Calculus Examples

$f (x) = \frac{6 x - 1}{2 x + 5}$

Step 1

Write $f (x) = \frac{6 x - 1}{2 x + 5}$ as an equation.

$y = \frac{6 x - 1}{2 x + 5}$

Step 2

Interchange the variables.

$x = \frac{6 y - 1}{2 y + 5}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{6 y - 1}{2 y + 5} = x$ .

$\frac{6 y - 1}{2 y + 5} = x$

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 $\frac{6 y - 1}{2 y + 5} = x$ by $2 y + 5$ to eliminate the fractions.

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Step 3.3.1

Multiply each term in $\frac{6 y - 1}{2 y + 5} = x$ by $2 y + 5$ .

$\frac{6 y - 1}{2 y + 5} (2 y + 5) = x (2 y + 5)$

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.

$\frac{6 y - 1}{2 y + 5} (2 y + 5) = x (2 y + 5)$

Step 3.3.2.1.2

Rewrite the expression.

$6 y - 1 = x (2 y + 5)$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Apply the distributive property.

$6 y - 1 = x (2 y) + x \cdot 5$

Step 3.3.3.2

Reorder.

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Step 3.3.3.2.1

Rewrite using the commutative property of multiplication.

$6 y - 1 = 2 x y + x \cdot 5$

Step 3.3.3.2.2

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

$6 y - 1 = 2 x y + 5 x$

Step 3.4

Solve the equation.

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Step 3.4.1

Subtract $2 x y$ from both sides of the equation.

$6 y - 1 - 2 x y = 5 x$

Step 3.4.2

Add $1$ to both sides of the equation.

$6 y - 2 x y = 5 x + 1$

Step 3.4.3

Factor $2 y$ out of $6 y - 2 x y$ .

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Step 3.4.3.1

Factor $2 y$ out of $6 y$ .

$2 y (3) - 2 x y = 5 x + 1$

Step 3.4.3.2

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

$2 y (3) + 2 y (- x) = 5 x + 1$

Step 3.4.3.3

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

$2 y (3 - x) = 5 x + 1$

Step 3.4.4

Divide each term in $2 y (3 - x) = 5 x + 1$ by $2 (3 - x)$ and simplify.

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Step 3.4.4.1

Divide each term in $2 y (3 - x) = 5 x + 1$ by $2 (3 - x)$ .

$\frac{2 y (3 - x)}{2 (3 - x)} = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$

Step 3.4.4.2

Simplify the left side.

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Step 3.4.4.2.1

Cancel the common factor of $2$ .

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Step 3.4.4.2.1.1

Cancel the common factor.

$\frac{2 y (3 - x)}{2 (3 - x)} = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$

Step 3.4.4.2.1.2

Rewrite the expression.

$\frac{y (3 - x)}{3 - x} = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$

Step 3.4.4.2.2

Cancel the common factor of $3 - x$ .

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Step 3.4.4.2.2.1

Cancel the common factor.

$\frac{y (3 - x)}{3 - x} = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$

Step 3.4.4.2.2.2

Divide $y$ by $1$ .

$y = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$

Step 4

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

$f^{- 1} (x) = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$

Step 5

Verify if $f^{- 1} (x) = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$ is the inverse of $f (x) = \frac{6 x - 1}{2 x + 5}$ .

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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 $f^{- 1} (\frac{6 x - 1}{2 x + 5})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{5 (\frac{6 x - 1}{2 x + 5})}{2 (3 - (\frac{6 x - 1}{2 x + 5}))} + \frac{1}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.3

Simplify terms.

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Step 5.2.3.1

Combine the numerators over the common denominator.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{5 (\frac{6 x - 1}{2 x + 5}) + 1}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.3.2

Combine $5$ and $\frac{6 x - 1}{2 x + 5}$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{5 (6 x - 1)}{2 x + 5} + 1}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.3.3

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

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{5 (6 x - 1)}{2 x + 5} + \frac{2 x + 5}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.3.4

Combine the numerators over the common denominator.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{5 (6 x - 1) + 2 x + 5}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.4

Simplify the numerator.

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Step 5.2.4.1

Apply the distributive property.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{5 (6 x) + 5 \cdot - 1 + 2 x + 5}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.4.2

Multiply $6$ by $5$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{30 x + 5 \cdot - 1 + 2 x + 5}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.4.3

Multiply $5$ by $- 1$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{30 x - 5 + 2 x + 5}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.4.4

Add $30 x$ and $2 x$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{32 x - 5 + 5}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.4.5

Add $- 5$ and $5$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{32 x + 0}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.4.6

Add $32 x$ and $0$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{\frac{32 x}{2 x + 5}}{2 (3 - (\frac{6 x - 1}{2 x + 5}))}$

Step 5.2.5

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{32 x}{2 x + 5} \cdot \frac{1}{2 (3 - \frac{6 x - 1}{2 x + 5})}$

Step 5.2.6

Simplify terms.

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Step 5.2.6.1

Cancel the common factor of $2$ .

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Step 5.2.6.1.1

Factor $2$ out of $32 x$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{2 (16 x)}{2 x + 5} \cdot \frac{1}{2 (3 - \frac{6 x - 1}{2 x + 5})}$

Step 5.2.6.1.2

Cancel the common factor.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{2 (16 x)}{2 x + 5} \cdot \frac{1}{2 (3 - \frac{6 x - 1}{2 x + 5})}$

Step 5.2.6.1.3

Rewrite the expression.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{2 x + 5} \cdot \frac{1}{3 - \frac{6 x - 1}{2 x + 5}}$

Step 5.2.6.2

Multiply $\frac{16 x}{2 x + 5}$ by $\frac{1}{3 - \frac{6 x - 1}{2 x + 5}}$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (3 - \frac{6 x - 1}{2 x + 5})}$

Step 5.2.7

Simplify the denominator.

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Step 5.2.7.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{2 x + 5}{2 x + 5}$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{3 (2 x + 5)}{2 x + 5} - \frac{6 x - 1}{2 x + 5})}$

Step 5.2.7.2

Combine the numerators over the common denominator.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{3 (2 x + 5) - (6 x - 1)}{2 x + 5})}$

Step 5.2.7.3

Rewrite $\frac{3 (2 x + 5) - (6 x - 1)}{2 x + 5}$ in a factored form.

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Step 5.2.7.3.1

Apply the distributive property.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{3 (2 x) + 3 \cdot 5 - (6 x - 1)}{2 x + 5})}$

Step 5.2.7.3.2

Multiply $2$ by $3$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{6 x + 3 \cdot 5 - (6 x - 1)}{2 x + 5})}$

Step 5.2.7.3.3

Multiply $3$ by $5$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{6 x + 15 - (6 x - 1)}{2 x + 5})}$

Step 5.2.7.3.4

Apply the distributive property.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{6 x + 15 - (6 x) + 1}{2 x + 5})}$

Step 5.2.7.3.5

Multiply $6$ by $- 1$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{6 x + 15 - 6 x + 1}{2 x + 5})}$

Step 5.2.7.3.6

Multiply $- 1$ by $- 1$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{6 x + 15 - 6 x + 1}{2 x + 5})}$

Step 5.2.7.3.7

Subtract $6 x$ from $6 x$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{0 + 15 + 1}{2 x + 5})}$

Step 5.2.7.3.8

Add $0$ and $15$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{15 + 1}{2 x + 5})}$

Step 5.2.7.3.9

Add $15$ and $1$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{16 x}{(2 x + 5) (\frac{16}{2 x + 5})}$

Step 5.2.8

Factor $\frac{16}{2 x + 5}$ out of $\frac{16 x}{(2 x + 5) \frac{16}{2 x + 5}}$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{2 x + 5}{16} \cdot \frac{16 x}{2 x + 5}$

Step 5.2.9

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

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Step 5.2.9.1

Cancel the common factor.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{2 x + 5}{16} \cdot \frac{16 x}{2 x + 5}$

Step 5.2.9.2

Rewrite the expression.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{1}{16} \cdot (16 x)$

Step 5.2.10

Cancel the common factor of $16$ .

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Step 5.2.10.1

Factor $16$ out of $16 x$ .

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{1}{16} \cdot (16 (x))$

Step 5.2.10.2

Cancel the common factor.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = \frac{1}{16} \cdot (16 x)$

Step 5.2.10.3

Rewrite the expression.

$f^{- 1} (\frac{6 x - 1}{2 x + 5}) = x$

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 $f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{6 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3

Simplify the numerator.

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Step 5.3.3.1

Apply the distributive property.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{6 (\frac{5 x}{2 (3 - x)}) + 6 (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

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 $6$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{2 (3) (\frac{5 x}{2 (3 - x)}) + 6 (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.2.2

Cancel the common factor.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{2 \cdot (3 (\frac{5 x}{2 (3 - x)})) + 6 (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.2.3

Rewrite the expression.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{3 (\frac{5 x}{3 - x}) + 6 (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.3

Combine $3$ and $\frac{5 x}{3 - x}$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x)}{3 - x} + 6 (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.4

Multiply $5$ by $3$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x}{3 - x} + 6 (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.5

Cancel the common factor of $2$ .

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Step 5.3.3.5.1

Factor $2$ out of $6$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x}{3 - x} + 2 (3) (\frac{1}{2 (3 - x)}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.5.2

Cancel the common factor.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x}{3 - x} + 2 \cdot (3 (\frac{1}{2 (3 - x)})) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.5.3

Rewrite the expression.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x}{3 - x} + 3 (\frac{1}{3 - x}) - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.6

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

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x}{3 - x} + \frac{3}{3 - x} - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.7

Combine the numerators over the common denominator.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3}{3 - x} - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.8

Factor $3$ out of $15 x + 3$ .

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Step 5.3.3.8.1

Factor $3$ out of $15 x$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x) + 3}{3 - x} - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.8.2

Factor $3$ out of $3$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x) + 3 (1)}{3 - x} - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.8.3

Factor $3$ out of $3 (5 x) + 3 (1)$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x + 1)}{3 - x} - 1}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.9

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

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x + 1)}{3 - x} - 1 \cdot \frac{3 - x}{3 - x}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.10

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

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x + 1)}{3 - x} + \frac{- (3 - x)}{3 - x}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.11

Combine the numerators over the common denominator.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x + 1) - (3 - x)}{3 - x}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.12

Reorder terms.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x + 1) - (3 - x)}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13

Rewrite $\frac{3 (5 x + 1) - (3 - x)}{- x + 3}$ in a factored form.

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Step 5.3.3.13.1

Apply the distributive property.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{3 (5 x) + 3 \cdot 1 - (3 - x)}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.2

Multiply $5$ by $3$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3 \cdot 1 - (3 - x)}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.3

Multiply $3$ by $1$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3 - (3 - x)}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.4

Apply the distributive property.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3 - 1 \cdot 3 + x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.5

Multiply $- 1$ by $3$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3 - 3 + x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.6

Multiply $- - x$ .

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Step 5.3.3.13.6.1

Multiply $- 1$ by $- 1$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3 - 3 + 1 x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.6.2

Multiply $x$ by $1$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{15 x + 3 - 3 + x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.7

Add $15 x$ and $x$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x + 3 - 3}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.8

Subtract $3$ from $3$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x + 0}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.3.13.9

Add $16 x$ and $0$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) + 5}$

Step 5.3.4

Simplify the denominator.

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Step 5.3.4.1

Apply the distributive property.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)}) + 2 (\frac{1}{2 (3 - x)}) + 5}$

Step 5.3.4.2

Cancel the common factor of $2$ .

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Step 5.3.4.2.1

Cancel the common factor.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{2 (\frac{5 x}{2 (3 - x)}) + 2 (\frac{1}{2 (3 - x)}) + 5}$

Step 5.3.4.2.2

Rewrite the expression.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x}{3 - x} + 2 (\frac{1}{2 (3 - x)}) + 5}$

Step 5.3.4.3

Cancel the common factor of $2$ .

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Step 5.3.4.3.1

Cancel the common factor.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x}{3 - x} + 2 (\frac{1}{2 (3 - x)}) + 5}$

Step 5.3.4.3.2

Rewrite the expression.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x}{3 - x} + \frac{1}{3 - x} + 5}$

Step 5.3.4.4

Combine the numerators over the common denominator.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 1}{3 - x} + 5}$

Step 5.3.4.5

To write $5$ as a fraction with a common denominator, multiply by $\frac{3 - x}{3 - x}$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 1}{3 - x} + \frac{5 (3 - x)}{3 - x}}$

Step 5.3.4.6

Combine the numerators over the common denominator.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 1 + 5 (3 - x)}{3 - x}}$

Step 5.3.4.7

Reorder terms.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 5 (3 - x) + 1}{- x + 3}}$

Step 5.3.4.8

Rewrite $\frac{5 x + 5 (3 - x) + 1}{- x + 3}$ in a factored form.

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Step 5.3.4.8.1

Apply the distributive property.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 5 \cdot 3 + 5 (- x) + 1}{- x + 3}}$

Step 5.3.4.8.2

Multiply $5$ by $3$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 15 + 5 (- x) + 1}{- x + 3}}$

Step 5.3.4.8.3

Multiply $- 1$ by $5$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{5 x + 15 - 5 x + 1}{- x + 3}}$

Step 5.3.4.8.4

Subtract $5 x$ from $5 x$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{0 + 15 + 1}{- x + 3}}$

Step 5.3.4.8.5

Add $0$ and $15$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{15 + 1}{- x + 3}}$

Step 5.3.4.8.6

Add $15$ and $1$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{\frac{16 x}{- x + 3}}{\frac{16}{- x + 3}}$

Step 5.3.5

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{16 x}{- x + 3} \cdot \frac{- x + 3}{16}$

Step 5.3.6

Cancel the common factor of $16$ .

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Step 5.3.6.1

Factor $16$ out of $16 x$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{16 (x)}{- x + 3} \cdot \frac{- x + 3}{16}$

Step 5.3.6.2

Cancel the common factor.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{16 x}{- x + 3} \cdot \frac{- x + 3}{16}$

Step 5.3.6.3

Rewrite the expression.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{x}{- x + 3} \cdot (- x + 3)$

Step 5.3.7

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

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{x (- x + 3)}{- x + 3}$

Step 5.3.8

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

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Step 5.3.8.1

Cancel the common factor.

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = \frac{x (- x + 3)}{- x + 3}$

Step 5.3.8.2

Divide $x$ by $1$ .

$f (\frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}) = x$

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$ is the inverse of $f (x) = \frac{6 x - 1}{2 x + 5}$ .

$f^{- 1} (x) = \frac{5 x}{2 (3 - x)} + \frac{1}{2 (3 - x)}$