Calculus Examples

$f (x) = \frac{x^{4} - 5 x^{2} + 4}{x^{4} - 24 x^{2} + 108}$

Step 1

Set the denominator in $\frac{x^{4} - 5 x^{2} + 4}{x^{4} - 24 x^{2} + 108}$ equal to $0$ to find where the expression is undefined.

$x^{4} - 24 x^{2} + 108 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 24 u + 108 = 0$

$u = x^{2}$

Step 2.2

Factor $u^{2} - 24 u + 108$ using the AC method.

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Step 2.2.1

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 $108$ and whose sum is $- 24$ .

$- 18, - 6$

Step 2.2.2

Write the factored form using these integers.

$(u - 18) (u - 6) = 0$

Step 2.3

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

$u - 18 = 0$

$u - 6 = 0$

Step 2.4

Set $u - 18$ equal to $0$ and solve for $u$ .

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Step 2.4.1

Set $u - 18$ equal to $0$ .

$u - 18 = 0$

Step 2.4.2

Add $18$ to both sides of the equation.

$u = 18$

Step 2.5

Set $u - 6$ equal to $0$ and solve for $u$ .

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Step 2.5.1

Set $u - 6$ equal to $0$ .

$u - 6 = 0$

Step 2.5.2

Add $6$ to both sides of the equation.

$u = 6$

Step 2.6

The final solution is all the values that make $(u - 18) (u - 6) = 0$ true.

$u = 18, 6$

Step 2.7

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 18$

${(x^{2})}^{1} = 6$

Step 2.8

Solve the first equation for $x$ .

$x^{2} = 18$

Step 2.9

Solve the equation for $x$ .

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Step 2.9.1

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

$x = \pm \sqrt{18}$

Step 2.9.2

Simplify $\pm \sqrt{18}$ .

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Step 2.9.2.1

Rewrite $18$ as $3^{2} \cdot 2$ .

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Step 2.9.2.1.1

Factor $9$ out of $18$ .

$x = \pm \sqrt{9 (2)}$

Step 2.9.2.1.2

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

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

Step 2.9.2.2

Pull terms out from under the radical.

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

Step 2.9.3

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

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Step 2.9.3.1

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

$x = 3 \sqrt{2}$

Step 2.9.3.2

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

$x = - 3 \sqrt{2}$

Step 2.9.3.3

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

$x = 3 \sqrt{2}, - 3 \sqrt{2}$

Step 2.10

Solve the second equation for $x$ .

${(x^{2})}^{1} = 6$

Step 2.11

Solve the equation for $x$ .

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Step 2.11.1

Remove parentheses.

$x^{2} = 6$

Step 2.11.2

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

$x = \pm \sqrt{6}$

Step 2.11.3

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

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Step 2.11.3.1

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

$x = \sqrt{6}$

Step 2.11.3.2

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

$x = - \sqrt{6}$

Step 2.11.3.3

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

$x = \sqrt{6}, - \sqrt{6}$

Step 2.12

The solution to $x^{4} - 24 x^{2} + 108 = 0$ is $x = 3 \sqrt{2}, - 3 \sqrt{2}, \sqrt{6}, - \sqrt{6}$ .

$x = 3 \sqrt{2}, - 3 \sqrt{2}, \sqrt{6}, - \sqrt{6}$

Step 3

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

Interval Notation:

$(- \infty, - 3 \sqrt{2}) \cup (- 3 \sqrt{2}, - \sqrt{6}) \cup (- \sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, 3 \sqrt{2}) \cup (3 \sqrt{2}, \infty)$

Set-Builder Notation:

${x | x \neq \sqrt{6}, - \sqrt{6}, 3 \sqrt{2}, - 3 \sqrt{2}}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - 1.39979505] \cup [- 0.04464939, \frac{1}{27}] \cup (1, \infty)$

Set-Builder Notation:

${y | y \leq - 1.39979505, - 0.04464939 \leq y \leq \frac{1}{27}, y > 1}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 3 \sqrt{2}) \cup (- 3 \sqrt{2}, - \sqrt{6}) \cup (- \sqrt{6}, \sqrt{6}) \cup (\sqrt{6}, 3 \sqrt{2}) \cup (3 \sqrt{2}, \infty), {x | x \neq \sqrt{6}, - \sqrt{6}, 3 \sqrt{2}, - 3 \sqrt{2}}$

Range: $(- \infty, - 1.39979505] \cup [- 0.04464939, \frac{1}{27}] \cup (1, \infty), {y | y \leq - 1.39979505, - 0.04464939 \leq y \leq \frac{1}{27}, y > 1}$

Step 6