Calculus Examples

$\ln (x \sqrt{x^{2} - 1})$

Step 1

Find the domain for $y = \ln (x \sqrt{x^{2} - 1})$ so that a list of $x$ values can be picked to find a list of points, which will help graphing the radical.

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Step 1.1

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

$x \sqrt{(x + 1) (x - 1)} > 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

To remove the radical on the left side of the inequality, square both sides of the inequality.

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

Step 1.2.2

Simplify each side of the inequality.

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Step 1.2.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(x + 1) (x - 1)}$ as ${((x + 1) (x - 1))}^{\frac{1}{2}}$ .

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

Step 1.2.2.2

Simplify the left side.

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Step 1.2.2.2.1

Simplify ${(x {((x + 1) (x - 1))}^{\frac{1}{2}})}^{2}$ .

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Step 1.2.2.2.1.1

Expand $(x + 1) (x - 1)$ using the FOIL Method.

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Step 1.2.2.2.1.1.1

Apply the distributive property.

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

Step 1.2.2.2.1.1.2

Apply the distributive property.

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

Step 1.2.2.2.1.1.3

Apply the distributive property.

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

Step 1.2.2.2.1.2

Simplify and combine like terms.

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Step 1.2.2.2.1.2.1

Simplify each term.

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Step 1.2.2.2.1.2.1.1

Multiply $x$ by $x$ .

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

Step 1.2.2.2.1.2.1.2

Move $- 1$ to the left of $x$ .

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

Step 1.2.2.2.1.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 1.2.2.2.1.2.1.4

Multiply $x$ by $1$ .

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

Step 1.2.2.2.1.2.1.5

Multiply $- 1$ by $1$ .

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

Step 1.2.2.2.1.2.2

Add $- x$ and $x$ .

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

Step 1.2.2.2.1.2.3

Add $x^{2}$ and $0$ .

${(x {(x^{2} - 1)}^{\frac{1}{2}})}^{2} > 0^{2}$

Step 1.2.2.2.1.3

Apply the product rule to $x {(x^{2} - 1)}^{\frac{1}{2}}$ .

$x^{2} {({(x^{2} - 1)}^{\frac{1}{2}})}^{2} > 0^{2}$

Step 1.2.2.2.1.4

Multiply the exponents in ${({(x^{2} - 1)}^{\frac{1}{2}})}^{2}$ .

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Step 1.2.2.2.1.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 1.2.2.2.1.4.2

Cancel the common factor of $2$ .

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Step 1.2.2.2.1.4.2.1

Cancel the common factor.

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

Step 1.2.2.2.1.4.2.2

Rewrite the expression.

$x^{2} {(x^{2} - 1)}^{1} > 0^{2}$

Step 1.2.2.2.1.5

Simplify.

$x^{2} (x^{2} - 1) > 0^{2}$

Step 1.2.2.2.1.6

Apply the distributive property.

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

Step 1.2.2.2.1.7

Multiply $x^{2}$ by $x^{2}$ by adding the exponents.

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Step 1.2.2.2.1.7.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 1.2.2.2.1.7.2

Add $2$ and $2$ .

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

Step 1.2.2.2.1.8

Move $- 1$ to the left of $x^{2}$ .

$x^{4} - 1 \cdot x^{2} > 0^{2}$

Step 1.2.2.2.1.9

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

$x^{4} - x^{2} > 0^{2}$

Step 1.2.2.3

Simplify the right side.

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Step 1.2.2.3.1

Raising $0$ to any positive power yields $0$ .

$x^{4} - x^{2} > 0$

Step 1.2.3

Solve for $x$ .

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Step 1.2.3.1

Convert the inequality to an equation.

$x^{4} - x^{2} = 0$

Step 1.2.3.2

Factor the left side of the equation.

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Step 1.2.3.2.1

Factor $x^{2}$ out of $x^{4} - x^{2}$ .

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Step 1.2.3.2.1.1

Factor $x^{2}$ out of $x^{4}$ .

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

Step 1.2.3.2.1.2

Factor $x^{2}$ out of $- x^{2}$ .

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

Step 1.2.3.2.1.3

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} \cdot - 1$ .

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

Step 1.2.3.2.2

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

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

Step 1.2.3.2.3

Factor.

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Step 1.2.3.2.3.1

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

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

Step 1.2.3.2.3.2

Remove unnecessary parentheses.

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

Step 1.2.3.3

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

$x^{2} = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 1.2.3.4

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

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Step 1.2.3.4.1

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Step 1.2.3.4.2

Solve $x^{2} = 0$ for $x$ .

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Step 1.2.3.4.2.1

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

$x = \pm \sqrt{0}$

Step 1.2.3.4.2.2

Simplify $\pm \sqrt{0}$ .

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Step 1.2.3.4.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 1.2.3.4.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 1.2.3.4.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 1.2.3.5

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

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Step 1.2.3.5.1

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

$x + 1 = 0$

Step 1.2.3.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.3.6

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

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Step 1.2.3.6.1

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

$x - 1 = 0$

Step 1.2.3.6.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.2.3.7

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

$x = 0, - 1, 1$

Step 1.2.4

Find the domain of $x \sqrt{(x + 1) (x - 1)}$ .

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Step 1.2.4.1

Set the radicand in $\sqrt{(x + 1) (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (x - 1) \geq 0$

Step 1.2.4.2

Solve for $x$ .

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Step 1.2.4.2.1

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

$x + 1 = 0$

$x - 1 = 0$

Step 1.2.4.2.2

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

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Step 1.2.4.2.2.1

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

$x + 1 = 0$

Step 1.2.4.2.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.4.2.3

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

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Step 1.2.4.2.3.1

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

$x - 1 = 0$

Step 1.2.4.2.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.2.4.2.4

The final solution is all the values that make $(x + 1) (x - 1) \geq 0$ true.

$x = - 1, 1$

Step 1.2.4.2.5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Step 1.2.4.2.6

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 1.2.4.2.6.1

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

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Step 1.2.4.2.6.1.1

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

$x = - 4$

Step 1.2.4.2.6.1.2

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

$((- 4) + 1) ((- 4) - 1) \geq 0$

Step 1.2.4.2.6.1.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.4.2.6.2

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

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Step 1.2.4.2.6.2.1

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

$x = 0$

Step 1.2.4.2.6.2.2

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

$((0) + 1) ((0) - 1) \geq 0$

Step 1.2.4.2.6.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.4.2.6.3

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

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Step 1.2.4.2.6.3.1

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

$x = 4$

Step 1.2.4.2.6.3.2

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

$((4) + 1) ((4) - 1) \geq 0$

Step 1.2.4.2.6.3.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.4.2.6.4

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

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 1.2.4.2.7

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 1$

Step 1.2.4.3

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

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

Step 1.2.5

The solution consists of all of the true intervals.

$x > 1$

Step 1.3

Set the radicand in $\sqrt{(x + 1) (x - 1)}$ greater than or equal to $0$ to find where the expression is defined.

$(x + 1) (x - 1) \geq 0$

Step 1.4

Solve for $x$ .

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Step 1.4.1

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

$x + 1 = 0$

$x - 1 = 0$

Step 1.4.2

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

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Step 1.4.2.1

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

$x + 1 = 0$

Step 1.4.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.4.3

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

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Step 1.4.3.1

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

$x - 1 = 0$

Step 1.4.3.2

Add $1$ to both sides of the equation.

$x = 1$

Step 1.4.4

The final solution is all the values that make $(x + 1) (x - 1) \geq 0$ true.

$x = - 1, 1$

Step 1.4.5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Step 1.4.6

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 1.4.6.1

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

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Step 1.4.6.1.1

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

$x = - 4$

Step 1.4.6.1.2

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

$((- 4) + 1) ((- 4) - 1) \geq 0$

Step 1.4.6.1.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.4.6.2

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

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Step 1.4.6.2.1

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

$x = 0$

Step 1.4.6.2.2

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

$((0) + 1) ((0) - 1) \geq 0$

Step 1.4.6.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.4.6.3

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

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Step 1.4.6.3.1

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

$x = 4$

Step 1.4.6.3.2

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

$((4) + 1) ((4) - 1) \geq 0$

Step 1.4.6.3.3

The left side $15$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.4.6.4

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

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 1.4.7

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 1$

Step 1.5

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

Interval Notation:

$(1, \infty)$

Set-Builder Notation:

${x | x > 1}$

Interval Notation:

$(1, \infty)$

Set-Builder Notation:

${x | x > 1}$

Step 2

To find the radical expression end point, substitute the $x$ value $1$ , which is the least value in the domain, into $f (x) = \ln (x \sqrt{(x + 1) (x - 1)})$ .

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Step 2.1

Replace the variable $x$ with $1$ in the expression.

$f (1) = \ln ((1) \sqrt{((1) + 1) ((1) - 1)})$

Step 2.2

Multiply $\sqrt{((1) + 1) ((1) - 1)}$ by $1$ .

$f (1) = \ln (\sqrt{((1) + 1) ((1) - 1)})$

Step 2.3

Add $1$ and $1$ .

$f (1) = \ln (\sqrt{2 (1 - 1)})$

Step 2.4

Subtract $1$ from $1$ .

$f (1) = \ln (\sqrt{2 \cdot 0})$

Step 2.5

Multiply $2$ by $0$ .

$f (1) = \ln (\sqrt{0})$

Step 2.6

Rewrite $0$ as $0^{2}$ .

$f (1) = \ln (\sqrt{0^{2}})$

Step 2.7

Pull terms out from under the radical, assuming positive real numbers.

$f (1) = \ln (0)$

Step 2.8

The natural logarithm of zero is undefined.

$f (1) = Undefined$

Undefined

Step 3

The radical expression end point is $(1, Undefined)$ .

$(1, Undefined)$

Step 4

Select a few $x$ values from the domain. It would be more useful to select the values so that they are next to the $x$ value of the radical expression end point.

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Step 4.1

Substitute the $x$ value $2$ into $f (x) = \ln (x \sqrt{(x + 1) (x - 1)})$ . In this case, the point is $(2, \ln (2 \sqrt{3}))$ .

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Step 4.1.1

Replace the variable $x$ with $2$ in the expression.

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Add $2$ and $1$ .

$f (2) = \ln (2 \sqrt{3 (2 - 1)})$

Step 4.1.2.2

Subtract $1$ from $2$ .

$f (2) = \ln (2 \sqrt{3 \cdot 1})$

Step 4.1.2.3

Multiply $3$ by $1$ .

$f (2) = \ln (2 \sqrt{3})$

Step 4.1.2.4

The final answer is $\ln (2 \sqrt{3})$ .

$y = \ln (2 \sqrt{3})$

Step 4.2

Substitute the $x$ value $3$ into $f (x) = \ln (x \sqrt{(x + 1) (x - 1)})$ . In this case, the point is $(3, \ln (6 \sqrt{2}))$ .

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Step 4.2.1

Replace the variable $x$ with $3$ in the expression.

$f (3) = \ln ((3) \sqrt{((3) + 1) ((3) - 1)})$

Step 4.2.2

Simplify the result.

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Step 4.2.2.1

Add $3$ and $1$ .

$f (3) = \ln (3 \sqrt{4 (3 - 1)})$

Step 4.2.2.2

Subtract $1$ from $3$ .

$f (3) = \ln (3 \sqrt{4 \cdot 2})$

Step 4.2.2.3

Multiply $4$ by $2$ .

$f (3) = \ln (3 \sqrt{8})$

Step 4.2.2.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 4.2.2.4.1

Factor $4$ out of $8$ .

$f (3) = \ln (3 \sqrt{4 (2)})$

Step 4.2.2.4.2

Rewrite $4$ as $2^{2}$ .

$f (3) = \ln (3 \sqrt{2^{2} \cdot 2})$

Step 4.2.2.5

Pull terms out from under the radical.

$f (3) = \ln (3 (2 \sqrt{2}))$

Step 4.2.2.6

Multiply $2$ by $3$ .

$f (3) = \ln (6 \sqrt{2})$

Step 4.2.2.7

The final answer is $\ln (6 \sqrt{2})$ .

$y = \ln (6 \sqrt{2})$

Step 4.3

The square root can be graphed using the points around the vertex $(1, Undefined), (2, 1.24), (3, 2.14)$

$\begin{array}{c} x & y \\ 1 & Undefined \\ 2 & 1.24 \\ 3 & 2.14 \end{array}$

Step 5