Precalculus Examples

$\arccos (x - π)$

Step 1

Interchange the variables.

$x = \arccos (y - π)$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\arccos (y - π) = x$ .

$\arccos (y - π) = x$

Step 2.2

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

$y - π = \cos (x)$

Step 2.3

Add $π$ to both sides of the equation.

$y = \cos (x) + π$

Step 3

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

$f^{- 1} (x) = \cos (x) + π$

Step 4

Verify if $f^{- 1} (x) = \cos (x) + π$ is the inverse of $f (x) = \arccos (x - π)$ .

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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 $f^{- 1} (\arccos (x - π))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\arccos (x - π)) = \cos (\arccos (x - π)) + π$

Step 4.2.3

The functions cosine and arccosine are inverses.

$f^{- 1} (\arccos (x - π)) = x - π + π$

Step 4.2.4

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

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Step 4.2.4.1

Add $- π$ and $π$ .

$f^{- 1} (\arccos (x - π)) = x + 0$

Step 4.2.4.2

Add $x$ and $0$ .

$f^{- 1} (\arccos (x - π)) = x$

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

$f (\cos (x) + π) = \arccos ((\cos (x) + π) - π)$

Step 4.3.3

Combine the opposite terms in $(\cos (x) + π) - π$ .

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Step 4.3.3.1

Subtract $π$ from $π$ .

$f (\cos (x) + π) = \arccos (\cos (x) + 0)$

Step 4.3.3.2

Add $\cos (x)$ and $0$ .

$f (\cos (x) + π) = \arccos (\cos (x))$

Step 4.4

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \cos (x) + π$ is the inverse of $f (x) = \arccos (x - π)$ .

$f^{- 1} (x) = \cos (x) + π$