Find the Domain and Range f(x)=- square root of x^3+8

$f (x) = - \sqrt{x^{3} + 8}$

Step 1

Set the radicand in $\sqrt{x^{3} + 8}$ greater than or equal to $0$ to find where the expression is defined.

$x^{3} + 8 \geq 0$

Step 2

Solve for $x$ .

Subtract $8$ from both sides of the inequality.

$x^{3} \geq - 8$

Add $8$ to both sides of the inequality.

$x^{3} + 8 \geq 0$

Convert the inequality to an equation.

$x^{3} + 8 = 0$

Factor the left side of the equation.

Rewrite $8$ as $2^{3}$ .

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

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 $b = 2$ .

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

Simplify.

Multiply $2$ by $- 1$ .

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

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

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

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

$x + 2 = 0$

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

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set $x^{2} - 2 x + 4$ equal to $0$ and solve for $x$ .

Set $x^{2} - 2 x + 4$ equal to $0$ .

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

Solve $x^{2} - 2 x + 4 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

Change the $\pm$ to $+$ .

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

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

Simplify the numerator.

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

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

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

Pull terms out from under the radical.

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

Move $2$ to the left of $i$ .

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

Multiply $2$ by $1$ .

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

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

The final solution is all the values that make $(x + 2) (x^{2} - 2 x + 4) = 0$ true.

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

The solution consists of all of the true intervals.

$x \geq - 2$

Step 3

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

Interval Notation:

$[- 2, \infty)$

Set-Builder Notation:

${x | x \geq - 2}$

Step 4

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[- 2 \sqrt{2}, 0]$

Set-Builder Notation:

${y | - 2 \sqrt{2} \leq y \leq 0}$

Step 5

Determine the domain and range.

Domain: $[- 2, \infty), {x | x \geq - 2}$

Range: $[- 2 \sqrt{2}, 0], {y | - 2 \sqrt{2} \leq y \leq 0}$

Step 6

Find the Domain and Range f(x)=- square root of x^3+8

Related Questions

Support LCMGCF.com by disabling your adblocker.