Precalculus Examples

$f (x) = \frac{1 - 2 x}{x + 3} + \sqrt{5 - 4 x - x^{2}}$

Step 1

Set the radicand in $\sqrt{5 - 4 x - x^{2}}$ greater than or equal to $0$ to find where the expression is defined.

$5 - 4 x - x^{2} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$5 - 4 x - x^{2} = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $- 1$ out of $5 - 4 x - x^{2}$ .

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Step 2.2.1.1

Reorder the expression.

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Step 2.2.1.1.1

Move $5$ .

$- 4 x - x^{2} + 5 = 0$

Step 2.2.1.1.2

Reorder $- 4 x$ and $- x^{2}$ .

$- x^{2} - 4 x + 5 = 0$

Step 2.2.1.2

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

$- (x^{2}) - 4 x + 5 = 0$

Step 2.2.1.3

Factor $- 1$ out of $- 4 x$ .

$- (x^{2}) - (4 x) + 5 = 0$

Step 2.2.1.4

Rewrite $5$ as $- 1 (- 5)$ .

$- (x^{2}) - (4 x) - 1 \cdot - 5 = 0$

Step 2.2.1.5

Factor $- 1$ out of $- (x^{2}) - (4 x)$ .

$- (x^{2} + 4 x) - 1 \cdot - 5 = 0$

Step 2.2.1.6

Factor $- 1$ out of $- (x^{2} + 4 x) - 1 (- 5)$ .

$- (x^{2} + 4 x - 5) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor $x^{2} + 4 x - 5$ 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 $- 5$ and whose sum is $4$ .

$- 1, 5$

Step 2.2.2.1.2

Write the factored form using these integers.

$- ((x - 1) (x + 5)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$- (x - 1) (x + 5) = 0$

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 $0$ .

$x - 1 = 0$

$x + 5 = 0$

Step 2.4

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

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Step 2.4.1

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

$x - 1 = 0$

Step 2.4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.5

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 2.5.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.6

The final solution is all the values that make $- (x - 1) (x + 5) = 0$ true.

$x = 1, - 5$

Step 2.7

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$x > 1$

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 $x < - 5$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 2.8.1.2

Replace $x$ with $- 8$ in the original inequality.

$5 - 4 \cdot - 8 - {(- 8)}^{2} \geq 0$

Step 2.8.1.3

The left side $- 27$ 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 $- 5 < x < 1$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval $- 5 < x < 1$ 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.

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

Step 2.8.2.3

The left side $5$ 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 $x > 1$ to see if it makes the inequality true.

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Step 2.8.3.1

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

$x = 4$

Step 2.8.3.2

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

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

Step 2.8.3.3

The left side $- 27$ 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.

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

$x < - 5$ False

$- 5 < x < 1$ True

$x > 1$ False

Step 2.9

The solution consists of all of the true intervals.

$- 5 \leq x \leq 1$

Step 3

Set the denominator in $\frac{1 - 2 x}{x + 3}$ equal to $0$ to find where the expression is undefined.

$x + 3 = 0$

Step 4

Subtract $3$ from both sides of the equation.

$x = - 3$

Step 5

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

Interval Notation:

$[- 5, - 3) \cup (- 3, 1]$

Set-Builder Notation:

${x | - 5 \leq x \leq 1, x \neq - 3}$

Step 6