Precalculus Examples

$\frac{\sqrt{4 - x}}{(x + 1) (x^{2} + 1)}$

Step 1

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

$4 - x \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Subtract $4$ from both sides of the inequality.

$- x \geq - 4$

Step 2.2

Divide each term in $- x \geq - 4$ by $- 1$ and simplify.

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Step 2.2.1

Divide each term in $- x \geq - 4$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{- 4}{- 1}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{- 4}{- 1}$

Step 2.2.2.2

Divide $x$ by $1$ .

$x \leq \frac{- 4}{- 1}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 4$ by $- 1$ .

$x \leq 4$

Step 3

Set the denominator in $\frac{\sqrt{4 - x}}{(x + 1) (x^{2} + 1)}$ equal to $0$ to find where the expression is undefined.

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

Step 4

Solve for $x$ .

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Step 4.1

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

$x + 1 = 0$

$x^{2} + 1 = 0$

Step 4.2

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

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Step 4.2.1

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

$x + 1 = 0$

Step 4.2.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4.3

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

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Step 4.3.1

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Step 4.3.2

Solve $x^{2} + 1 = 0$ for $x$ .

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Step 4.3.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 4.3.2.2

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

$x = \pm \sqrt{- 1}$

Step 4.3.2.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i$

Step 4.3.2.4

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

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Step 4.3.2.4.1

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

$x = i$

Step 4.3.2.4.2

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

$x = - i$

Step 4.3.2.4.3

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

$x = i, - i$

Step 4.4

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

$x = - 1, i, - i$

Step 5

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

Interval Notation:

$(- \infty, - 1) \cup (- 1, 4]$

Set-Builder Notation:

${x | x \leq 4, x \neq - 1}$

Step 6