Calculus Examples

$f (x) = \frac{1}{4 - x^{2}}$

Step 1

Find the domain to determine if the expression is continuous.

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Step 1.1

Set the denominator in $\frac{1}{4 - x^{2}}$ equal to $0$ to find where the expression is undefined.

$4 - x^{2} = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Subtract $4$ from both sides of the equation.

$- x^{2} = - 4$

Step 1.2.2

Divide each term in $- x^{2} = - 4$ by $- 1$ and simplify.

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Step 1.2.2.1

Divide each term in $- x^{2} = - 4$ by $- 1$ .

$\frac{- x^{2}}{- 1} = \frac{- 4}{- 1}$

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x^{2}}{1} = \frac{- 4}{- 1}$

Step 1.2.2.2.2

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

$x^{2} = \frac{- 4}{- 1}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Divide $- 4$ by $- 1$ .

$x^{2} = 4$

Step 1.2.3

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

$x = \pm \sqrt{4}$

Step 1.2.4

Simplify $\pm \sqrt{4}$ .

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Step 1.2.4.1

Rewrite $4$ as $2^{2}$ .

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

Step 1.2.4.2

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

$x = \pm 2$

Step 1.2.5

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

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Step 1.2.5.1

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

$x = 2$

Step 1.2.5.2

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

$x = - 2$

Step 1.2.5.3

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

$x = 2, - 2$

Step 1.3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 2}$

Interval Notation:

$(- \infty, - 2) \cup (- 2, 2) \cup (2, \infty)$

Set-Builder Notation:

${x | x \neq 2, - 2}$

Step 2

Since the domain is not all real numbers, $\frac{1}{4 - x^{2}}$ is not continuous over all real numbers.

Not continuous

Step 3