Finite Math Examples

$g (x) = \frac{{(x)}^{42 x^{4} - x^{3} + x^{2} - x + 8}}{x^{2} - 5 x - 14}$

Step 1

Find the domain to determine if the expression is continuous.

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Step 1.1

Set the denominator in $\frac{{(x)}^{42 x^{4} - x^{3} + x^{2} - x + 8}}{x^{2} - 5 x - 14}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 5 x - 14 = 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Factor $x^{2} - 5 x - 14$ using the AC method.

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Step 1.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 14$ and whose sum is $- 5$ .

$- 7, 2$

Step 1.2.1.2

Write the factored form using these integers.

$(x - 7) (x + 2) = 0$

Step 1.2.2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 7 = 0$

$x + 2 = 0$

Step 1.2.3

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 1.2.3.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 1.2.3.2

Add $7$ to both sides of the equation.

$x = 7$

Step 1.2.4

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 1.2.4.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 1.2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 1.2.5

The final solution is all the values that make $(x - 7) (x + 2) = 0$ true.

$x = 7, - 2$

Step 1.3

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

Interval Notation:

$(- \infty, - 2) \cup (- 2, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq - 2, 7}$

Interval Notation:

$(- \infty, - 2) \cup (- 2, 7) \cup (7, \infty)$

Set-Builder Notation:

${x | x \neq - 2, 7}$

Step 2

Since the domain is not all real numbers, $\frac{{(x)}^{42 x^{4} - x^{3} + x^{2} - x + 8}}{x^{2} - 5 x - 14}$ is not continuous over all real numbers.

Not continuous

Step 3