Finite Math Examples

$y = \sqrt{12 + x - x^{2}}$

Step 1

Set the radicand in $\sqrt{12 + x - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$12 + x - x^{2} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$12 + x - x^{2} = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $- 1$ out of $12 + x - x^{2}$ .

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Step 2.2.1.1

Reorder the expression.

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Step 2.2.1.1.1

Move $12$ .

$x - x^{2} + 12 = 0$

Step 2.2.1.1.2

Reorder $x$ and $- x^{2}$ .

$- x^{2} + x + 12 = 0$

Step 2.2.1.2

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

$- (x^{2}) + x + 12 = 0$

Step 2.2.1.3

Factor $- 1$ out of $x$ .

$- (x^{2}) - 1 (- x) + 12 = 0$

Step 2.2.1.4

Rewrite $12$ as $- 1 (- 12)$ .

$- (x^{2}) - 1 (- x) - 1 \cdot - 12 = 0$

Step 2.2.1.5

Factor $- 1$ out of $- (x^{2}) - 1 (- x)$ .

$- (x^{2} - x) - 1 \cdot - 12 = 0$

Step 2.2.1.6

Factor $- 1$ out of $- (x^{2} - x) - 1 (- 12)$ .

$- (x^{2} - x - 12) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $x^{2} - x - 12$ using the AC method.

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Step 2.2.2.1.1

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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Step 2.2.2.1.2

Write the factored form using these integers.

$- ((x - 4) (x + 3)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$- (x - 4) (x + 3) = 0$

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 $0$ .

$x - 4 = 0$

$x + 3 = 0$

Step 2.4

Set $x - 4$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Step 2.4.2

Add $4$ to both sides of the equation.

$x = 4$

Step 2.5

Set $x + 3$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Step 2.5.2

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 2.6

The final solution is all the values that make $- (x - 4) (x + 3) = 0$ true.

$x = 4, - 3$

Step 2.7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 4$

$x > 4$

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.

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Step 2.8.1

Test a value on the interval $x < - 3$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.8.1.2

Replace $x$ with $- 6$ in the original inequality.

$12 - 6 - {(- 6)}^{2} \geq 0$

Step 2.8.1.3

The left side $- 30$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.2

Test a value on the interval $- 3 < x < 4$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 3 < x < 4$ 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.

$12 + 0 - {(0)}^{2} \geq 0$

Step 2.8.2.3

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

True

Step 2.8.3

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 2.8.3.2

Replace $x$ with $6$ in the original inequality.

$12 + 6 - {(6)}^{2} \geq 0$

Step 2.8.3.3

The left side $- 18$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.4

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

$x < - 3$ False

$- 3 < x < 4$ True

$x > 4$ False

$x < - 3$ False

$- 3 < x < 4$ True

$x > 4$ False

Step 2.9

The solution consists of all of the true intervals.

$- 3 \leq x \leq 4$

Step 3

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

Interval Notation:

$[- 3, 4]$

Set-Builder Notation:

${x | - 3 \leq x \leq 4}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \frac{7}{2}]$

Set-Builder Notation:

${y | 0 \leq y \leq \frac{7}{2}}$

Step 5

Determine the domain and range.

Domain: $[- 3, 4], {x | - 3 \leq x \leq 4}$

Range: $[0, \frac{7}{2}], {y | 0 \leq y \leq \frac{7}{2}}$

Step 6