Calculus Examples

$f (x) = \frac{\sqrt{x^{2} - 25}}{\ln (x + 9)}$

Step 1

Find the domain to determine if the expression is continuous.

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Step 1.1

Set the argument in $\ln (x + 9)$ greater than $0$ to find where the expression is defined.

$x + 9 > 0$

Step 1.2

Subtract $9$ from both sides of the inequality.

$x > - 9$

Step 1.3

Set the radicand in $\sqrt{x^{2} - 25}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 25 \geq 0$

Step 1.4

Solve for $x$ .

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Step 1.4.1

Add $25$ to both sides of the inequality.

$x^{2} \geq 25$

Step 1.4.2

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

$\sqrt{x^{2}} \geq \sqrt{25}$

Step 1.4.3

Simplify the equation.

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Step 1.4.3.1

Simplify the left side.

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Step 1.4.3.1.1

Pull terms out from under the radical.

$| x | \geq \sqrt{25}$

Step 1.4.3.2

Simplify the right side.

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Step 1.4.3.2.1

Simplify $\sqrt{25}$ .

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Step 1.4.3.2.1.1

Rewrite $25$ as $5^{2}$ .

$| x | \geq \sqrt{5^{2}}$

Step 1.4.3.2.1.2

Pull terms out from under the radical.

$| x | \geq | 5 |$

Step 1.4.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $5$ is $5$ .

$| x | \geq 5$

Step 1.4.4

Write $| x | \geq 5$ as a piecewise.

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Step 1.4.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 1.4.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq 5$

Step 1.4.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 1.4.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 5$

Step 1.4.4.5

Write as a piecewise.

${\begin{cases} x \geq 5 & x \geq 0 \\ - x \geq 5 & x < 0 \end{cases}$

Step 1.4.5

Find the intersection of $x \geq 5$ and $x \geq 0$ .

$x \geq 5$

Step 1.4.6

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

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Step 1.4.6.1

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 1.4.6.2

Simplify the left side.

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Step 1.4.6.2.1

Dividing two negative values results in a positive value.

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

Step 1.4.6.2.2

Divide $x$ by $1$ .

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

Step 1.4.6.3

Simplify the right side.

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Step 1.4.6.3.1

Divide $5$ by $- 1$ .

$x \leq - 5$

Step 1.4.7

Find the union of the solutions.

$x \leq - 5$ or $x \geq 5$

Step 1.5

Set the denominator in $\frac{\sqrt{x^{2} - 25}}{\ln (x + 9)}$ equal to $0$ to find where the expression is undefined.

$\ln (x + 9) = 0$

Step 1.6

Solve for $x$ .

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Step 1.6.1

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x + 9)} = e^{0}$

Step 1.6.2

Rewrite $\ln (x + 9) = 0$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{0} = x + 9$

Step 1.6.3

Solve for $x$ .

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Step 1.6.3.1

Rewrite the equation as $x + 9 = e^{0}$ .

$x + 9 = e^{0}$

Step 1.6.3.2

Anything raised to $0$ is $1$ .

$x + 9 = 1$

Step 1.6.3.3

Move all terms not containing $x$ to the right side of the equation.

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Step 1.6.3.3.1

Subtract $9$ from both sides of the equation.

$x = 1 - 9$

Step 1.6.3.3.2

Subtract $9$ from $1$ .

$x = - 8$

Step 1.7

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

Interval Notation:

$(- 9, - 8) \cup (- 8, - 5] \cup [5, \infty)$

Set-Builder Notation:

${x | - 9 < x \leq - 5, x \geq 5, x \neq - 8}$

Interval Notation:

$(- 9, - 8) \cup (- 8, - 5] \cup [5, \infty)$

Set-Builder Notation:

${x | - 9 < x \leq - 5, x \geq 5, x \neq - 8}$

Step 2

Since the domain is not all real numbers, $\frac{\sqrt{x^{2} - 25}}{\ln (x + 9)}$ is not continuous over all real numbers.

Not continuous

Step 3