Calculus Examples

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

Step 1

Find the domain to determine if the expression is continuous.

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Step 1.1

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

$x - 2 \geq 0$

Step 1.2

Add $2$ to both sides of the inequality.

$x \geq 2$

Step 1.3

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

$x^{2} - 9 = 0$

Step 1.4

Solve for $x$ .

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Step 1.4.1

Add $9$ to both sides of the equation.

$x^{2} = 9$

Step 1.4.2

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

$x = \pm \sqrt{9}$

Step 1.4.3

Simplify $\pm \sqrt{9}$ .

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Step 1.4.3.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Step 1.4.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

Step 1.4.4

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

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Step 1.4.4.1

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

$x = 3$

Step 1.4.4.2

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

$x = - 3$

Step 1.4.4.3

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

$x = 3, - 3$

Step 1.5

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

Interval Notation:

$[2, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \geq 2, x \neq 3}$

Interval Notation:

$[2, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \geq 2, x \neq 3}$

Step 2

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

Not continuous

Step 3