = 4 x$ by $2 + 3 x$ and simplify.

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Step 3.4.3.1

Divide each term in $y (2 + 3 x) = 4 x$ by $2 + 3 x$ .

$\frac{y (2 + 3 x)}{2 + 3 x} = \frac{4 x}{2 + 3 x}$

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $2 + 3 x$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

$\frac{y (2 + 3 x)}{2 + 3 x} = \frac{4 x}{2 + 3 x}$

Step 3.4.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{4 x}{2 + 3 x}$

Step 4

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

$f^{- 1} (x) = \frac{4 x}{2 + 3 x}$

Step 5

Verify if $f^{- 1} (x) = \frac{4 x}{2 + 3 x}$ is the inverse of $f (x) = \frac{2 x}{4 - 3 x}$ .

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Step 5.1

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

Step 5.2

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

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Step 5.2.1

Set up the composite result function.

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

Step 5.2.2

Evaluate $f^{- 1} (\frac{2 x}{4 - 3 x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{4 (\frac{2 x}{4 - 3 x})}{2 + 3 (\frac{2 x}{4 - 3 x})}$

Step 5.2.3

Combine $4$ and $\frac{2 x}{4 - 3 x}$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{2 + 3 (\frac{2 x}{4 - 3 x})}$

Step 5.2.4

Simplify the denominator.

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Step 5.2.4.1

Multiply $3 \frac{2 x}{4 - 3 x}$ .

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Step 5.2.4.1.1

Combine $3$ and $\frac{2 x}{4 - 3 x}$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{2 + \frac{3 (2 x)}{4 - 3 x}}$

Step 5.2.4.1.2

Multiply $2$ by $3$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{2 + \frac{6 x}{4 - 3 x}}$

Step 5.2.4.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{4 - 3 x}{4 - 3 x}$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{2 (4 - 3 x)}{4 - 3 x} + \frac{6 x}{4 - 3 x}}$

Step 5.2.4.3

Combine the numerators over the common denominator.

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{2 (4 - 3 x) + 6 x}{4 - 3 x}}$

Step 5.2.4.4

Rewrite $\frac{2 (4 - 3 x) + 6 x}{4 - 3 x}$ in a factored form.

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Step 5.2.4.4.1

Factor $2$ out of $2 (4 - 3 x) + 6 x$ .

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Step 5.2.4.4.1.1

Factor $2$ out of $6 x$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{2 (4 - 3 x) + 2 (3 x)}{4 - 3 x}}$

Step 5.2.4.4.1.2

Factor $2$ out of $2 (4 - 3 x) + 2 (3 x)$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{2 (4 - 3 x + 3 x)}{4 - 3 x}}$

Step 5.2.4.4.2

Add $- 3 x$ and $3 x$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{2 (4 + 0)}{4 - 3 x}}$

Step 5.2.4.4.3

Add $4$ and $0$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{2 \cdot 4}{4 - 3 x}}$

Step 5.2.4.5

Multiply $2$ by $4$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{4 (2 x)}{4 - 3 x}}{\frac{8}{4 - 3 x}}$

Step 5.2.5

Multiply $4$ by $2$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{\frac{8 x}{4 - 3 x}}{\frac{8}{4 - 3 x}}$

Step 5.2.6

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{8 x}{4 - 3 x} \cdot \frac{4 - 3 x}{8}$

Step 5.2.7

Cancel the common factor of $8$ .

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Step 5.2.7.1

Factor $8$ out of $8 x$ .

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{8 (x)}{4 - 3 x} \cdot \frac{4 - 3 x}{8}$

Step 5.2.7.2

Cancel the common factor.

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{8 x}{4 - 3 x} \cdot \frac{4 - 3 x}{8}$

Step 5.2.7.3

Rewrite the expression.

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{x}{4 - 3 x} \cdot (4 - 3 x)$

Step 5.2.8

Cancel the common factor of $4 - 3 x$ .

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Step 5.2.8.1

Cancel the common factor.

$f^{- 1} (\frac{2 x}{4 - 3 x}) = \frac{x}{4 - 3 x} \cdot (4 - 3 x)$

Step 5.2.8.2

Rewrite the expression.

$f^{- 1} (\frac{2 x}{4 - 3 x}) = x$

Step 5.3

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

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Step 5.3.1

Set up the composite result function.

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

Step 5.3.2

Evaluate $f (\frac{4 x}{2 + 3 x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{2 (\frac{4 x}{2 + 3 x})}{4 - 3 \frac{4 x}{2 + 3 x}}$

Step 5.3.3

Combine $2$ and $\frac{4 x}{2 + 3 x}$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{4 - 3 \frac{4 x}{2 + 3 x}}$

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Multiply $- 3 \frac{4 x}{2 + 3 x}$ .

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Step 5.3.4.1.1

Combine $- 3$ and $\frac{4 x}{2 + 3 x}$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{4 + \frac{- 3 (4 x)}{2 + 3 x}}$

Step 5.3.4.1.2

Multiply $4$ by $- 3$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{4 + \frac{- 12 x}{2 + 3 x}}$

Step 5.3.4.2

Move the negative in front of the fraction.

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{4 - \frac{12 x}{2 + 3 x}}$

Step 5.3.4.3

To write $4$ as a fraction with a common denominator, multiply by $\frac{2 + 3 x}{2 + 3 x}$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{4 (2 + 3 x)}{2 + 3 x} - \frac{12 x}{2 + 3 x}}$

Step 5.3.4.4

Combine the numerators over the common denominator.

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{4 (2 + 3 x) - 12 x}{2 + 3 x}}$

Step 5.3.4.5

Rewrite $\frac{4 (2 + 3 x) - 12 x}{2 + 3 x}$ in a factored form.

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Step 5.3.4.5.1

Factor $4$ out of $4 (2 + 3 x) - 12 x$ .

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Step 5.3.4.5.1.1

Factor $4$ out of $- 12 x$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{4 (2 + 3 x) + 4 (- 3 x)}{2 + 3 x}}$

Step 5.3.4.5.1.2

Factor $4$ out of $4 (2 + 3 x) + 4 (- 3 x)$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{4 (2 + 3 x - 3 x)}{2 + 3 x}}$

Step 5.3.4.5.2

Subtract $3 x$ from $3 x$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{4 (2 + 0)}{2 + 3 x}}$

Step 5.3.4.5.3

Add $2$ and $0$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{4 \cdot 2}{2 + 3 x}}$

Step 5.3.4.6

Multiply $4$ by $2$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{2 (4 x)}{2 + 3 x}}{\frac{8}{2 + 3 x}}$

Step 5.3.5

Multiply $2$ by $4$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{\frac{8 x}{2 + 3 x}}{\frac{8}{2 + 3 x}}$

Step 5.3.6

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{4 x}{2 + 3 x}) = \frac{8 x}{2 + 3 x} \cdot \frac{2 + 3 x}{8}$

Step 5.3.7

Cancel the common factor of $8$ .

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Step 5.3.7.1

Factor $8$ out of $8 x$ .

$f (\frac{4 x}{2 + 3 x}) = \frac{8 (x)}{2 + 3 x} \cdot \frac{2 + 3 x}{8}$

Step 5.3.7.2

Cancel the common factor.

$f (\frac{4 x}{2 + 3 x}) = \frac{8 x}{2 + 3 x} \cdot \frac{2 + 3 x}{8}$

Step 5.3.7.3

Rewrite the expression.

$f (\frac{4 x}{2 + 3 x}) = \frac{x}{2 + 3 x} \cdot (2 + 3 x)$

Step 5.3.8

Cancel the common factor of $2 + 3 x$ .

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Step 5.3.8.1

Cancel the common factor.

$f (\frac{4 x}{2 + 3 x}) = \frac{x}{2 + 3 x} \cdot (2 + 3 x)$

Step 5.3.8.2

Rewrite the expression.

$f (\frac{4 x}{2 + 3 x}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{4 x}{2 + 3 x}$ is the inverse of $f (x) = \frac{2 x}{4 - 3 x}$ .

$f^{- 1} (x) = \frac{4 x}{2 + 3 x}$