Calculus Examples

$f (x) = \ln (x^{2} - 2 x + 1)$

Step 1

Set the argument in $\ln (x^{2} - 2 x + 1)$ greater than $0$ to find where the expression is defined.

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

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

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

Step 2.2

Factor using the perfect square rule.

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Step 2.2.1

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

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

Step 2.2.2

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 2.2.3

Rewrite the polynomial.

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

Step 2.2.4

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 2.3

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

$x - 1 = 0$

Step 2.4

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

Use each root to create test intervals.

$x < 1$

$x > 1$

Step 2.6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 2.6.1

Test a value on the interval $x < 1$ to see if it makes the inequality true.

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Step 2.6.1.1

Choose a value on the interval $x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 2.6.1.2

Replace $x$ with $0$ in the original inequality.

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

Step 2.6.1.3

The left side $1$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.2

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 2.6.2.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 2.6.2.2

Replace $x$ with $4$ in the original inequality.

${(4)}^{2} - 2 \cdot 4 + 1 > 0$

Step 2.6.2.3

The left side $9$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.3

Compare the intervals to determine which ones satisfy the original inequality.

$x < 1$ True

$x > 1$ True

$x < 1$ True

$x > 1$ True

Step 2.7

The solution consists of all of the true intervals.

$x < 1$ or $x > 1$

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