Precalculus Examples

$f (x) = 5 \arcsin (5 x - 1) + 2$

Step 1

Write $f (x) = 5 \arcsin (5 x - 1) + 2$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $5 \arcsin (5 y - 1) + 2 = x$ .

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

Step 3.2

Subtract $2$ from both sides of the equation.

$5 \arcsin (5 y - 1) = x - 2$

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $5$ .

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Step 3.3.2.1.1

Cancel the common factor.

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

Step 3.3.2.1.2

Divide $\arcsin (5 y - 1)$ by $1$ .

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

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Move the negative in front of the fraction.

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

Step 3.4

Take the inverse arcsine of both sides of the equation to extract $y$ from inside the arcsine.

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

Step 3.5

Add $1$ to both sides of the equation.

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

Step 3.6

Divide each term in $5 y = \sin (\frac{x}{5} - \frac{2}{5}) + 1$ by $5$ and simplify.

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Step 3.6.1

Divide each term in $5 y = \sin (\frac{x}{5} - \frac{2}{5}) + 1$ by $5$ .

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

Step 3.6.2

Simplify the left side.

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Step 3.6.2.1

Cancel the common factor of $5$ .

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Step 3.6.2.1.1

Cancel the common factor.

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

Step 3.6.2.1.2

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Step 5.2.3

Combine the numerators over the common denominator.

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

Step 5.2.4

Simplify each term.

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Step 5.2.4.1

Combine the numerators over the common denominator.

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

Step 5.2.4.2

Combine the opposite terms in $5 \arcsin (5 x - 1) + 2 - 2$ .

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Step 5.2.4.2.1

Subtract $2$ from $2$ .

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

Step 5.2.4.2.2

Add $5 \arcsin (5 x - 1)$ and $0$ .

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

Step 5.2.4.3

Cancel the common factor of $5$ .

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Step 5.2.4.3.1

Cancel the common factor.

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

Step 5.2.4.3.2

Divide $\arcsin (5 x - 1)$ by $1$ .

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

Step 5.2.4.4

The functions sine and arcsine are inverses.

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

Step 5.2.5

Simplify terms.

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Step 5.2.5.1

Combine the opposite terms in $5 x - 1 + 1$ .

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Step 5.2.5.1.1

Add $- 1$ and $1$ .

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

Step 5.2.5.1.2

Add $5 x$ and $0$ .

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

Step 5.2.5.2

Cancel the common factor of $5$ .

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Step 5.2.5.2.1

Cancel the common factor.

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

Step 5.2.5.2.2

Divide $x$ by $1$ .

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

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

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

Step 5.3.3

Simplify each term.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Apply the distributive property.

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

Step 5.3.3.1.2

Cancel the common factor of $5$ .

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Step 5.3.3.1.2.1

Cancel the common factor.

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

Step 5.3.3.1.2.2

Rewrite the expression.

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

Step 5.3.3.1.3

Cancel the common factor of $5$ .

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Step 5.3.3.1.3.1

Cancel the common factor.

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

Step 5.3.3.1.3.2

Rewrite the expression.

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

Step 5.3.3.2

Combine the opposite terms in $\sin (\frac{x}{5} - \frac{2}{5}) + 1 - 1$ .

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Step 5.3.3.2.1

Subtract $1$ from $1$ .

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

Step 5.3.3.2.2

Add $\sin (\frac{x}{5} - \frac{2}{5})$ and $0$ .

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

Step 5.4

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

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