Calculus Examples

$6 k^{3} - 5 k^{2} < 4 k$

Step 1

Subtract $4 k$ from both sides of the inequality.

$6 k^{3} - 5 k^{2} - 4 k < 0$

Step 2

Convert the inequality to an equation.

$6 k^{3} - 5 k^{2} - 4 k = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Factor $k$ out of $6 k^{3} - 5 k^{2} - 4 k$ .

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Step 3.1.1

Factor $k$ out of $6 k^{3}$ .

$k (6 k^{2}) - 5 k^{2} - 4 k = 0$

Step 3.1.2

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

$k (6 k^{2}) + k (- 5 k) - 4 k = 0$

Step 3.1.3

Factor $k$ out of $- 4 k$ .

$k (6 k^{2}) + k (- 5 k) + k \cdot - 4 = 0$

Step 3.1.4

Factor $k$ out of $k (6 k^{2}) + k (- 5 k)$ .

$k (6 k^{2} - 5 k) + k \cdot - 4 = 0$

Step 3.1.5

Factor $k$ out of $k (6 k^{2} - 5 k) + k \cdot - 4$ .

$k (6 k^{2} - 5 k - 4) = 0$

Step 3.2

Factor.

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Step 3.2.1

Factor by grouping.

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Step 3.2.1.1

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 $a \cdot c = 6 \cdot - 4 = - 24$ and whose sum is $b = - 5$ .

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Step 3.2.1.1.1

Factor $- 5$ out of $- 5 k$ .

$k (6 k^{2} - 5 k - 4) = 0$

Step 3.2.1.1.2

Rewrite $- 5$ as $3$ plus $- 8$

$k (6 k^{2} + (3 - 8) k - 4) = 0$

Step 3.2.1.1.3

Apply the distributive property.

$k (6 k^{2} + 3 k - 8 k - 4) = 0$

Step 3.2.1.2

Factor out the greatest common factor from each group.

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Step 3.2.1.2.1

Group the first two terms and the last two terms.

$k ((6 k^{2} + 3 k) - 8 k - 4) = 0$

Step 3.2.1.2.2

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

$k (3 k (2 k + 1) - 4 (2 k + 1)) = 0$

Step 3.2.1.3

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

$k ((2 k + 1) (3 k - 4)) = 0$

Step 3.2.2

Remove unnecessary parentheses.

$k (2 k + 1) (3 k - 4) = 0$

Step 4

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

$k = 0$

$2 k + 1 = 0$

$3 k - 4 = 0$

Step 5

Set $k$ equal to $0$ .

$k = 0$

Step 6

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

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Step 6.1

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

$2 k + 1 = 0$

Step 6.2

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

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Step 6.2.1

Subtract $1$ from both sides of the equation.

$2 k = - 1$

Step 6.2.2

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

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Step 6.2.2.1

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

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

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

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

Step 6.2.2.2.1.2

Divide $k$ by $1$ .

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

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Move the negative in front of the fraction.

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

Step 7

Set $3 k - 4$ equal to $0$ and solve for $k$ .

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Step 7.1

Set $3 k - 4$ equal to $0$ .

$3 k - 4 = 0$

Step 7.2

Solve $3 k - 4 = 0$ for $k$ .

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Step 7.2.1

Add $4$ to both sides of the equation.

$3 k = 4$

Step 7.2.2

Divide each term in $3 k = 4$ by $3$ and simplify.

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Step 7.2.2.1

Divide each term in $3 k = 4$ by $3$ .

$\frac{3 k}{3} = \frac{4}{3}$

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Cancel the common factor of $3$ .

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Step 7.2.2.2.1.1

Cancel the common factor.

$\frac{3 k}{3} = \frac{4}{3}$

Step 7.2.2.2.1.2

Divide $k$ by $1$ .

$k = \frac{4}{3}$

Step 8

The final solution is all the values that make $k (2 k + 1) (3 k - 4) = 0$ true.

$k = 0, - \frac{1}{2}, \frac{4}{3}$

Step 9

Use each root to create test intervals.

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

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

$0 < k < \frac{4}{3}$

$k > \frac{4}{3}$

Step 10

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 10.1

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

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Step 10.1.1

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

$k = - 3$

Step 10.1.2

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

$6 {(- 3)}^{3} - 5 {(- 3)}^{2} < 4 (- 3)$

Step 10.1.3

The left side $- 207$ is less than the right side $- 12$ , which means that the given statement is always true.

True

Step 10.2

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

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Step 10.2.1

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

$k = - 0.25$

Step 10.2.2

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

$6 {(- 0.25)}^{3} - 5 {(- 0.25)}^{2} < 4 (- 0.25)$

Step 10.2.3

The left side $- 0.40625$ is not less than the right side $- 1$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $0 < k < \frac{4}{3}$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $0 < k < \frac{4}{3}$ and see if this value makes the original inequality true.

$k = 1$

Step 10.3.2

Replace $k$ with $1$ in the original inequality.

$6 {(1)}^{3} - 5 {(1)}^{2} < 4 (1)$

Step 10.3.3

The left side $1$ is less than the right side $4$ , which means that the given statement is always true.

True

Step 10.4

Test a value on the interval $k > \frac{4}{3}$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $k > \frac{4}{3}$ and see if this value makes the original inequality true.

$k = 4$

Step 10.4.2

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

$6 {(4)}^{3} - 5 {(4)}^{2} < 4 (4)$

Step 10.4.3

The left side $304$ is not less than the right side $16$ , which means that the given statement is false.

False

Step 10.5

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

$k < - \frac{1}{2}$ True

$- \frac{1}{2} < k < 0$ False

$0 < k < \frac{4}{3}$ True

$k > \frac{4}{3}$ False

$k < - \frac{1}{2}$ True

$- \frac{1}{2} < k < 0$ False

$0 < k < \frac{4}{3}$ True

$k > \frac{4}{3}$ False

Step 11

The solution consists of all of the true intervals.

$k < - \frac{1}{2}$ or $0 < k < \frac{4}{3}$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$k < - \frac{1}{2} or 0 < k < \frac{4}{3}$

Interval Notation:

$(- \infty, - \frac{1}{2}) \cup (0, \frac{4}{3})$

Step 13