Trigonometry Examples

$f (x) - \frac{3}{5} x + 9$

Step 1

Interchange the variables.

$f = y (x) - \frac{3}{5} x + 9$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $y x - \frac{3}{5} x + 9 = f$ .

$y x - \frac{3}{5} x + 9 = f$

Step 2.2

Simplify each term.

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Step 2.2.1

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

$y x - \frac{x \cdot 3}{5} + 9 = f$

Step 2.2.2

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

$y x - \frac{3 x}{5} + 9 = f$

Step 2.3

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

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Step 2.3.1

Add $\frac{3 x}{5}$ to both sides of the equation.

$y x + 9 = f + \frac{3 x}{5}$

Step 2.3.2

Subtract $9$ from both sides of the equation.

$y x = f + \frac{3 x}{5} - 9$

Step 2.4

Divide each term in $y x = f + \frac{3 x}{5} - 9$ by $x$ and simplify.

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Step 2.4.1

Divide each term in $y x = f + \frac{3 x}{5} - 9$ by $x$ .

$\frac{y x}{x} = \frac{f}{x} + \frac{\frac{3 x}{5}}{x} + \frac{- 9}{x}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $x$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{y x}{x} = \frac{f}{x} + \frac{\frac{3 x}{5}}{x} + \frac{- 9}{x}$

Step 2.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{f}{x} + \frac{\frac{3 x}{5}}{x} + \frac{- 9}{x}$

Step 2.4.3

Simplify the right side.

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Step 2.4.3.1

Simplify each term.

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Step 2.4.3.1.1

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{f}{x} + \frac{3 x}{5} \cdot \frac{1}{x} + \frac{- 9}{x}$

Step 2.4.3.1.2

Cancel the common factor of $x$ .

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Step 2.4.3.1.2.1

Factor $x$ out of $3 x$ .

$y = \frac{f}{x} + \frac{x \cdot 3}{5} \cdot \frac{1}{x} + \frac{- 9}{x}$

Step 2.4.3.1.2.2

Cancel the common factor.

$y = \frac{f}{x} + \frac{x \cdot 3}{5} \cdot \frac{1}{x} + \frac{- 9}{x}$

Step 2.4.3.1.2.3

Rewrite the expression.

$y = \frac{f}{x} + \frac{3}{5} + \frac{- 9}{x}$

Step 2.4.3.1.3

Move the negative in front of the fraction.

$y = \frac{f}{x} + \frac{3}{5} - \frac{9}{x}$

Step 3

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

$f^{- 1} (f) = \frac{f}{x} + \frac{3}{5} - \frac{9}{x}$

Step 4

Verify if $f^{- 1} (f) = \frac{f}{x} + \frac{3}{5} - \frac{9}{x}$ is the inverse of $f (f) = f (x) - \frac{3}{5} x + 9$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} (f (x) - \frac{3}{5} x + 9)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x) - \frac{3}{5} x + 9}{x} + \frac{3}{5} - \frac{9}{x}$

Step 4.2.3

Simplify terms.

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Step 4.2.3.1

Combine the numerators over the common denominator.

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x) - \frac{3}{5} x + 9 - 9}{x} + \frac{3}{5}$

Step 4.2.3.2

Combine the opposite terms in $f (x) - \frac{3}{5} x + 9 - 9$ .

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Step 4.2.3.2.1

Subtract $9$ from $9$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x) - \frac{3}{5} x + 0}{x} + \frac{3}{5}$

Step 4.2.3.2.2

Add $f (x) - \frac{3}{5} x$ and $0$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x) - \frac{3}{5} x}{x} + \frac{3}{5}$

Step 4.2.4

Simplify each term.

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Step 4.2.4.1

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

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x) - \frac{x \cdot 3}{5}}{x} + \frac{3}{5}$

Step 4.2.4.2

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

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x) - \frac{3 x}{5}}{x} + \frac{3}{5}$

Step 4.2.5

Reorder $f (x)$ and $- \frac{3 x}{5}$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- \frac{3 x}{5} + f (x)}{x} + \frac{3}{5}$

Step 4.2.6

Simplify each term.

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Step 4.2.6.1

Simplify the numerator.

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Step 4.2.6.1.1

To write $f (x)$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- \frac{3 x}{5} + f (x) \cdot \frac{5}{5}}{x} + \frac{3}{5}$

Step 4.2.6.1.2

Combine $f (x)$ and $\frac{5}{5}$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- \frac{3 x}{5} + \frac{f (x) \cdot 5}{5}}{x} + \frac{3}{5}$

Step 4.2.6.1.3

Combine the numerators over the common denominator.

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{\frac{- 3 x + f (x) \cdot 5}{5}}{x} + \frac{3}{5}$

Step 4.2.6.1.4

Move $5$ to the left of $f (x)$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{\frac{- 3 x + 5 f (x)}{5}}{x} + \frac{3}{5}$

Step 4.2.6.2

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- 3 x + 5 f (x)}{5} \cdot \frac{1}{x} + \frac{3}{5}$

Step 4.2.6.3

Multiply $\frac{- 3 x + 5 f (x)}{5}$ by $\frac{1}{x}$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- 3 x + 5 f (x)}{5 x} + \frac{3}{5}$

Step 4.2.7

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

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- 3 x + 5 f (x)}{5 x} + \frac{3}{5} \cdot \frac{x}{x}$

Step 4.2.8

Simplify terms.

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Step 4.2.8.1

Multiply $\frac{3}{5}$ by $\frac{x}{x}$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- 3 x + 5 f (x)}{5 x} + \frac{3 x}{5 x}$

Step 4.2.8.2

Combine the numerators over the common denominator.

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{- 3 x + 5 f (x) + 3 x}{5 x}$

Step 4.2.9

Simplify the numerator.

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Step 4.2.9.1

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

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{5 f (x) + 0}{5 x}$

Step 4.2.9.2

Add $5 f (x)$ and $0$ .

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{5 f (x)}{5 x}$

Step 4.2.10

Cancel the common factor of $5$ .

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Step 4.2.10.1

Cancel the common factor.

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{5 f (x)}{5 x}$

Step 4.2.10.2

Rewrite the expression.

$f^{- 1} (f (x) - \frac{3}{5} x + 9) = \frac{f (x)}{x}$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) (x) - \frac{3}{5} x + 9$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = \frac{f}{x} \cdot x + \frac{3}{5} x - \frac{9}{x} \cdot x - \frac{3}{5} x + 9$

Step 4.3.3.2

Simplify.

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Step 4.3.3.2.1

Cancel the common factor of $x$ .

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Step 4.3.3.2.1.1

Cancel the common factor.

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = \frac{f}{x} \cdot x + \frac{3}{5} x - \frac{9}{x} \cdot x - \frac{3}{5} x + 9$

Step 4.3.3.2.1.2

Rewrite the expression.

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3}{5} x - \frac{9}{x} \cdot x - \frac{3}{5} x + 9$

Step 4.3.3.2.2

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

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3 x}{5} - \frac{9}{x} \cdot x - \frac{3}{5} x + 9$

Step 4.3.3.2.3

Cancel the common factor of $x$ .

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Step 4.3.3.2.3.1

Move the leading negative in $- \frac{9}{x}$ into the numerator.

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3 x}{5} + \frac{- 9}{x} \cdot x - \frac{3}{5} x + 9$

Step 4.3.3.2.3.2

Cancel the common factor.

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3 x}{5} + \frac{- 9}{x} \cdot x - \frac{3}{5} x + 9$

Step 4.3.3.2.3.3

Rewrite the expression.

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3 x}{5} - 9 - \frac{3}{5} x + 9$

Step 4.3.3.3

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

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3 x}{5} - 9 - \frac{x \cdot 3}{5} + 9$

Step 4.3.3.4

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

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + \frac{3 x}{5} - 9 - \frac{3 x}{5} + 9$

Step 4.3.4

Combine the opposite terms in $f + \frac{3 x}{5} - 9 - \frac{3 x}{5} + 9$ .

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Step 4.3.4.1

Subtract $\frac{3 x}{5}$ from $\frac{3 x}{5}$ .

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + 0 - 9 + 9$

Step 4.3.4.2

Add $f$ and $0$ .

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f - 9 + 9$

Step 4.3.4.3

Add $- 9$ and $9$ .

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f + 0$

Step 4.3.4.4

Add $f$ and $0$ .

$f (\frac{f}{x} + \frac{3}{5} - \frac{9}{x}) = f$

Step 4.4

Since $f^{- 1} (f (f)) = f$ and $f (f^{- 1} (f)) = f$ , then $f^{- 1} (f) = \frac{f}{x} + \frac{3}{5} - \frac{9}{x}$ is the inverse of $f (f) = f (x) - \frac{3}{5} x + 9$ .

$f^{- 1} (f) = \frac{f}{x} + \frac{3}{5} - \frac{9}{x}$