Find the Domain square root of x^3+1

$\sqrt{x^{3} + 1}$

Step 1

Set the radicand in $\sqrt{x^{3} + 1}$ greater than or equal to $0$ to find where the expression is defined.

$x^{3} + 1 \geq 0$

Step 2

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$x^{3} \geq - 1$

Add $1$ to both sides of the inequality.

$x^{3} + 1 \geq 0$

Convert the inequality to an equation.

$x^{3} + 1 = 0$

Factor the left side of the equation.

Rewrite $1$ as $1^{3}$ .

$x^{3} + 1^{3} = 0$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

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

Simplify.

Multiply $- 1$ by $1$ .

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

One to any power is one.

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

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

$x + 1 = 0$

$x^{2} - x + 1 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

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

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

$x^{2} - x + 1 = 0$

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Change the $\pm$ to $+$ .

$x = \frac{1 + i \sqrt{3}}{2}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

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

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

Change the $\pm$ to $-$ .

$x = \frac{1 - i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

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

$x = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Identify the leading coefficient.

The leading term in a polynomial is the term with the highest degree.

$x^{3}$

The leading coefficient in a polynomial is the coefficient of the leading term.

$1$

Since there are no real x-intercepts and the leading coefficient is positive, the parabola opens up and $x^{3} + 1$ is always greater than $0$ .

All real numbers

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Find the Domain square root of x^3+1

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