Trigonometry Examples

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $\frac{1}{5} \cdot (y - 2) = x$ .

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

Step 3.2

Multiply both sides of the equation by $5$ .

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

Step 3.3

Simplify the left side.

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Step 3.3.1

Simplify $5 (\frac{1}{5} \cdot (y - 2))$ .

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Step 3.3.1.1

Apply the distributive property.

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

Step 3.3.1.2

Combine $\frac{1}{5}$ and $y$ .

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

Step 3.3.1.3

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

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

Step 3.3.1.4

Move the negative in front of the fraction.

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

Step 3.3.1.5

Apply the distributive property.

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

Step 3.3.1.6

Cancel the common factor of $5$ .

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Step 3.3.1.6.1

Cancel the common factor.

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

Step 3.3.1.6.2

Rewrite the expression.

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

Step 3.3.1.7

Cancel the common factor of $5$ .

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Step 3.3.1.7.1

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

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

Step 3.3.1.7.2

Cancel the common factor.

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

Step 3.3.1.7.3

Rewrite the expression.

$y - 2 = 5 x$

Step 3.4

Add $2$ to both sides of the equation.

$y = 5 x + 2$

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Simplify each term.

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Step 5.2.3.1

Apply the distributive property.

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

Step 5.2.3.2

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

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

Step 5.2.3.3

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

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

Step 5.2.3.4

Move the negative in front of the fraction.

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

Step 5.2.3.5

Apply the distributive property.

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

Step 5.2.3.6

Cancel the common factor of $5$ .

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Step 5.2.3.6.1

Cancel the common factor.

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

Step 5.2.3.6.2

Rewrite the expression.

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

Step 5.2.3.7

Cancel the common factor of $5$ .

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Step 5.2.3.7.1

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

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

Step 5.2.3.7.2

Cancel the common factor.

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

Step 5.2.3.7.3

Rewrite the expression.

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

Step 5.2.4

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

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Step 5.2.4.1

Add $- 2$ and $2$ .

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

Step 5.2.4.2

Add $x$ and $0$ .

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

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

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

Step 5.3.3

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

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Step 5.3.3.1

Subtract $2$ from $2$ .

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

Step 5.3.3.2

Add $5 x$ and $0$ .

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

Step 5.3.4

Cancel the common factor of $5$ .

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Step 5.3.4.1

Factor $5$ out of $5 x$ .

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

Step 5.3.4.2

Cancel the common factor.

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

Step 5.3.4.3

Rewrite the expression.

$f (5 x + 2) = x$

Step 5.4

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

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