Calculus Examples

$\frac{x + \frac{1}{x} + 17}{x + \frac{1}{x} + 2}$

Step 1

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

$x = 0$

Step 2

Set the denominator in $\frac{x + \frac{1}{x} + 17}{x + \frac{1}{x} + 2}$ equal to $0$ to find where the expression is undefined.

$x + \frac{1}{x} + 2 = 0$

Step 3

Solve for $x$ .

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Step 3.1

Find the LCD of the terms in the equation.

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Step 3.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, x, 1, 1$

Step 3.1.2

The LCM of one and any expression is the expression.

$x$

Step 3.2

Multiply each term in $x + \frac{1}{x} + 2 = 0$ by $x$ to eliminate the fractions.

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Step 3.2.1

Multiply each term in $x + \frac{1}{x} + 2 = 0$ by $x$ .

$x \cdot x + \frac{1}{x} x + 2 x = 0 x$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Multiply $x$ by $x$ .

$x^{2} + \frac{1}{x} x + 2 x = 0 x$

Step 3.2.2.1.2

Cancel the common factor of $x$ .

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Step 3.2.2.1.2.1

Cancel the common factor.

$x^{2} + \frac{1}{x} x + 2 x = 0 x$

Step 3.2.2.1.2.2

Rewrite the expression.

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

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Multiply $0$ by $x$ .

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

Step 3.3

Solve the equation.

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Step 3.3.1

Factor using the perfect square rule.

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Step 3.3.1.1

Rearrange terms.

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

Step 3.3.1.2

Rewrite $1$ as $1^{2}$ .

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

Step 3.3.1.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 3.3.1.4

Rewrite the polynomial.

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

Step 3.3.1.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

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

Step 3.3.2

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

$x + 1 = 0$

Step 3.3.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

Set-Builder Notation:

${x | x \neq 0, - 1}$

Step 5