Calculus Examples

$f (x) = \frac{1 + \cos (6 x)}{1 + x^{3}}$

Step 1

Set the denominator in $\frac{1 + \cos (6 x)}{1 + x^{3}}$ equal to $0$ to find where the expression is undefined.

$1 + x^{3} = 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $1$ from both sides of the equation.

$x^{3} = - 1$

Step 2.2

Add $1$ to both sides of the equation.

$x^{3} + 1 = 0$

Step 2.3

Factor the left side of the equation.

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Step 2.3.1

Rewrite $1$ as $1^{3}$ .

$x^{3} + 1^{3} = 0$

Step 2.3.2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

$(x + 1) (x^{2} - x \cdot 1 + 1^{2}) = 0$

Step 2.3.3

Simplify.

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Step 2.3.3.1

Multiply $- 1$ by $1$ .

$(x + 1) (x^{2} - x + 1^{2}) = 0$

Step 2.3.3.2

One to any power is one.

$(x + 1) (x^{2} - x + 1) = 0$

Step 2.4

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

$x + 1 = 0$

$x^{2} - x + 1 = 0$

Step 2.5

Set $x + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6

Set $x^{2} - x + 1$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x^{2} - x + 1$ equal to $0$ .

$x^{2} - x + 1 = 0$

Step 2.6.2

Solve $x^{2} - x + 1 = 0$ for $x$ .

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Step 2.6.2.1

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2.6.2.2

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Step 2.6.2.3

Simplify.

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Step 2.6.2.3.1

Simplify the numerator.

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Step 2.6.2.3.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.6.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.6.2.3.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.6.2.3.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.6.2.3.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.6.2.3.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.3.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.3.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 2.6.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 2.6.2.4.1

Simplify the numerator.

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Step 2.6.2.4.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.6.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.6.2.4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.6.2.4.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.6.2.4.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.6.2.4.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.6.2.4.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.4.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 2.6.2.4.3

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{3}}{2}$

Step 2.6.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 2.6.2.5.1

Simplify the numerator.

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Step 2.6.2.5.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Step 2.6.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 1$ .

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Step 2.6.2.5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Step 2.6.2.5.1.2.2

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Step 2.6.2.5.1.3

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Step 2.6.2.5.1.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Step 2.6.2.5.1.5

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Step 2.6.2.5.2

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Step 2.6.2.5.3

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{3}}{2}$

Step 2.6.2.6

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 2.7

The final solution is all the values that make $(x + 1) (x^{2} - x + 1) = 0$ true.

$x = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 3

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, \infty)$

Set-Builder Notation:

${x | x \neq - 1}$

Step 4

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

Interval Notation:

$(- \infty, - 1] \cup [1, \infty)$

Set-Builder Notation:

${y | y \leq - 1, y \geq 1}$

Step 5

Determine the domain and range.

Domain: $(- \infty, - 1) \cup (- 1, \infty), {x | x \neq - 1}$

Range: $(- \infty, - 1] \cup [1, \infty), {y | y \leq - 1, y \geq 1}$

Step 6