Precalculus Examples

$f (x) = \sqrt{2 x^{3} - 5 x^{2} - 3 x}$

Step 1

Set the radicand in $\sqrt{2 x^{3} - 5 x^{2} - 3 x}$ greater than or equal to $0$ to find where the expression is defined.

$2 x^{3} - 5 x^{2} - 3 x \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$2 x^{3} - 5 x^{2} - 3 x = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $x$ out of $2 x^{3} - 5 x^{2} - 3 x$ .

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Step 2.2.1.1

Factor $x$ out of $2 x^{3}$ .

$x (2 x^{2}) - 5 x^{2} - 3 x = 0$

Step 2.2.1.2

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

$x (2 x^{2}) + x (- 5 x) - 3 x = 0$

Step 2.2.1.3

Factor $x$ out of $- 3 x$ .

$x (2 x^{2}) + x (- 5 x) + x \cdot - 3 = 0$

Step 2.2.1.4

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

$x (2 x^{2} - 5 x) + x \cdot - 3 = 0$

Step 2.2.1.5

Factor $x$ out of $x (2 x^{2} - 5 x) + x \cdot - 3$ .

$x (2 x^{2} - 5 x - 3) = 0$

Step 2.2.2

Factor.

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Step 2.2.2.1

Factor by grouping.

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Step 2.2.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = - 5$ .

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Step 2.2.2.1.1.1

Factor $- 5$ out of $- 5 x$ .

$x (2 x^{2} - 5 x - 3) = 0$

Step 2.2.2.1.1.2

Rewrite $- 5$ as $1$ plus $- 6$

$x (2 x^{2} + (1 - 6) x - 3) = 0$

Step 2.2.2.1.1.3

Apply the distributive property.

$x (2 x^{2} + 1 x - 6 x - 3) = 0$

Step 2.2.2.1.1.4

Multiply $x$ by $1$ .

$x (2 x^{2} + x - 6 x - 3) = 0$

Step 2.2.2.1.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1.2.1

Group the first two terms and the last two terms.

$x ((2 x^{2} + x) - 6 x - 3) = 0$

Step 2.2.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$x (x (2 x + 1) - 3 (2 x + 1)) = 0$

Step 2.2.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 x + 1$ .

$x ((2 x + 1) (x - 3)) = 0$

Step 2.2.2.2

Remove unnecessary parentheses.

$x (2 x + 1) (x - 3) = 0$

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 $0$ .

$x = 0$

$2 x + 1 = 0$

$x - 3 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $2 x + 1$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Step 2.5.2

Solve $2 x + 1 = 0$ for $x$ .

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Step 2.5.2.1

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Step 2.5.2.2

Divide each term in $2 x = - 1$ by $2$ and simplify.

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Step 2.5.2.2.1

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 2.5.2.2.2

Simplify the left side.

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Step 2.5.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.5.2.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Step 2.5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Step 2.5.2.2.3

Simplify the right side.

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Step 2.5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 2.6

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

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Step 2.6.1

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

$x - 3 = 0$

Step 2.6.2

Add $3$ to both sides of the equation.

$x = 3$

Step 2.7

The final solution is all the values that make $x (2 x + 1) (x - 3) = 0$ true.

$x = 0, - \frac{1}{2}, 3$

Step 2.8

Use each root to create test intervals.

$x < - \frac{1}{2}$

$- \frac{1}{2} < x < 0$

$0 < x < 3$

$x > 3$

Step 2.9

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 2.9.1

Test a value on the interval $x < - \frac{1}{2}$ to see if it makes the inequality true.

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Step 2.9.1.1

Choose a value on the interval $x < - \frac{1}{2}$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.9.1.2

Replace $x$ with $- 3$ in the original inequality.

$2 {(- 3)}^{3} - 5 {(- 3)}^{2} - 3 \cdot - 3 \geq 0$

Step 2.9.1.3

The left side $- 90$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.9.2

Test a value on the interval $- \frac{1}{2} < x < 0$ to see if it makes the inequality true.

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Step 2.9.2.1

Choose a value on the interval $- \frac{1}{2} < x < 0$ and see if this value makes the original inequality true.

$x = - 0.25$

Step 2.9.2.2

Replace $x$ with $- 0.25$ in the original inequality.

$2 {(- 0.25)}^{3} - 5 {(- 0.25)}^{2} - 3 \cdot - 0.25 \geq 0$

Step 2.9.2.3

The left side $0.40625$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.3

Test a value on the interval $0 < x < 3$ to see if it makes the inequality true.

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Step 2.9.3.1

Choose a value on the interval $0 < x < 3$ and see if this value makes the original inequality true.

$x = 2$

Step 2.9.3.2

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

$2 {(2)}^{3} - 5 {(2)}^{2} - 3 \cdot 2 \geq 0$

Step 2.9.3.3

The left side $- 10$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.9.4

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

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Step 2.9.4.1

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

$x = 6$

Step 2.9.4.2

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

$2 {(6)}^{3} - 5 {(6)}^{2} - 3 \cdot 6 \geq 0$

Step 2.9.4.3

The left side $234$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.5

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

$x < - \frac{1}{2}$ False

$- \frac{1}{2} < x < 0$ True

$0 < x < 3$ False

$x > 3$ True

$x < - \frac{1}{2}$ False

$- \frac{1}{2} < x < 0$ True

$0 < x < 3$ False

$x > 3$ True

Step 2.10

The solution consists of all of the true intervals.

$- \frac{1}{2} \leq x \leq 0$ or $x \geq 3$

Step 3

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

Interval Notation:

$[- \frac{1}{2}, 0] \cup [3, \infty)$

Set-Builder Notation:

${x | - \frac{1}{2} \leq x \leq 0, x \geq 3}$

Step 4