Calculus Examples

$y = 4 x^{2}$ , $y = 4 \sqrt{x}$

Step 1

Solve by substitution to find the intersection between the curves.

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Step 1.1

Eliminate the equal sides of each equation and combine.

$4 x^{2} = 4 \sqrt{x}$

Step 1.2

Solve $4 x^{2} = 4 \sqrt{x}$ for $x$ .

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Step 1.2.1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$4 \sqrt{x} = 4 x^{2}$

Step 1.2.2

To remove the radical on the left side of the equation, square both sides of the equation.

${(4 \sqrt{x})}^{2} = {(4 x^{2})}^{2}$

Step 1.2.3

Simplify each side of the equation.

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Step 1.2.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(4 x^{\frac{1}{2}})}^{2} = {(4 x^{2})}^{2}$

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Simplify ${(4 x^{\frac{1}{2}})}^{2}$ .

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Step 1.2.3.2.1.1

Apply the product rule to $4 x^{\frac{1}{2}}$ .

$4^{2} {(x^{\frac{1}{2}})}^{2} = {(4 x^{2})}^{2}$

Step 1.2.3.2.1.2

Raise $4$ to the power of $2$ .

$16 {(x^{\frac{1}{2}})}^{2} = {(4 x^{2})}^{2}$

Step 1.2.3.2.1.3

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

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Step 1.2.3.2.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 x^{\frac{1}{2} \cdot 2} = {(4 x^{2})}^{2}$

Step 1.2.3.2.1.3.2

Cancel the common factor of $2$ .

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Step 1.2.3.2.1.3.2.1

Cancel the common factor.

$16 x^{\frac{1}{2} \cdot 2} = {(4 x^{2})}^{2}$

Step 1.2.3.2.1.3.2.2

Rewrite the expression.

$16 x^{1} = {(4 x^{2})}^{2}$

Step 1.2.3.2.1.4

Simplify.

$16 x = {(4 x^{2})}^{2}$

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Simplify ${(4 x^{2})}^{2}$ .

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Step 1.2.3.3.1.1

Apply the product rule to $4 x^{2}$ .

$16 x = 4^{2} {(x^{2})}^{2}$

Step 1.2.3.3.1.2

Raise $4$ to the power of $2$ .

$16 x = 16 {(x^{2})}^{2}$

Step 1.2.3.3.1.3

Multiply the exponents in ${(x^{2})}^{2}$ .

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Step 1.2.3.3.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$16 x = 16 x^{2 \cdot 2}$

Step 1.2.3.3.1.3.2

Multiply $2$ by $2$ .

$16 x = 16 x^{4}$

Step 1.2.4

Solve for $x$ .

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Step 1.2.4.1

Subtract $16 x^{4}$ from both sides of the equation.

$16 x - 16 x^{4} = 0$

Step 1.2.4.2

Factor the left side of the equation.

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Step 1.2.4.2.1

Factor $16 x$ out of $16 x - 16 x^{4}$ .

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Step 1.2.4.2.1.1

Factor $16 x$ out of $16 x$ .

$16 x (1) - 16 x^{4} = 0$

Step 1.2.4.2.1.2

Factor $16 x$ out of $- 16 x^{4}$ .

$16 x (1) + 16 x (- x^{3}) = 0$

Step 1.2.4.2.1.3

Factor $16 x$ out of $16 x (1) + 16 x (- x^{3})$ .

$16 x (1 - x^{3}) = 0$

Step 1.2.4.2.2

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

$16 x (1^{3} - x^{3}) = 0$

Step 1.2.4.2.3

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

$16 x ((1 - x) (1^{2} + 1 x + x^{2})) = 0$

Step 1.2.4.2.4

Factor.

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Step 1.2.4.2.4.1

Simplify.

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Step 1.2.4.2.4.1.1

One to any power is one.

$16 x ((1 - x) (1 + 1 x + x^{2})) = 0$

Step 1.2.4.2.4.1.2

Multiply $x$ by $1$ .

$16 x ((1 - x) (1 + x + x^{2})) = 0$

Step 1.2.4.2.4.2

Remove unnecessary parentheses.

$16 x (1 - x) (1 + x + x^{2}) = 0$

Step 1.2.4.3

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

$x = 0$

$1 - x = 0$

$1 + x + x^{2} = 0 + y = 4 \sqrt{x}$

Step 1.2.4.4

Set $x$ equal to $0$ .

$x = 0$

Step 1.2.4.5

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

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Step 1.2.4.5.1

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

$1 - x = 0$

Step 1.2.4.5.2

Solve $1 - x = 0$ for $x$ .

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Step 1.2.4.5.2.1

Subtract $1$ from both sides of the equation.

$- x = - 1$

Step 1.2.4.5.2.2

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

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Step 1.2.4.5.2.2.1

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Step 1.2.4.5.2.2.2

Simplify the left side.

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Step 1.2.4.5.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Step 1.2.4.5.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Step 1.2.4.5.2.2.3

Simplify the right side.

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Step 1.2.4.5.2.2.3.1

Divide $- 1$ by $- 1$ .

$x = 1$

Step 1.2.4.6

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

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Step 1.2.4.6.1

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

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

Step 1.2.4.6.2

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

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Step 1.2.4.6.2.1

Use the quadratic formula to find the solutions.

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

Step 1.2.4.6.2.2

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

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1} + y = 4 \sqrt{x}$

Step 1.2.4.6.2.3

Simplify.

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Step 1.2.4.6.2.3.1

Simplify the numerator.

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Step 1.2.4.6.2.3.1.1

One to any power is one.

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

Step 1.2.4.6.2.3.1.2

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

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Step 1.2.4.6.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.2.4.6.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 1.2.4.6.2.3.1.3

Subtract $4$ from $1$ .

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

Step 1.2.4.6.2.3.1.4

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

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

Step 1.2.4.6.2.3.1.5

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

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

Step 1.2.4.6.2.3.1.6

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

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

Step 1.2.4.6.2.3.2

Multiply $2$ by $1$ .

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

Step 1.2.4.6.2.4

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

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Step 1.2.4.6.2.4.1

Simplify the numerator.

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Step 1.2.4.6.2.4.1.1

One to any power is one.

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

Step 1.2.4.6.2.4.1.2

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

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Step 1.2.4.6.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.2.4.6.2.4.1.2.2

Multiply $- 4$ by $1$ .

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

Step 1.2.4.6.2.4.1.3

Subtract $4$ from $1$ .

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

Step 1.2.4.6.2.4.1.4

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

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

Step 1.2.4.6.2.4.1.5

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

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

Step 1.2.4.6.2.4.1.6

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

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

Step 1.2.4.6.2.4.2

Multiply $2$ by $1$ .

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

Step 1.2.4.6.2.4.3

Change the $\pm$ to $+$ .

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

Step 1.2.4.6.2.4.4

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

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

Step 1.2.4.6.2.4.5

Factor $- 1$ out of $i \sqrt{3}$ .

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

Step 1.2.4.6.2.4.6

Factor $- 1$ out of $- 1 (1) - (- i \sqrt{3})$ .

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

Step 1.2.4.6.2.4.7

Move the negative in front of the fraction.

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

Step 1.2.4.6.2.5

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

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Step 1.2.4.6.2.5.1

Simplify the numerator.

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Step 1.2.4.6.2.5.1.1

One to any power is one.

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

Step 1.2.4.6.2.5.1.2

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

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Step 1.2.4.6.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 1.2.4.6.2.5.1.2.2

Multiply $- 4$ by $1$ .

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

Step 1.2.4.6.2.5.1.3

Subtract $4$ from $1$ .

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

Step 1.2.4.6.2.5.1.4

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

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

Step 1.2.4.6.2.5.1.5

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

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

Step 1.2.4.6.2.5.1.6

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

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

Step 1.2.4.6.2.5.2

Multiply $2$ by $1$ .

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

Step 1.2.4.6.2.5.3

Change the $\pm$ to $-$ .

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

Step 1.2.4.6.2.5.4

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

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

Step 1.2.4.6.2.5.5

Factor $- 1$ out of $- i \sqrt{3}$ .

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

Step 1.2.4.6.2.5.6

Factor $- 1$ out of $- 1 (1) - (i \sqrt{3})$ .

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

Step 1.2.4.6.2.5.7

Move the negative in front of the fraction.

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

Step 1.2.4.6.2.6

The final answer is the combination of both solutions.

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

Step 1.2.4.7

The final solution is all the values that make $16 x (1 - x) (1 + x + x^{2}) = 0$ true.

$x = 0, 1, - \frac{1 - i \sqrt{3}}{2}, - \frac{1 + i \sqrt{3}}{2}$

Step 1.3

Evaluate $y$ when $x = 0$ .

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Step 1.3.1

Substitute $0$ for $x$ .

$y = 4 \sqrt{0}$

Step 1.3.2

Simplify $4 \sqrt{0}$ .

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Step 1.3.2.1

Remove parentheses.

$y = 4 \sqrt{0}$

Step 1.3.2.2

Rewrite $0$ as $0^{2}$ .

$y = 4 \sqrt{0^{2}}$

Step 1.3.2.3

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

$y = 4 \cdot 0$

Step 1.3.2.4

Multiply $4$ by $0$ .

$y = 0$

Step 1.4

Evaluate $y$ when $x = 1$ .

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Step 1.4.1

Substitute $1$ for $x$ .

$y = 4 \sqrt{1}$

Step 1.4.2

Simplify $4 \sqrt{1}$ .

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Step 1.4.2.1

Remove parentheses.

$y = 4 \sqrt{1}$

Step 1.4.2.2

Any root of $1$ is $1$ .

$y = 4 \cdot 1$

Step 1.4.2.3

Multiply $4$ by $1$ .

$y = 4$

Step 1.5

Evaluate $y$ when $x = - \frac{1 - i \sqrt{3}}{2}$ .

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Step 1.5.1

Substitute $- \frac{1 - i \sqrt{3}}{2}$ for $x$ .

$y = 4 \sqrt{- \frac{1 - i \sqrt{3}}{2}}$

Step 1.5.2

Substitute $- \frac{1 - i \sqrt{3}}{2}$ for $x$ in $y = 4 \sqrt{- \frac{1 - i \sqrt{3}}{2}}$ and solve for $y$ .

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Step 1.5.2.1

Remove parentheses.

$y = 4 \sqrt{- \frac{1 - i \sqrt{3}}{2}}$

Step 1.5.2.2

Simplify $4 \sqrt{- \frac{1 - i \sqrt{3}}{2}}$ .

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Step 1.5.2.2.1

Rewrite $- \frac{1 - i \sqrt{3}}{2}$ as $i^{2} \frac{1^{2} - i \sqrt{3}}{2}$ .

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Step 1.5.2.2.1.1

Rewrite $- 1$ as $i^{2}$ .

$y = 4 \sqrt{i^{2} \frac{1 - i \sqrt{3}}{2}}$

Step 1.5.2.2.1.2

Rewrite $1$ as $1^{2}$ .

$y = 4 \sqrt{i^{2} \frac{1^{2} - i \sqrt{3}}{2}}$

Step 1.5.2.2.2

Pull terms out from under the radical.

$y = 4 (i \sqrt{\frac{1^{2} - i \sqrt{3}}{2}})$

Step 1.5.2.2.3

One to any power is one.

$y = 4 (i \sqrt{\frac{1 - i \sqrt{3}}{2}})$

Step 1.5.2.2.4

Rewrite $\sqrt{\frac{1 - i \sqrt{3}}{2}}$ as $\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}}$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}})$

Step 1.5.2.2.5

Multiply $\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = 4 (i (\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}))$

Step 1.5.2.2.6

Simplify terms.

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Step 1.5.2.2.6.1

Combine and simplify the denominator.

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Step 1.5.2.2.6.1.1

Multiply $\frac{\sqrt{1 - i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{\sqrt{2} \sqrt{2}})$

Step 1.5.2.2.6.1.2

Raise $\sqrt{2}$ to the power of $1$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}})$

Step 1.5.2.2.6.1.3

Raise $\sqrt{2}$ to the power of $1$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}})$

Step 1.5.2.2.6.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{1 + 1}})$

Step 1.5.2.2.6.1.5

Add $1$ and $1$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{2}})$

Step 1.5.2.2.6.1.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 1.5.2.2.6.1.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}})$

Step 1.5.2.2.6.1.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}})$

Step 1.5.2.2.6.1.6.3

Combine $\frac{1}{2}$ and $2$ .

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{2^{\frac{2}{2}}})$

Step 1.5.2.2.6.1.6.4

Cancel the common factor of $2$ .

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Step 1.5.2.2.6.1.6.4.1

Cancel the common factor.

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{2^{\frac{2}{2}}})$

Step 1.5.2.2.6.1.6.4.2

Rewrite the expression.

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{2^{1}})$

Step 1.5.2.2.6.1.6.5

Evaluate the exponent.

$y = 4 (i \frac{\sqrt{1 - i \sqrt{3}} \sqrt{2}}{2})$

Step 1.5.2.2.6.2

Combine using the product rule for radicals.

$y = 4 (i \frac{\sqrt{(1 - i \sqrt{3}) \cdot 2}}{2})$

Step 1.5.2.2.6.3

Combine $i$ and $\frac{\sqrt{(1 - i \sqrt{3}) \cdot 2}}{2}$ .

$y = 4 \frac{i \sqrt{(1 - i \sqrt{3}) \cdot 2}}{2}$

Step 1.5.2.2.7

Simplify the numerator.

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Step 1.5.2.2.7.1

Apply the distributive property.

$y = 4 \frac{i \sqrt{1 \cdot 2 - i \sqrt{3} \cdot 2}}{2}$

Step 1.5.2.2.7.2

Multiply $2$ by $1$ .

$y = 4 \frac{i \sqrt{2 - i \sqrt{3} \cdot 2}}{2}$

Step 1.5.2.2.7.3

Multiply $2$ by $- 1$ .

$y = 4 \frac{i \sqrt{2 - 2 i \sqrt{3}}}{2}$

Step 1.5.2.2.8

Cancel the common factor of $2$ .

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Step 1.5.2.2.8.1

Factor $2$ out of $4$ .

$y = 2 (2) \frac{i \sqrt{2 - 2 i \sqrt{3}}}{2}$

Step 1.5.2.2.8.2

Cancel the common factor.

$y = 2 \cdot 2 \frac{i \sqrt{2 - 2 i \sqrt{3}}}{2}$

Step 1.5.2.2.8.3

Rewrite the expression.

$y = 2 (i \sqrt{2 - 2 i \sqrt{3}})$

$y = 2 i \sqrt{2 - 2 i \sqrt{3}}$

Step 1.6

Evaluate $y$ when $x = - \frac{1 + i \sqrt{3}}{2}$ .

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Step 1.6.1

Substitute $- \frac{1 + i \sqrt{3}}{2}$ for $x$ .

$y = 4 \sqrt{- \frac{1 + i \sqrt{3}}{2}}$

Step 1.6.2

Substitute $- \frac{1 + i \sqrt{3}}{2}$ for $x$ in $y = 4 \sqrt{- \frac{1 + i \sqrt{3}}{2}}$ and solve for $y$ .

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Step 1.6.2.1

Remove parentheses.

$y = 4 \sqrt{- \frac{1 + i \sqrt{3}}{2}}$

Step 1.6.2.2

Simplify $4 \sqrt{- \frac{1 + i \sqrt{3}}{2}}$ .

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Step 1.6.2.2.1

Rewrite $- \frac{1 + i \sqrt{3}}{2}$ as $i^{2} \frac{1^{2} + i \sqrt{3}}{2}$ .

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Step 1.6.2.2.1.1

Rewrite $- 1$ as $i^{2}$ .

$y = 4 \sqrt{i^{2} \frac{1 + i \sqrt{3}}{2}}$

Step 1.6.2.2.1.2

Rewrite $1$ as $1^{2}$ .

$y = 4 \sqrt{i^{2} \frac{1^{2} + i \sqrt{3}}{2}}$

Step 1.6.2.2.2

Pull terms out from under the radical.

$y = 4 (i \sqrt{\frac{1^{2} + i \sqrt{3}}{2}})$

Step 1.6.2.2.3

One to any power is one.

$y = 4 (i \sqrt{\frac{1 + i \sqrt{3}}{2}})$

Step 1.6.2.2.4

Rewrite $\sqrt{\frac{1 + i \sqrt{3}}{2}}$ as $\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}}$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}})$

Step 1.6.2.2.5

Multiply $\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = 4 (i (\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}))$

Step 1.6.2.2.6

Simplify terms.

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Step 1.6.2.2.6.1

Combine and simplify the denominator.

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Step 1.6.2.2.6.1.1

Multiply $\frac{\sqrt{1 + i \sqrt{3}}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{\sqrt{2} \sqrt{2}})$

Step 1.6.2.2.6.1.2

Raise $\sqrt{2}$ to the power of $1$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}})$

Step 1.6.2.2.6.1.3

Raise $\sqrt{2}$ to the power of $1$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}})$

Step 1.6.2.2.6.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{1 + 1}})$

Step 1.6.2.2.6.1.5

Add $1$ and $1$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{{\sqrt{2}}^{2}})$

Step 1.6.2.2.6.1.6

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 1.6.2.2.6.1.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{{(2^{\frac{1}{2}})}^{2}})$

Step 1.6.2.2.6.1.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}})$

Step 1.6.2.2.6.1.6.3

Combine $\frac{1}{2}$ and $2$ .

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{2^{\frac{2}{2}}})$

Step 1.6.2.2.6.1.6.4

Cancel the common factor of $2$ .

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Step 1.6.2.2.6.1.6.4.1

Cancel the common factor.

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{2^{\frac{2}{2}}})$

Step 1.6.2.2.6.1.6.4.2

Rewrite the expression.

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{2^{1}})$

Step 1.6.2.2.6.1.6.5

Evaluate the exponent.

$y = 4 (i \frac{\sqrt{1 + i \sqrt{3}} \sqrt{2}}{2})$

Step 1.6.2.2.6.2

Combine using the product rule for radicals.

$y = 4 (i \frac{\sqrt{(1 + i \sqrt{3}) \cdot 2}}{2})$

Step 1.6.2.2.6.3

Combine $i$ and $\frac{\sqrt{(1 + i \sqrt{3}) \cdot 2}}{2}$ .

$y = 4 \frac{i \sqrt{(1 + i \sqrt{3}) \cdot 2}}{2}$

Step 1.6.2.2.7

Simplify the numerator.

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Step 1.6.2.2.7.1

Apply the distributive property.

$y = 4 \frac{i \sqrt{1 \cdot 2 + i \sqrt{3} \cdot 2}}{2}$

Step 1.6.2.2.7.2

Multiply $2$ by $1$ .

$y = 4 \frac{i \sqrt{2 + i \sqrt{3} \cdot 2}}{2}$

Step 1.6.2.2.7.3

Move $2$ to the left of $i \sqrt{3}$ .

$y = 4 \frac{i \sqrt{2 + 2 i \sqrt{3}}}{2}$

Step 1.6.2.2.8

Cancel the common factor of $2$ .

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Step 1.6.2.2.8.1

Factor $2$ out of $4$ .

$y = 2 (2) \frac{i \sqrt{2 + 2 i \sqrt{3}}}{2}$

Step 1.6.2.2.8.2

Cancel the common factor.

$y = 2 \cdot 2 \frac{i \sqrt{2 + 2 i \sqrt{3}}}{2}$

Step 1.6.2.2.8.3

Rewrite the expression.

$y = 2 (i \sqrt{2 + 2 i \sqrt{3}})$

$y = 2 i \sqrt{2 + 2 i \sqrt{3}}$

Step 1.7

List all of the solutions.

$y = 0, x = 0$

$y = 4, x = 1$

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

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

$y = 0, x = 0$

$y = 4, x = 1$

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

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

Step 2

The area between the given curves is unbounded.

Unbounded area

Step 3