Calculus Examples

$f (x) = \sqrt{x^{2} - 5 x - 14}$

Step 1

Set the radicand in $\sqrt{x^{2} - 5 x - 14}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 5 x - 14 \geq 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Convert the inequality to an equation.

$x^{2} - 5 x - 14 = 0$

Step 2.2

Factor $x^{2} - 5 x - 14$ using the AC method.

Tap for more steps...

Step 2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 14$ and whose sum is $- 5$ .

$- 7, 2$

Step 2.2.2

Write the factored form using these integers.

$(x - 7) (x + 2) = 0$

Step 2.3

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

$x - 7 = 0$

$x + 2 = 0$

Step 2.4

Set $x - 7$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.4.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.5

Set $x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.5.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.6

The final solution is all the values that make $(x - 7) (x + 2) = 0$ true.

$x = 7, - 2$

Step 2.7

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 7$

$x > 7$

Step 2.8

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

Tap for more steps...

Step 2.8.1

Test a value on the interval $x < - 2$ to see if it makes the inequality true.

Tap for more steps...

Step 2.8.1.1

Choose a value on the interval $x < - 2$ and see if this value makes the original inequality true.

$x = - 4$

Step 2.8.1.2

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

${(- 4)}^{2} - 5 \cdot - 4 - 14 \geq 0$

Step 2.8.1.3

The left side $22$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.2

Test a value on the interval $- 2 < x < 7$ to see if it makes the inequality true.

Tap for more steps...

Step 2.8.2.1

Choose a value on the interval $- 2 < x < 7$ and see if this value makes the original inequality true.

$x = 0$

Step 2.8.2.2

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

${(0)}^{2} - 5 \cdot 0 - 14 \geq 0$

Step 2.8.2.3

The left side $- 14$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.3

Test a value on the interval $x > 7$ to see if it makes the inequality true.

Tap for more steps...

Step 2.8.3.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 2.8.3.2

Replace $x$ with $10$ in the original inequality.

${(10)}^{2} - 5 \cdot 10 - 14 \geq 0$

Step 2.8.3.3

The left side $36$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

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

$x < - 2$ True

$- 2 < x < 7$ False

$x > 7$ True

$x < - 2$ True

$- 2 < x < 7$ False

$x > 7$ True

Step 2.9

The solution consists of all of the true intervals.

$x \leq - 2$ or $x \geq 7$

Step 3

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

Interval Notation:

$(- \infty, - 2] \cup [7, \infty)$

Set-Builder Notation:

${x | x \leq - 2, x \geq 7}$

Step 4