Calculus Examples

$f (x) = \sqrt{x^{2} + x}$

Step 1

Find the domain to determine if the expression is continuous.

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Step 1.1

Set the radicand in $\sqrt{x^{2} + x}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} + x \geq 0$

Step 1.2

Solve for $x$ .

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Step 1.2.1

Convert the inequality to an equation.

$x^{2} + x = 0$

Step 1.2.2

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

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Step 1.2.2.1

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

$x \cdot x + x = 0$

Step 1.2.2.2

Raise $x$ to the power of $1$ .

$x \cdot x + x = 0$

Step 1.2.2.3

Factor $x$ out of $x^{1}$ .

$x \cdot x + x \cdot 1 = 0$

Step 1.2.2.4

Factor $x$ out of $x \cdot x + x \cdot 1$ .

$x (x + 1) = 0$

Step 1.2.3

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

$x = 0$

$x + 1 = 0$

Step 1.2.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.5

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

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Step 1.2.5.1

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

$x + 1 = 0$

Step 1.2.5.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 1.2.6

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

$x = 0, - 1$

Step 1.2.7

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$x > 0$

Step 1.2.8

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 1.2.8.1

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

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Step 1.2.8.1.1

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

$x = - 4$

Step 1.2.8.1.2

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

${(- 4)}^{2} - 4 \geq 0$

Step 1.2.8.1.3

The left side $12$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 1.2.8.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 1.2.8.2.2

Replace $x$ with $- 0.5$ in the original inequality.

${(- 0.5)}^{2} - 0.5 \geq 0$

Step 1.2.8.2.3

The left side $- 0.25$ is less than the right side $0$ , which means that the given statement is false.

False

Step 1.2.8.3

Test a value on the interval $x > 0$ to see if it makes the inequality true.

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Step 1.2.8.3.1

Choose a value on the interval $x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 1.2.8.3.2

Replace $x$ with $2$ in the original inequality.

${(2)}^{2} + 2 \geq 0$

Step 1.2.8.3.3

The left side $6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 1.2.8.4

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

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

$x < - 1$ True

$- 1 < x < 0$ False

$x > 0$ True

Step 1.2.9

The solution consists of all of the true intervals.

$x \leq - 1$ or $x \geq 0$

Step 1.3

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

Interval Notation:

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

Set-Builder Notation:

${x | x \leq - 1, x \geq 0}$

Interval Notation:

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

Set-Builder Notation:

${x | x \leq - 1, x \geq 0}$

Step 2

Since the domain is not all real numbers, $\sqrt{x^{2} + x}$ is not continuous over all real numbers.

Not continuous

Step 3