Calculus Examples

$f (x) = - x^{2} - 5$

Step 1

Write $f (x) = - x^{2} - 5$ as an equation.

$y = - x^{2} - 5$

Step 2

Interchange the variables.

$x = - y^{2} - 5$

Step 3

Solve for $y$ .

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Step 3.1

Rewrite the equation as $- y^{2} - 5 = x$ .

$- y^{2} - 5 = x$

Step 3.2

Add $5$ to both sides of the equation.

$- y^{2} = x + 5$

Step 3.3

Divide each term in $- y^{2} = x + 5$ by $- 1$ and simplify.

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Step 3.3.1

Divide each term in $- y^{2} = x + 5$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{x}{- 1} + \frac{5}{- 1}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{x}{- 1} + \frac{5}{- 1}$

Step 3.3.2.2

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

$y^{2} = \frac{x}{- 1} + \frac{5}{- 1}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

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

$y^{2} = - 1 \cdot x + \frac{5}{- 1}$

Step 3.3.3.1.2

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

$y^{2} = - x + \frac{5}{- 1}$

Step 3.3.3.1.3

Divide $5$ by $- 1$ .

$y^{2} = - x - 5$

Step 3.4

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

$y = \pm \sqrt{- x - 5}$

Step 3.5

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

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Step 3.5.1

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

$y = \sqrt{- x - 5}$

Step 3.5.2

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

$y = - \sqrt{- x - 5}$

Step 3.5.3

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

$y = \sqrt{- x - 5}$

$y = - \sqrt{- x - 5}$

$y = \sqrt{- x - 5}$

$y = - \sqrt{- x - 5}$

$y = \sqrt{- x - 5}$

$y = - \sqrt{- x - 5}$

Step 4

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

$f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$

Step 5

Verify if $f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$ is the inverse of $f (x) = - x^{2} - 5$ .

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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 $f (x) = - x^{2} - 5$ and $f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$ and compare them.

Step 5.2

Find the range of $f (x) = - x^{2} - 5$ .

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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:

$(- \infty, - 5]$

Step 5.3

Find the domain of $\sqrt{- x - 5}$ .

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Step 5.3.1

Set the radicand in $\sqrt{- x - 5}$ greater than or equal to $0$ to find where the expression is defined.

$- x - 5 \geq 0$

Step 5.3.2

Solve for $x$ .

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Step 5.3.2.1

Add $5$ to both sides of the inequality.

$- x \geq 5$

Step 5.3.2.2

Divide each term in $- x \geq 5$ by $- 1$ and simplify.

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Step 5.3.2.2.1

Divide each term in $- x \geq 5$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{5}{- 1}$

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.

$\frac{x}{1} \leq \frac{5}{- 1}$

Step 5.3.2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{5}{- 1}$

Step 5.3.2.2.3

Simplify the right side.

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Step 5.3.2.2.3.1

Divide $5$ by $- 1$ .

$x \leq - 5$

Step 5.3.3

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

$(- \infty, - 5]$

Step 5.4

Find the domain of $f (x) = - x^{2} - 5$ .

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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 $f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$ is the range of $f (x) = - x^{2} - 5$ and the range of $f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$ is the domain of $f (x) = - x^{2} - 5$ , then $f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$ is the inverse of $f (x) = - x^{2} - 5$ .

$f^{- 1} (x) = \sqrt{- x - 5}, - \sqrt{- x - 5}$

Step 6