Precalculus Examples

$f (x) = \frac{2 x + 1}{x}$

Step 1

Write $f (x) = \frac{2 x + 1}{x}$ as an equation.

$y = \frac{2 x + 1}{x}$

Step 2

Interchange the variables.

$x = \frac{2 y + 1}{y}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{2 y + 1}{y} = x$ .

$\frac{2 y + 1}{y} = x$

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.

$y, 1$

Step 3.2.2

The LCM of one and any expression is the expression.

$y$

Step 3.3

Multiply each term in $\frac{2 y + 1}{y} = x$ by $y$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{2 y + 1}{y} = x$ by $y$ .

$\frac{2 y + 1}{y} y = x y$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $y$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{2 y + 1}{y} y = x y$

Step 3.3.2.1.2

Rewrite the expression.

$2 y + 1 = x y$

Step 3.4

Solve the equation.

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Step 3.4.1

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

$2 y + 1 - x y = 0$

Step 3.4.2

Subtract $1$ from both sides of the equation.

$2 y - x y = - 1$

Step 3.4.3

Factor $y$ out of $2 y - x y$ .

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Step 3.4.3.1

Factor $y$ out of $2 y$ .

$y \cdot 2 - x y = - 1$

Step 3.4.3.2

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

$y \cdot 2 + y (- x) = - 1$

Step 3.4.3.3

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

$y (2 - x) = - 1$

Step 3.4.4

Divide each term in $y (2 - x) = - 1$ by $2 - x$ and simplify.

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Step 3.4.4.1

Divide each term in $y (2 - x) = - 1$ by $2 - x$ .

$\frac{y (2 - x)}{2 - x} = \frac{- 1}{2 - x}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Cancel the common factor of $2 - x$ .

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Step 3.4.4.2.1.1

Cancel the common factor.

$\frac{y (2 - x)}{2 - x} = \frac{- 1}{2 - x}$

Step 3.4.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 1}{2 - x}$

Step 3.4.4.3

Simplify the right side.

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Step 3.4.4.3.1

Move the negative in front of the fraction.

$y = - \frac{1}{2 - x}$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify the denominator.

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Step 5.2.3.1

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

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

Step 5.2.3.2

Combine the numerators over the common denominator.

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

Step 5.2.3.3

Rewrite $\frac{2 x - (2 x + 1)}{x}$ in a factored form.

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Step 5.2.3.3.1

Apply the distributive property.

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

Step 5.2.3.3.2

Multiply $2$ by $- 1$ .

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

Step 5.2.3.3.3

Multiply $- 1$ by $1$ .

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

Step 5.2.3.3.4

Subtract $2 x$ from $2 x$ .

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

Step 5.2.3.3.5

Subtract $1$ from $0$ .

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

Step 5.2.3.4

Move the negative in front of the fraction.

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

Step 5.2.4

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

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Step 5.2.4.1

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

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

Step 5.2.4.2

Move the negative in front of the fraction.

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

Step 5.2.5

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.2.6

Multiply $x$ by $1$ .

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

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

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

Step 5.3.3

Multiply the numerator by the reciprocal of the denominator.

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

Step 5.3.4

Simplify each term.

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Step 5.3.4.1

Multiply $2 (- \frac{1}{2 - x})$ .

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Step 5.3.4.1.1

Multiply $- 1$ by $2$ .

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

Step 5.3.4.1.2

Combine $- 2$ and $\frac{1}{2 - x}$ .

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

Step 5.3.4.2

Move the negative in front of the fraction.

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

Step 5.3.5

Simplify terms.

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Step 5.3.5.1

Write $1$ as a fraction with a common denominator.

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

Step 5.3.5.2

Combine the numerators over the common denominator.

$f (- \frac{1}{2 - x}) = \frac{- 2 + 2 - x}{2 - x} \cdot (- (2 - x))$

Step 5.3.5.3

Rewrite using the commutative property of multiplication.

$f (- \frac{1}{2 - x}) = - \frac{- 2 + 2 - x}{2 - x} \cdot (2 - x)$

Step 5.3.5.4

Cancel the common factor of $2 - x$ .

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Step 5.3.5.4.1

Move the leading negative in $- \frac{- 2 + 2 - x}{2 - x}$ into the numerator.

$f (- \frac{1}{2 - x}) = \frac{- (- 2 + 2 - x)}{2 - x} \cdot (2 - x)$

Step 5.3.5.4.2

Cancel the common factor.

$f (- \frac{1}{2 - x}) = \frac{- (- 2 + 2 - x)}{2 - x} \cdot (2 - x)$

Step 5.3.5.4.3

Rewrite the expression.

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

Step 5.3.5.5

Simplify by adding numbers.

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Step 5.3.5.5.1

Add $- 2$ and $2$ .

$f (- \frac{1}{2 - x}) = - (0 - x)$

Step 5.3.5.5.2

Subtract $x$ from $0$ .

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

Step 5.4

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

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