Calculus Examples

$\frac{4 - 3 x - x^{2}}{x^{2} - 25} > 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$4 - 3 x - x^{2} = 0$

$x^{2} - 25 = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $- 1$ out of $4 - 3 x - x^{2}$ .

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Step 2.1.1

Reorder the expression.

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Step 2.1.1.1

Move $4$ .

$- 3 x - x^{2} + 4 = 0$

Step 2.1.1.2

Reorder $- 3 x$ and $- x^{2}$ .

$- x^{2} - 3 x + 4 = 0$

Step 2.1.2

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

$- (x^{2}) - 3 x + 4 = 0$

Step 2.1.3

Factor $- 1$ out of $- 3 x$ .

$- (x^{2}) - (3 x) + 4 = 0$

Step 2.1.4

Rewrite $4$ as $- 1 (- 4)$ .

$- (x^{2}) - (3 x) - 1 \cdot - 4 = 0$

Step 2.1.5

Factor $- 1$ out of $- (x^{2}) - (3 x)$ .

$- (x^{2} + 3 x) - 1 \cdot - 4 = 0$

Step 2.1.6

Factor $- 1$ out of $- (x^{2} + 3 x) - 1 (- 4)$ .

$- (x^{2} + 3 x - 4) = 0$

Step 2.2

Factor.

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Step 2.2.1

Factor $x^{2} + 3 x - 4$ using the AC method.

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Step 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Step 2.2.1.2

Write the factored form using these integers.

$- ((x - 1) (x + 4)) = 0$

Step 2.2.2

Remove unnecessary parentheses.

$- (x - 1) (x + 4) = 0$

Step 3

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

$x - 1 = 0$

$x + 4 = 0$

Step 4

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

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Step 4.1

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

$x - 1 = 0$

Step 4.2

Add $1$ to both sides of the equation.

$x = 1$

Step 5

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

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Step 5.1

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

$x + 4 = 0$

Step 5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 6

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

$x = 1, - 4$

Step 7

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 8

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Step 9

Simplify $\pm \sqrt{25}$ .

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Step 9.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 9.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 10

The complete solution is the result of both the positive and negative portions of the solution.

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Step 10.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Step 10.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

Step 10.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$

Step 11

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 1, - 4$

$x = 5, - 5$

Step 12

Consolidate the solutions.

$x = 1, - 4, 5, - 5$

Step 13

Find the domain of $\frac{4 - 3 x - x^{2}}{x^{2} - 25}$ .

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Step 13.1

Set the denominator in $\frac{4 - 3 x - x^{2}}{x^{2} - 25}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 25 = 0$

Step 13.2

Solve for $x$ .

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Step 13.2.1

Add $25$ to both sides of the equation.

$x^{2} = 25$

Step 13.2.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{25}$

Step 13.2.3

Simplify $\pm \sqrt{25}$ .

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Step 13.2.3.1

Rewrite $25$ as $5^{2}$ .

$x = \pm \sqrt{5^{2}}$

Step 13.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 5$

Step 13.2.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 13.2.4.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 5$

Step 13.2.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 5$

Step 13.2.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 5, - 5$

Step 13.3

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

$(- \infty, - 5) \cup (- 5, 5) \cup (5, \infty)$

Step 14

Use each root to create test intervals.

$x < - 5$

$- 5 < x < - 4$

$- 4 < x < 1$

$1 < x < 5$

$x > 5$

Step 15

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 15.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 15.1.1

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

$x = - 8$

Step 15.1.2

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

$\frac{4 - 3 \cdot - 8 - {(- 8)}^{2}}{{(- 8)}^{2} - 25} > 0$

Step 15.1.3

The left side $- 0.\overline{923076}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 15.2

Test a value on the interval $- 5 < x < - 4$ to see if it makes the inequality true.

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Step 15.2.1

Choose a value on the interval $- 5 < x < - 4$ and see if this value makes the original inequality true.

$x = - 4.5$

Step 15.2.2

Replace $x$ with $- 4.5$ in the original inequality.

$\frac{4 - 3 \cdot - 4.5 - {(- 4.5)}^{2}}{{(- 4.5)}^{2} - 25} > 0$

Step 15.2.3

The left side $0.57894736$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 15.3

Test a value on the interval $- 4 < x < 1$ to see if it makes the inequality true.

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Step 15.3.1

Choose a value on the interval $- 4 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 15.3.2

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

$\frac{4 - 3 \cdot 0 - {(0)}^{2}}{{(0)}^{2} - 25} > 0$

Step 15.3.3

The left side $- 0.16$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 15.4

Test a value on the interval $1 < x < 5$ to see if it makes the inequality true.

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Step 15.4.1

Choose a value on the interval $1 < x < 5$ and see if this value makes the original inequality true.

$x = 3$

Step 15.4.2

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

$\frac{4 - 3 \cdot 3 - {(3)}^{2}}{{(3)}^{2} - 25} > 0$

Step 15.4.3

The left side $0.875$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 15.5

Test a value on the interval $x > 5$ to see if it makes the inequality true.

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Step 15.5.1

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

$x = 8$

Step 15.5.2

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

$\frac{4 - 3 \cdot 8 - {(8)}^{2}}{{(8)}^{2} - 25} > 0$

Step 15.5.3

The left side $- 2.\overline{153846}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 15.6

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

$x < - 5$ False

$- 5 < x < - 4$ True

$- 4 < x < 1$ False

$1 < x < 5$ True

$x > 5$ False

$x < - 5$ False

$- 5 < x < - 4$ True

$- 4 < x < 1$ False

$1 < x < 5$ True

$x > 5$ False

Step 16

The solution consists of all of the true intervals.

$- 5 < x < - 4$ or $1 < x < 5$

Step 17

Convert the inequality to interval notation.

$(- 5, - 4) \cup (1, 5)$

Step 18