Precalculus Examples

$y = \arctan (x + \frac{π}{2})$

Step 1

Interchange the variables.

$x = \arctan (y + \frac{π}{2})$

Step 2

Solve for $y$ .

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Step 2.1

Rewrite the equation as $\arctan (y + \frac{π}{2}) = x$ .

$\arctan (y + \frac{π}{2}) = x$

Step 2.2

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

$y + \frac{π}{2} = \tan (x)$

Step 2.3

Subtract $\frac{π}{2}$ from both sides of the equation.

$y = \tan (x) - \frac{π}{2}$

Step 3

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

$f^{- 1} (x) = \tan (x) - \frac{π}{2}$

Step 4

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

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

$f^{- 1} (\arctan (x + \frac{π}{2})) = \tan (\arctan (x + \frac{π}{2})) - \frac{π}{2}$

Step 4.2.3

The functions tangent and arctangent are inverses.

$f^{- 1} (\arctan (x + \frac{π}{2})) = x + \frac{π}{2} - \frac{π}{2}$

Step 4.2.4

Combine the opposite terms in $x + \frac{π}{2} - \frac{π}{2}$ .

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Step 4.2.4.1

Combine the numerators over the common denominator.

$f^{- 1} (\arctan (x + \frac{π}{2})) = x + \frac{π - π}{2}$

Step 4.2.4.2

Subtract $π$ from $π$ .

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

Step 4.2.5

Simplify the expression.

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Step 4.2.5.1

Divide $0$ by $2$ .

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

Step 4.2.5.2

Add $x$ and $0$ .

$f^{- 1} (\arctan (x + \frac{π}{2})) = x$

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

$f (\tan (x) - \frac{π}{2}) = \arctan ((\tan (x) - \frac{π}{2}) + \frac{π}{2})$

Step 4.3.3

Combine the opposite terms in $(\tan (x) - \frac{π}{2}) + \frac{π}{2}$ .

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Step 4.3.3.1

Add $- \frac{π}{2}$ and $\frac{π}{2}$ .

$f (\tan (x) - \frac{π}{2}) = \arctan (\tan (x) + 0)$

Step 4.3.3.2

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

$f (\tan (x) - \frac{π}{2}) = \arctan (\tan (x))$

Step 4.4

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

$f^{- 1} (x) = \tan (x) - \frac{π}{2}$