Precalculus Examples

$f (x) = \frac{1}{2 x + 5}$

Step 1

Write $f (x) = \frac{1}{2 x + 5}$ as an equation.

$y = \frac{1}{2 x + 5}$

Step 2

Interchange the variables.

$x = \frac{1}{2 y + 5}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{1}{2 y + 5} = x$ .

$\frac{1}{2 y + 5} = x$

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.

$2 y + 5, 1$

Step 3.2.2

Remove parentheses.

$2 y + 5, 1$

Step 3.2.3

The LCM of one and any expression is the expression.

$2 y + 5$

Step 3.3

Multiply each term in $\frac{1}{2 y + 5} = x$ by $2 y + 5$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{1}{2 y + 5} = x$ by $2 y + 5$ .

$\frac{1}{2 y + 5} (2 y + 5) = x (2 y + 5)$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $2 y + 5$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{1}{2 y + 5} (2 y + 5) = x (2 y + 5)$

Step 3.3.2.1.2

Rewrite the expression.

$1 = x (2 y + 5)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$1 = x (2 y) + x \cdot 5$

Step 3.3.3.2

Reorder.

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Step 3.3.3.2.1

Rewrite using the commutative property of multiplication.

$1 = 2 x y + x \cdot 5$

Step 3.3.3.2.2

Move $5$ to the left of $x$ .

$1 = 2 x y + 5 x$

Step 3.4

Solve the equation.

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Step 3.4.1

Rewrite the equation as $2 x y + 5 x = 1$ .

$2 x y + 5 x = 1$

Step 3.4.2

Subtract $5 x$ from both sides of the equation.

$2 x y = 1 - 5 x$

Step 3.4.3

Divide each term in $2 x y = 1 - 5 x$ by $2 x$ and simplify.

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Step 3.4.3.1

Divide each term in $2 x y = 1 - 5 x$ by $2 x$ .

$\frac{2 x y}{2 x} = \frac{1}{2 x} + \frac{- 5 x}{2 x}$

Step 3.4.3.2

Simplify the left side.

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Step 3.4.3.2.1

Cancel the common factor of $2$ .

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Step 3.4.3.2.1.1

Cancel the common factor.

$\frac{2 x y}{2 x} = \frac{1}{2 x} + \frac{- 5 x}{2 x}$

Step 3.4.3.2.1.2

Rewrite the expression.

$\frac{x y}{x} = \frac{1}{2 x} + \frac{- 5 x}{2 x}$

Step 3.4.3.2.2

Cancel the common factor of $x$ .

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Step 3.4.3.2.2.1

Cancel the common factor.

$\frac{x y}{x} = \frac{1}{2 x} + \frac{- 5 x}{2 x}$

Step 3.4.3.2.2.2

Divide $y$ by $1$ .

$y = \frac{1}{2 x} + \frac{- 5 x}{2 x}$

Step 3.4.3.3

Simplify the right side.

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Step 3.4.3.3.1

Simplify each term.

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Step 3.4.3.3.1.1

Cancel the common factor of $x$ .

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Step 3.4.3.3.1.1.1

Cancel the common factor.

$y = \frac{1}{2 x} + \frac{- 5 x}{2 x}$

Step 3.4.3.3.1.1.2

Rewrite the expression.

$y = \frac{1}{2 x} + \frac{- 5}{2}$

Step 3.4.3.3.1.2

Move the negative in front of the fraction.

$y = \frac{1}{2 x} - \frac{5}{2}$

Step 4

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

$f^{- 1} (x) = \frac{1}{2 x} - \frac{5}{2}$

Step 5

Verify if $f^{- 1} (x) = \frac{1}{2 x} - \frac{5}{2}$ is the inverse of $f (x) = \frac{1}{2 x + 5}$ .

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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 $f^{- 1} (\frac{1}{2 x + 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{1}{2 (\frac{1}{2 x + 5})} - \frac{5}{2}$

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Combine $2$ and $\frac{1}{2 x + 5}$ .

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{1}{\frac{2}{2 x + 5}} - \frac{5}{2}$

Step 5.2.3.2

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{1}{2 x + 5}) = 1 (\frac{2 x + 5}{2}) - \frac{5}{2}$

Step 5.2.3.3

Multiply $\frac{2 x + 5}{2}$ by $1$ .

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{2 x + 5}{2} - \frac{5}{2}$

Step 5.2.4

Simplify terms.

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Step 5.2.4.1

Combine the numerators over the common denominator.

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{2 x + 5 - 5}{2}$

Step 5.2.4.2

Combine the opposite terms in $2 x + 5 - 5$ .

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Step 5.2.4.2.1

Subtract $5$ from $5$ .

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{2 x + 0}{2}$

Step 5.2.4.2.2

Add $2 x$ and $0$ .

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{2 x}{2}$

Step 5.2.4.3

Cancel the common factor of $2$ .

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Step 5.2.4.3.1

Cancel the common factor.

$f^{- 1} (\frac{1}{2 x + 5}) = \frac{2 x}{2}$

Step 5.2.4.3.2

Divide $x$ by $1$ .

$f^{- 1} (\frac{1}{2 x + 5}) = x$

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 $f (\frac{1}{2 x} - \frac{5}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{2 (\frac{1}{2 x} - \frac{5}{2}) + 5}$

Step 5.3.3

Simplify the denominator.

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Step 5.3.3.1

Apply the distributive property.

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{2 (\frac{1}{2 x}) + 2 (- \frac{5}{2}) + 5}$

Step 5.3.3.2

Cancel the common factor of $2$ .

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Step 5.3.3.2.1

Factor $2$ out of $2 x$ .

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{2 (\frac{1}{2 (x)}) + 2 (- \frac{5}{2}) + 5}$

Step 5.3.3.2.2

Cancel the common factor.

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{2 (\frac{1}{2 x}) + 2 (- \frac{5}{2}) + 5}$

Step 5.3.3.2.3

Rewrite the expression.

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{\frac{1}{x} + 2 (- \frac{5}{2}) + 5}$

Step 5.3.3.3

Cancel the common factor of $2$ .

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Step 5.3.3.3.1

Move the leading negative in $- \frac{5}{2}$ into the numerator.

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{\frac{1}{x} + 2 (\frac{- 5}{2}) + 5}$

Step 5.3.3.3.2

Cancel the common factor.

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{\frac{1}{x} + 2 (\frac{- 5}{2}) + 5}$

Step 5.3.3.3.3

Rewrite the expression.

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{\frac{1}{x} - 5 + 5}$

Step 5.3.3.4

Add $- 5$ and $5$ .

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{\frac{1}{x} + 0}$

Step 5.3.3.5

Add $\frac{1}{x}$ and $0$ .

$f (\frac{1}{2 x} - \frac{5}{2}) = \frac{1}{\frac{1}{x}}$

Step 5.3.4

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{1}{2 x} - \frac{5}{2}) = 1 x$

Step 5.3.5

Multiply $x$ by $1$ .

$f (\frac{1}{2 x} - \frac{5}{2}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{1}{2 x} - \frac{5}{2}$ is the inverse of $f (x) = \frac{1}{2 x + 5}$ .

$f^{- 1} (x) = \frac{1}{2 x} - \frac{5}{2}$