Precalculus Examples

$f (x) = 16 - {(- x)}^{2}$

Step 1

Write $f (x) = 16 - {(- x)}^{2}$ as an equation.

$y = 16 - {(- x)}^{2}$

Step 2

Interchange the variables.

$x = 16 - {(- y)}^{2}$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $16 - {(- y)}^{2} = x$ .

$16 - {(- y)}^{2} = x$

Step 3.2

Simplify each term.

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Step 3.2.1

Apply the product rule to $- y$ .

$16 - ({(- 1)}^{2} y^{2}) = x$

Step 3.2.2

Multiply $- 1$ by ${(- 1)}^{2}$ by adding the exponents.

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Step 3.2.2.1

Move ${(- 1)}^{2}$ .

$16 + {(- 1)}^{2} \cdot - 1 y^{2} = x$

Step 3.2.2.2

Multiply ${(- 1)}^{2}$ by $- 1$ .

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Step 3.2.2.2.1

Raise $- 1$ to the power of $1$ .

$16 + {(- 1)}^{2} \cdot {(- 1)}^{1} y^{2} = x$

Step 3.2.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$16 + {(- 1)}^{2 + 1} y^{2} = x$

Step 3.2.2.3

Add $2$ and $1$ .

$16 + {(- 1)}^{3} y^{2} = x$

Step 3.2.3

Raise $- 1$ to the power of $3$ .

$16 - y^{2} = x$

Step 3.3

Subtract $16$ from both sides of the equation.

$- y^{2} = x - 16$

Step 3.4

Divide each term in $- y^{2} = x - 16$ by $- 1$ and simplify.

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Step 3.4.1

Divide each term in $- y^{2} = x - 16$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{x}{- 1} + \frac{- 16}{- 1}$

Step 3.4.2

Simplify the left side.

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Step 3.4.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{x}{- 1} + \frac{- 16}{- 1}$

Step 3.4.2.2

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{x}{- 1} + \frac{- 16}{- 1}$

Step 3.4.3

Simplify the right side.

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Step 3.4.3.1

Simplify each term.

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Step 3.4.3.1.1

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$y^{2} = - 1 \cdot x + \frac{- 16}{- 1}$

Step 3.4.3.1.2

Rewrite $- 1 \cdot x$ as $- x$ .

$y^{2} = - x + \frac{- 16}{- 1}$

Step 3.4.3.1.3

Divide $- 16$ by $- 1$ .

$y^{2} = - x + 16$

Step 3.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- x + 16}$

Step 3.6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 3.6.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{- x + 16}$

Step 3.6.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{- x + 16}$

Step 3.6.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{- x + 16}$

$y = - \sqrt{- x + 16}$

$y = \sqrt{- x + 16}$

$y = - \sqrt{- x + 16}$

$y = \sqrt{- x + 16}$

$y = - \sqrt{- x + 16}$

Step 4

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

$f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$

Step 5

Verify if $f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$ is the inverse of $f (x) = 16 - {(- x)}^{2}$ .

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Step 5.1

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 16 - {(- x)}^{2}$ and $f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$ and compare them.

Step 5.2

Find the range of $f (x) = 16 - {(- x)}^{2}$ .

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Step 5.2.1

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, 16]$

Step 5.3

Find the domain of $\sqrt{- x + 16}$ .

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Step 5.3.1

Set the radicand in $\sqrt{- x + 16}$ greater than or equal to $0$ to find where the expression is defined.

$- x + 16 \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Subtract $16$ from both sides of the inequality.

$- x \geq - 16$

Step 5.3.2.2

Divide each term in $- x \geq - 16$ by $- 1$ and simplify.

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Step 5.3.2.2.1

Divide each term in $- x \geq - 16$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 16}{- 1}$

Step 5.3.2.2.2

Simplify the left side.

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Step 5.3.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 16}{- 1}$

Step 5.3.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 16}{- 1}$

Step 5.3.2.2.3

Simplify the right side.

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Step 5.3.2.2.3.1

Divide $- 16$ by $- 1$ .

$x \leq 16$

Step 5.3.3

The domain is all values of $x$ that make the expression defined.

$(- \infty, 16]$

Step 5.4

Find the domain of $f (x) = 16 - {(- x)}^{2}$ .

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Step 5.4.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Step 5.5

Since the domain of $f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$ is the range of $f (x) = 16 - {(- x)}^{2}$ and the range of $f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$ is the domain of $f (x) = 16 - {(- x)}^{2}$ , then $f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$ is the inverse of $f (x) = 16 - {(- x)}^{2}$ .

$f^{- 1} (x) = \sqrt{- x + 16}, - \sqrt{- x + 16}$

Step 6