Trigonometry Examples

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

Step 1

Write $f (x) = - \frac{4}{3} x + \frac{8}{3}$ as an equation.

$y = - \frac{4}{3} x + \frac{8}{3}$

Step 2

Interchange the variables.

$x = - \frac{4}{3} y + \frac{8}{3}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $- \frac{4}{3} y + \frac{8}{3} = x$ .

$- \frac{4}{3} y + \frac{8}{3} = x$

Step 3.2

Simplify each term.

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Step 3.2.1

Combine $y$ and $\frac{4}{3}$ .

$- \frac{y \cdot 4}{3} + \frac{8}{3} = x$

Step 3.2.2

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

$- \frac{4 y}{3} + \frac{8}{3} = x$

Step 3.3

Subtract $\frac{8}{3}$ from both sides of the equation.

$- \frac{4 y}{3} = x - \frac{8}{3}$

Step 3.4

Multiply both sides of the equation by $- \frac{3}{4}$ .

$- \frac{3}{4} (- \frac{4 y}{3}) = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5

Simplify both sides of the equation.

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Step 3.5.1

Simplify the left side.

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Step 3.5.1.1

Simplify $- \frac{3}{4} (- \frac{4 y}{3})$ .

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Step 3.5.1.1.1

Cancel the common factor of $3$ .

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Step 3.5.1.1.1.1

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$\frac{- 3}{4} (- \frac{4 y}{3}) = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.1.2

Move the leading negative in $- \frac{4 y}{3}$ into the numerator.

$\frac{- 3}{4} \cdot \frac{- 4 y}{3} = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.1.3

Factor $3$ out of $- 3$ .

$\frac{3 (- 1)}{4} \cdot \frac{- 4 y}{3} = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.1.4

Cancel the common factor.

$\frac{3 \cdot - 1}{4} \cdot \frac{- 4 y}{3} = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.1.5

Rewrite the expression.

$\frac{- 1}{4} (- 4 y) = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.2

Cancel the common factor of $4$ .

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Step 3.5.1.1.2.1

Factor $4$ out of $- 4 y$ .

$\frac{- 1}{4} (4 (- y)) = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.2.2

Cancel the common factor.

$\frac{- 1}{4} (4 (- y)) = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.2.3

Rewrite the expression.

$- - y = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.3

Multiply.

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Step 3.5.1.1.3.1

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.1.1.3.2

Multiply $y$ by $1$ .

$y = - \frac{3}{4} (x - \frac{8}{3})$

Step 3.5.2

Simplify the right side.

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Step 3.5.2.1

Simplify $- \frac{3}{4} (x - \frac{8}{3})$ .

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Step 3.5.2.1.1

Simplify terms.

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Step 3.5.2.1.1.1

Apply the distributive property.

$y = - \frac{3}{4} x - \frac{3}{4} (- \frac{8}{3})$

Step 3.5.2.1.1.2

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

$y = - \frac{x \cdot 3}{4} - \frac{3}{4} (- \frac{8}{3})$

Step 3.5.2.1.1.3

Cancel the common factor of $3$ .

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Step 3.5.2.1.1.3.1

Move the leading negative in $- \frac{3}{4}$ into the numerator.

$y = - \frac{x \cdot 3}{4} + \frac{- 3}{4} (- \frac{8}{3})$

Step 3.5.2.1.1.3.2

Move the leading negative in $- \frac{8}{3}$ into the numerator.

$y = - \frac{x \cdot 3}{4} + \frac{- 3}{4} \cdot \frac{- 8}{3}$

Step 3.5.2.1.1.3.3

Factor $3$ out of $- 3$ .

$y = - \frac{x \cdot 3}{4} + \frac{3 (- 1)}{4} \cdot \frac{- 8}{3}$

Step 3.5.2.1.1.3.4

Cancel the common factor.

$y = - \frac{x \cdot 3}{4} + \frac{3 \cdot - 1}{4} \cdot \frac{- 8}{3}$

Step 3.5.2.1.1.3.5

Rewrite the expression.

$y = - \frac{x \cdot 3}{4} + \frac{- 1}{4} \cdot - 8$

Step 3.5.2.1.1.4

Cancel the common factor of $4$ .

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Step 3.5.2.1.1.4.1

Factor $4$ out of $- 8$ .

$y = - \frac{x \cdot 3}{4} + \frac{- 1}{4} \cdot (4 (- 2))$

Step 3.5.2.1.1.4.2

Cancel the common factor.

$y = - \frac{x \cdot 3}{4} + \frac{- 1}{4} \cdot (4 \cdot - 2)$

Step 3.5.2.1.1.4.3

Rewrite the expression.

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

Step 3.5.2.1.1.5

Multiply $- 1$ by $- 2$ .

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

Step 3.5.2.1.2

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

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

Step 4

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

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

Step 5

Verify if $f^{- 1} (x) = - \frac{3 x}{4} + 2$ is the inverse of $f (x) = - \frac{4}{3} x + \frac{8}{3}$ .

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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 $f^{- 1} (- \frac{4}{3} x + \frac{8}{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Simplify the numerator.

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Step 5.2.3.1.1

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

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

Step 5.2.3.1.2

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

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

Step 5.2.3.1.3

Combine the numerators over the common denominator.

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

Step 5.2.3.1.4

Factor $4$ out of $- 4 x + 8$ .

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Step 5.2.3.1.4.1

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

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

Step 5.2.3.1.4.2

Factor $4$ out of $8$ .

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

Step 5.2.3.1.4.3

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

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

Step 5.2.3.2

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

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

Step 5.2.3.3

Multiply $3$ by $4$ .

$f^{- 1} (- \frac{4}{3} x + \frac{8}{3}) = - \frac{\frac{12 (- x + 2)}{3}}{4} + 2$

Step 5.2.3.4

Reduce the expression by cancelling the common factors.

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Step 5.2.3.4.1

Reduce the expression $\frac{12 (- x + 2)}{3}$ by cancelling the common factors.

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Step 5.2.3.4.1.1

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

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

Step 5.2.3.4.1.2

Factor $3$ out of $3$ .

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

Step 5.2.3.4.1.3

Cancel the common factor.

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

Step 5.2.3.4.1.4

Rewrite the expression.

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

Step 5.2.3.4.2

Divide $4 (- x + 2)$ by $1$ .

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

Step 5.2.3.5

Cancel the common factor of $4$ .

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Step 5.2.3.5.1

Cancel the common factor.

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

Step 5.2.3.5.2

Divide $- x + 2$ by $1$ .

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

Step 5.2.3.6

Apply the distributive property.

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

Step 5.2.3.7

Multiply $- - x$ .

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Step 5.2.3.7.1

Multiply $- 1$ by $- 1$ .

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

Step 5.2.3.7.2

Multiply $x$ by $1$ .

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

Step 5.2.3.8

Multiply $- 1$ by $2$ .

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

Step 5.2.4

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

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Step 5.2.4.1

Add $- 2$ and $2$ .

$f^{- 1} (- \frac{4}{3} x + \frac{8}{3}) = x + 0$

Step 5.2.4.2

Add $x$ and $0$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Apply the distributive property.

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

Step 5.3.3.2

Cancel the common factor of $4$ .

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Step 5.3.3.2.1

Move the leading negative in $- \frac{4}{3}$ into the numerator.

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

Step 5.3.3.2.2

Move the leading negative in $- \frac{3 x}{4}$ into the numerator.

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

Step 5.3.3.2.3

Factor $4$ out of $- 4$ .

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

Step 5.3.3.2.4

Cancel the common factor.

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

Step 5.3.3.2.5

Rewrite the expression.

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

Step 5.3.3.3

Cancel the common factor of $3$ .

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Step 5.3.3.3.1

Factor $3$ out of $- 3 x$ .

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

Step 5.3.3.3.2

Cancel the common factor.

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

Step 5.3.3.3.3

Rewrite the expression.

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

Step 5.3.3.4

Multiply $- 1$ by $- 1$ .

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

Step 5.3.3.5

Multiply $x$ by $1$ .

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

Step 5.3.3.6

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

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Step 5.3.3.6.1

Multiply $2$ by $- 1$ .

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

Step 5.3.3.6.2

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

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

Step 5.3.3.6.3

Multiply $- 2$ by $4$ .

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

Step 5.3.3.7

Move the negative in front of the fraction.

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

Step 5.3.4

Combine the opposite terms in $x - \frac{8}{3} + \frac{8}{3}$ .

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Step 5.3.4.1

Add $- \frac{8}{3}$ and $\frac{8}{3}$ .

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

Step 5.3.4.2

Add $x$ and $0$ .

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

Step 5.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{3 x}{4} + 2$ is the inverse of $f (x) = - \frac{4}{3} x + \frac{8}{3}$ .

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