Trigonometry Examples

$(x - 3) - (- 0.5 x)$

Step 1

Interchange the variables.

$x = (y - 3) - (- 0.5 y)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $y - 3 - (- 0.5 y) = x$ .

$y - 3 - (- 0.5 y) = x$

Step 2.2

Simplify $y - 3 - (- 0.5 y)$ .

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Step 2.2.1

Multiply $- 0.5$ by $- 1$ .

$y - 3 + 0.5 y = x$

Step 2.2.2

Add $y$ and $0.5 y$ .

$1.5 y - 3 = x$

Step 2.3

Add $3$ to both sides of the equation.

$1.5 y = x + 3$

Step 2.4

Divide each term in $1.5 y = x + 3$ by $1.5$ and simplify.

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Step 2.4.1

Divide each term in $1.5 y = x + 3$ by $1.5$ .

$\frac{1.5 y}{1.5} = \frac{x}{1.5} + \frac{3}{1.5}$

Step 2.4.2

Simplify the left side.

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Step 2.4.2.1

Cancel the common factor of $1.5$ .

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Step 2.4.2.1.1

Cancel the common factor.

$\frac{1.5 y}{1.5} = \frac{x}{1.5} + \frac{3}{1.5}$

Step 2.4.2.1.2

Divide $y$ by $1$ .

$y = \frac{x}{1.5} + \frac{3}{1.5}$

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 by $1$ .

$y = \frac{1 x}{1.5} + \frac{3}{1.5}$

Step 2.4.3.1.2

Factor $1.5$ out of $1.5$ .

$y = \frac{1 x}{1.5 (1)} + \frac{3}{1.5}$

Step 2.4.3.1.3

Separate fractions.

$y = \frac{1}{1.5} \cdot \frac{x}{1} + \frac{3}{1.5}$

Step 2.4.3.1.4

Divide $1$ by $1.5$ .

$y = 0.\overline{6} \frac{x}{1} + \frac{3}{1.5}$

Step 2.4.3.1.5

Divide $x$ by $1$ .

$y = 0.\overline{6} x + \frac{3}{1.5}$

Step 2.4.3.1.6

Divide $3$ by $1.5$ .

$y = 0.\overline{6} x + 2$

Step 3

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

$f^{- 1} (x) = 0.\overline{6} x + 2$

Step 4

Verify if $f^{- 1} (x) = 0.\overline{6} x + 2$ is the inverse of $f (x) = (x - 3) - (- 0.5 x)$ .

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Step 4.1

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

Step 4.2

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

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Step 4.2.1

Set up the composite result function.

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

Step 4.2.2

Evaluate $f^{- 1} ((x - 3) - (- 0.5 x))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = 0.\overline{6} ((x - 3) - (- 0.5 x)) + 2$

Step 4.2.3

Simplify each term.

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Step 4.2.3.1

Multiply $- 0.5$ by $- 1$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = 0.\overline{6} (x - 3 + 0.5 x) + 2$

Step 4.2.3.2

Add $x$ and $0.5 x$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = 0.\overline{6} (1.5 x - 3) + 2$

Step 4.2.3.3

Apply the distributive property.

$f^{- 1} ((x - 3) - (- 0.5 x)) = 0.\overline{6} (1.5 x) + 0.\overline{6} \cdot - 3 + 2$

Step 4.2.3.4

Multiply $0.\overline{6} (1.5 x)$ .

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Step 4.2.3.4.1

Multiply $1.5$ by $0.\overline{6}$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = 1 x + 0.\overline{6} \cdot - 3 + 2$

Step 4.2.3.4.2

Multiply $x$ by $1$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = x + 0.\overline{6} \cdot - 3 + 2$

Step 4.2.3.5

Multiply $0.\overline{6}$ by $- 3$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = x - 2 + 2$

Step 4.2.4

Add $- 2$ and $2$ .

$f^{- 1} ((x - 3) - (- 0.5 x)) = x 0$

Step 4.3

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

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Step 4.3.1

Set up the composite result function.

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

Step 4.3.2

Evaluate $f (0.\overline{6} x + 2)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (0.\overline{6} x + 2) = ((0.\overline{6} x + 2) - 3) - (- 0.5 (0.\overline{6} x + 2))$

Step 4.3.3

Simplify each term.

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Step 4.3.3.1

Apply the distributive property.

$f (0.\overline{6} x + 2) = (0.\overline{6} x + 2) - 3 - (- 0.5 (0.\overline{6} x) - 0.5 \cdot 2)$

Step 4.3.3.2

Multiply $0.\overline{6}$ by $- 0.5$ .

$f (0.\overline{6} x + 2) = (0.\overline{6} x + 2) - 3 - (- 0.\overline{3} x - 0.5 \cdot 2)$

Step 4.3.3.3

Multiply $- 0.5$ by $2$ .

$f (0.\overline{6} x + 2) = (0.\overline{6} x + 2) - 3 - (- 0.\overline{3} x - 1)$

Step 4.3.3.4

Apply the distributive property.

$f (0.\overline{6} x + 2) = (0.\overline{6} x + 2) - 3 - (- 0.\overline{3} x) + 1$

Step 4.3.3.5

Multiply $- 0.\overline{3}$ by $- 1$ .

$f (0.\overline{6} x + 2) = (0.\overline{6} x + 2) - 3 + 0.\overline{3} x + 1$

Step 4.3.3.6

Multiply $- 1$ by $- 1$ .

$f (0.\overline{6} x + 2) = (0.\overline{6} x + 2) - 3 + 0.\overline{3} x + 1$

Step 4.3.4

Simplify by adding terms.

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Step 4.3.4.1

Add $0.\overline{6} x$ and $0.\overline{3} x$ .

$f (0.\overline{6} x + 2) = 1 x + 2 - 3 + 1$

Step 4.3.4.2

Subtract $3$ from $2$ .

$f (0.\overline{6} x + 2) = 1 x - 1 + 1$

Step 4.3.4.3

Combine the opposite terms in $1 x - 1 + 1$ .

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Step 4.3.4.3.1

Add $- 1$ and $1$ .

$f (0.\overline{6} x + 2) = 1 x + 0$

Step 4.3.4.3.2

Add $1 x$ and $0$ .

$f (0.\overline{6} x + 2) = 1 x$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 0.\overline{6} x + 2$ is the inverse of $f (x) = (x - 3) - (- 0.5 x)$ .

$f^{- 1} (x) = 0.\overline{6} x + 2$