Calculus Examples

$f (x) = 2 \sin^{3} (x) + 3 \sin (x) + 4$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $2 \sin^{3} (x) + 3 \sin (x) + 4$ with respect to $x$ is $\frac{d}{d x} [2 \sin^{3} (x)] + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$ .

$\frac{d}{d x} [2 \sin^{3} (x)] + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.2

Evaluate $\frac{d}{d x} [2 \sin^{3} (x)]$ .

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Step 1.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 \sin^{3} (x)$ with respect to $x$ is $2 \frac{d}{d x} [\sin^{3} (x)]$ .

$2 \frac{d}{d x} [\sin^{3} (x)] + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = \sin (x)$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

$2 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [\sin (x)]) + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$2 (3 u^{2} \frac{d}{d x} [\sin (x)]) + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.2.2.3

Replace all occurrences of $u$ with $\sin (x)$ .

$2 (3 \sin^{2} (x) \frac{d}{d x} [\sin (x)]) + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.2.3

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$2 (3 \sin^{2} (x) \cos (x)) + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.2.4

Multiply $3$ by $2$ .

$6 \sin^{2} (x) \cos (x) + \frac{d}{d x} [3 \sin (x)] + \frac{d}{d x} [4]$

Step 1.3

Evaluate $\frac{d}{d x} [3 \sin (x)]$ .

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Step 1.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \sin (x)$ with respect to $x$ is $3 \frac{d}{d x} [\sin (x)]$ .

$6 \sin^{2} (x) \cos (x) + 3 \frac{d}{d x} [\sin (x)] + \frac{d}{d x} [4]$

Step 1.3.2

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$6 \sin^{2} (x) \cos (x) + 3 \cos (x) + \frac{d}{d x} [4]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$6 \sin^{2} (x) \cos (x) + 3 \cos (x) + 0$

Step 1.4.2

Add $6 \sin^{2} (x) \cos (x) + 3 \cos (x)$ and $0$ .

$6 \sin^{2} (x) \cos (x) + 3 \cos (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $6 \sin^{2} (x) \cos (x) + 3 \cos (x)$ with respect to $x$ is $\frac{d}{d x} [6 \sin^{2} (x) \cos (x)] + \frac{d}{d x} [3 \cos (x)]$ .

$f'' (x) = \frac{d}{d x} (6 \sin^{2} (x) \cos (x)) + \frac{d}{d x} (3 \cos (x))$

Step 2.2

Evaluate $\frac{d}{d x} [6 \sin^{2} (x) \cos (x)]$ .

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Step 2.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \sin^{2} (x) \cos (x)$ with respect to $x$ is $6 \frac{d}{d x} [\sin^{2} (x) \cos (x)]$ .

$f'' (x) = 6 \frac{d}{d x} (\sin^{2} (x) \cos (x)) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = \sin^{2} (x)$ and $g (x) = \cos (x)$ .

$f'' (x) = 6 (\sin^{2} (x) \frac{d}{d x} (\cos (x)) + \cos (x) \frac{d}{d x} (\sin^{2} (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.3

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$f'' (x) = 6 (\sin^{2} (x) (- \sin (x)) + \cos (x) \frac{d}{d x} (\sin^{2} (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.4

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sin (x)$ .

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Step 2.2.4.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

$f'' (x) = 6 (\sin^{2} (x) (- \sin (x)) + \cos (x) (\frac{d}{d u} (u^{2}) \frac{d}{d x} (\sin (x)))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.4.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

$f'' (x) = 6 (\sin^{2} (x) (- \sin (x)) + \cos (x) (2 u \frac{d}{d x} (\sin (x)))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.4.3

Replace all occurrences of $u$ with $\sin (x)$ .

$f'' (x) = 6 (\sin^{2} (x) (- \sin (x)) + \cos (x) (2 \sin (x) \frac{d}{d x} (\sin (x)))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.5

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f'' (x) = 6 (\sin^{2} (x) (- \sin (x)) + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.6

Multiply $\sin^{2} (x)$ by $\sin (x)$ by adding the exponents.

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Step 2.2.6.1

Move $\sin (x)$ .

$f'' (x) = 6 (\sin (x) \sin^{2} (x) \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.6.2

Multiply $\sin (x)$ by $\sin^{2} (x)$ .

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Step 2.2.6.2.1

Raise $\sin (x)$ to the power of $1$ .

$f'' (x) = 6 (\sin (x) \sin^{2} (x) \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.6.2.2

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

$f'' (x) = 6 ({\sin (x)}^{1 + 2} \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.6.3

Add $1$ and $2$ .

$f'' (x) = 6 (\sin^{3} (x) \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.7

Move $- 1$ to the left of $\sin^{3} (x)$ .

$f'' (x) = 6 (- 1 \cdot \sin^{3} (x) + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.8

Rewrite $- 1 \sin^{3} (x)$ as $- \sin^{3} (x)$ .

$f'' (x) = 6 (- \sin^{3} (x) + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.9

Raise $\cos (x)$ to the power of $1$ .

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) (\cos (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.10

Raise $\cos (x)$ to the power of $1$ .

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) (\cos (x) \cos (x))) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.11

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

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) {\cos (x)}^{1 + 1}) + \frac{d}{d x} (3 \cos (x))$

Step 2.2.12

Add $1$ and $1$ .

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) \cos^{2} (x)) + \frac{d}{d x} (3 \cos (x))$

Step 2.3

Evaluate $\frac{d}{d x} [3 \cos (x)]$ .

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Step 2.3.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \cos (x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos (x)]$ .

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) \cos^{2} (x)) + 3 \frac{d}{d x} (\cos (x))$

Step 2.3.2

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) \cos^{2} (x)) + 3 (- \sin (x))$

Step 2.3.3

Multiply $- 1$ by $3$ .

$f'' (x) = 6 (- \sin^{3} (x) + 2 \sin (x) \cos^{2} (x)) - 3 \sin (x)$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$f'' (x) = 6 (- \sin^{3} (x)) + 6 (2 \sin (x) \cos^{2} (x)) - 3 \sin (x)$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $- 1$ by $6$ .

$f'' (x) = - 6 \sin^{3} (x) + 6 (2 \sin (x) \cos^{2} (x)) - 3 \sin (x)$

Step 2.4.2.2

Multiply $2$ by $6$ .

$f'' (x) = - 6 \sin^{3} (x) + 12 \sin (x) \cos^{2} (x) - 3 \sin (x)$

Step 2.4.3

Reorder terms.

$f'' (x) = 12 \cos^{2} (x) \sin (x) - 6 \sin^{3} (x) - 3 \sin (x)$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$6 \sin^{2} (x) \cos (x) + 3 \cos (x) = 0$

Step 4

Factor $3 \cos (x)$ out of $6 \sin^{2} (x) \cos (x) + 3 \cos (x)$ .

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Step 4.1

Factor $3 \cos (x)$ out of $6 \sin^{2} (x) \cos (x)$ .

$3 \cos (x) (2 \sin^{2} (x)) + 3 \cos (x) = 0$

Step 4.2

Factor $3 \cos (x)$ out of $3 \cos (x)$ .

$3 \cos (x) (2 \sin^{2} (x)) + 3 \cos (x) \cdot 1 = 0$

Step 4.3

Factor $3 \cos (x)$ out of $3 \cos (x) (2 \sin^{2} (x)) + 3 \cos (x) \cdot 1$ .

$3 \cos (x) (2 \sin^{2} (x) + 1) = 0$

Step 5

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

$\cos (x) = 0$

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

Step 6

Set $\cos (x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\cos (x)$ equal to $0$ .

$\cos (x) = 0$

Step 6.2

Solve $\cos (x) = 0$ for $x$ .

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Step 6.2.1

Take the inverse cosine of both sides of the equation to extract $x$ from inside the cosine.

$x = \arccos (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 6.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - \frac{π}{2}$

Step 6.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 6.2.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = 2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 6.2.4.2

Combine fractions.

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Step 6.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

$x = \frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 6.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 6.2.4.3

Simplify the numerator.

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Step 6.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 6.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 6.2.5

The solution to the equation $x = \frac{π}{2}$ .

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 7

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

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Step 7.1

Set $2 \sin^{2} (x) + 1$ equal to $0$ .

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

Step 7.2

Solve $2 \sin^{2} (x) + 1 = 0$ for $x$ .

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Step 7.2.1

Subtract $1$ from both sides of the equation.

$2 \sin^{2} (x) = - 1$

Step 7.2.2

Divide each term in $2 \sin^{2} (x) = - 1$ by $2$ and simplify.

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Step 7.2.2.1

Divide each term in $2 \sin^{2} (x) = - 1$ by $2$ .

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

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Cancel the common factor of $2$ .

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Step 7.2.2.2.1.1

Cancel the common factor.

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

Step 7.2.2.2.1.2

Divide $\sin^{2} (x)$ by $1$ .

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

Step 7.2.2.3

Simplify the right side.

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Step 7.2.2.3.1

Move the negative in front of the fraction.

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

Step 7.2.3

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

$\sin (x) = \pm \sqrt{- \frac{1}{2}}$

Step 7.2.4

Simplify $\pm \sqrt{- \frac{1}{2}}$ .

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Step 7.2.4.1

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

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Step 7.2.4.1.1

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

$\sin (x) = \pm \sqrt{i^{2} \frac{1}{2}}$

Step 7.2.4.1.2

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

$\sin (x) = \pm \sqrt{i^{2} \frac{1^{2}}{2}}$

Step 7.2.4.2

Pull terms out from under the radical.

$\sin (x) = \pm i \sqrt{\frac{1^{2}}{2}}$

Step 7.2.4.3

One to any power is one.

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

Step 7.2.4.4

Rewrite $\sqrt{\frac{1}{2}}$ as $\frac{\sqrt{1}}{\sqrt{2}}$ .

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

Step 7.2.4.5

Any root of $1$ is $1$ .

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

Step 7.2.4.6

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Step 7.2.4.7

Combine and simplify the denominator.

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Step 7.2.4.7.1

Multiply $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$\sin (x) = \pm i \frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}$

Step 7.2.4.7.2

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

$\sin (x) = \pm i \frac{\sqrt{2}}{{\sqrt{2}}^{1} \sqrt{2}}$

Step 7.2.4.7.3

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

$\sin (x) = \pm i \frac{\sqrt{2}}{{\sqrt{2}}^{1} {\sqrt{2}}^{1}}$

Step 7.2.4.7.4

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

$\sin (x) = \pm i \frac{\sqrt{2}}{{\sqrt{2}}^{1 + 1}}$

Step 7.2.4.7.5

Add $1$ and $1$ .

$\sin (x) = \pm i \frac{\sqrt{2}}{{\sqrt{2}}^{2}}$

Step 7.2.4.7.6

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

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Step 7.2.4.7.6.1

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

$\sin (x) = \pm i \frac{\sqrt{2}}{{(2^{\frac{1}{2}})}^{2}}$

Step 7.2.4.7.6.2

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

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

Step 7.2.4.7.6.3

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

$\sin (x) = \pm i \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 7.2.4.7.6.4

Cancel the common factor of $2$ .

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Step 7.2.4.7.6.4.1

Cancel the common factor.

$\sin (x) = \pm i \frac{\sqrt{2}}{2^{\frac{2}{2}}}$

Step 7.2.4.7.6.4.2

Rewrite the expression.

$\sin (x) = \pm i \frac{\sqrt{2}}{2^{1}}$

Step 7.2.4.7.6.5

Evaluate the exponent.

$\sin (x) = \pm i \frac{\sqrt{2}}{2}$

Step 7.2.4.8

Combine $i$ and $\frac{\sqrt{2}}{2}$ .

$\sin (x) = \pm \frac{i \sqrt{2}}{2}$

Step 7.2.5

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

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Step 7.2.5.1

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

$\sin (x) = \frac{i \sqrt{2}}{2}$

Step 7.2.5.2

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

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

Step 7.2.5.3

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

$\sin (x) = \frac{i \sqrt{2}}{2}, - \frac{i \sqrt{2}}{2}$

Step 7.2.6

Set up each of the solutions to solve for $x$ .

$\sin (x) = \frac{i \sqrt{2}}{2}$

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

Step 7.2.7

Solve for $x$ in $\sin (x) = \frac{i \sqrt{2}}{2}$ .

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Step 7.2.7.1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

$x = \arcsin (\frac{i \sqrt{2}}{2})$

Step 7.2.7.2

The inverse sine of $\arcsin (\frac{i \sqrt{2}}{2})$ is undefined.

Undefined

Step 7.2.8

Solve for $x$ in $\sin (x) = - \frac{i \sqrt{2}}{2}$ .

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Step 7.2.8.1

Take the inverse sine of both sides of the equation to extract $x$ from inside the sine.

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

Step 7.2.8.2

The inverse sine of $\arcsin (- \frac{i \sqrt{2}}{2})$ is undefined.

Undefined

Step 7.2.9

List all of the solutions.

No solution

Step 8

The final solution is all the values that make $3 \cos (x) (2 \sin^{2} (x) + 1) = 0$ true.

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 9

Evaluate the second derivative at $x = \frac{π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 \cos^{2} (\frac{π}{2}) \sin (\frac{π}{2}) - 6 \sin^{3} (\frac{π}{2}) - 3 \sin (\frac{π}{2})$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$12 \cdot 0^{2} \sin (\frac{π}{2}) - 6 \sin^{3} (\frac{π}{2}) - 3 \sin (\frac{π}{2})$

Step 10.1.2

Raising $0$ to any positive power yields $0$ .

$12 \cdot 0 \sin (\frac{π}{2}) - 6 \sin^{3} (\frac{π}{2}) - 3 \sin (\frac{π}{2})$

Step 10.1.3

Multiply $12$ by $0$ .

$0 \sin (\frac{π}{2}) - 6 \sin^{3} (\frac{π}{2}) - 3 \sin (\frac{π}{2})$

Step 10.1.4

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$0 \cdot 1 - 6 \sin^{3} (\frac{π}{2}) - 3 \sin (\frac{π}{2})$

Step 10.1.5

Multiply $0$ by $1$ .

$0 - 6 \sin^{3} (\frac{π}{2}) - 3 \sin (\frac{π}{2})$

Step 10.1.6

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$0 - 6 \cdot 1^{3} - 3 \sin (\frac{π}{2})$

Step 10.1.7

One to any power is one.

$0 - 6 \cdot 1 - 3 \sin (\frac{π}{2})$

Step 10.1.8

Multiply $- 6$ by $1$ .

$0 - 6 - 3 \sin (\frac{π}{2})$

Step 10.1.9

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$0 - 6 - 3 \cdot 1$

Step 10.1.10

Multiply $- 3$ by $1$ .

$0 - 6 - 3$

Step 10.2

Simplify by subtracting numbers.

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Step 10.2.1

Subtract $6$ from $0$ .

$- 6 - 3$

Step 10.2.2

Subtract $3$ from $- 6$ .

$- 9$

Step 11

$x = \frac{π}{2}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{2}$ is a local maximum

Step 12

Find the y-value when $x = \frac{π}{2}$ .

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Step 12.1

Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = 2 \sin^{3} (\frac{π}{2}) + 3 \sin (\frac{π}{2}) + 4$

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2}) = 2 \cdot 1^{3} + 3 \sin (\frac{π}{2}) + 4$

Step 12.2.1.2

One to any power is one.

$f (\frac{π}{2}) = 2 \cdot 1 + 3 \sin (\frac{π}{2}) + 4$

Step 12.2.1.3

Multiply $2$ by $1$ .

$f (\frac{π}{2}) = 2 + 3 \sin (\frac{π}{2}) + 4$

Step 12.2.1.4

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{π}{2}) = 2 + 3 \cdot 1 + 4$

Step 12.2.1.5

Multiply $3$ by $1$ .

$f (\frac{π}{2}) = 2 + 3 + 4$

Step 12.2.2

Simplify by adding numbers.

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Step 12.2.2.1

Add $2$ and $3$ .

$f (\frac{π}{2}) = 5 + 4$

Step 12.2.2.2

Add $5$ and $4$ .

$f (\frac{π}{2}) = 9$

Step 12.2.3

The final answer is $9$ .

$y = 9$

Step 13

Evaluate the second derivative at $x = \frac{3 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$12 \cos^{2} (\frac{3 π}{2}) \sin (\frac{3 π}{2}) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$12 \cos^{2} (\frac{π}{2}) \sin (\frac{3 π}{2}) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.2

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$12 \cdot 0^{2} \sin (\frac{3 π}{2}) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.3

Raising $0$ to any positive power yields $0$ .

$12 \cdot 0 \sin (\frac{3 π}{2}) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.4

Multiply $12$ by $0$ .

$0 \sin (\frac{3 π}{2}) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.5

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$0 (- \sin (\frac{π}{2})) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.6

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$0 (- 1 \cdot 1) - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.7

Multiply $0 (- 1 \cdot 1)$ .

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Step 14.1.7.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.7.2

Multiply $0$ by $- 1$ .

$0 - 6 \sin^{3} (\frac{3 π}{2}) - 3 \sin (\frac{3 π}{2})$

Step 14.1.8

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$0 - 6 {(- \sin (\frac{π}{2}))}^{3} - 3 \sin (\frac{3 π}{2})$

Step 14.1.9

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$0 - 6 {(- 1 \cdot 1)}^{3} - 3 \sin (\frac{3 π}{2})$

Step 14.1.10

Multiply $- 1$ by $1$ .

$0 - 6 {(- 1)}^{3} - 3 \sin (\frac{3 π}{2})$

Step 14.1.11

Raise $- 1$ to the power of $3$ .

$0 - 6 \cdot - 1 - 3 \sin (\frac{3 π}{2})$

Step 14.1.12

Multiply $- 6$ by $- 1$ .

$0 + 6 - 3 \sin (\frac{3 π}{2})$

Step 14.1.13

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$0 + 6 - 3 (- \sin (\frac{π}{2}))$

Step 14.1.14

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$0 + 6 - 3 (- 1 \cdot 1)$

Step 14.1.15

Multiply $- 3 (- 1 \cdot 1)$ .

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Step 14.1.15.1

Multiply $- 1$ by $1$ .

$0 + 6 - 3 \cdot - 1$

Step 14.1.15.2

Multiply $- 3$ by $- 1$ .

$0 + 6 + 3$

Step 14.2

Simplify by adding numbers.

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Step 14.2.1

Add $0$ and $6$ .

$6 + 3$

Step 14.2.2

Add $6$ and $3$ .

$9$

Step 15

$x = \frac{3 π}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{3 π}{2}$ is a local minimum

Step 16

Find the y-value when $x = \frac{3 π}{2}$ .

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Step 16.1

Replace the variable $x$ with $\frac{3 π}{2}$ in the expression.

$f (\frac{3 π}{2}) = 2 \sin^{3} (\frac{3 π}{2}) + 3 \sin (\frac{3 π}{2}) + 4$

Step 16.2

Simplify the result.

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Step 16.2.1

Simplify each term.

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Step 16.2.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$f (\frac{3 π}{2}) = 2 {(- \sin (\frac{π}{2}))}^{3} + 3 \sin (\frac{3 π}{2}) + 4$

Step 16.2.1.2

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{3 π}{2}) = 2 {(- 1 \cdot 1)}^{3} + 3 \sin (\frac{3 π}{2}) + 4$

Step 16.2.1.3

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = 2 {(- 1)}^{3} + 3 \sin (\frac{3 π}{2}) + 4$

Step 16.2.1.4

Raise $- 1$ to the power of $3$ .

$f (\frac{3 π}{2}) = 2 \cdot - 1 + 3 \sin (\frac{3 π}{2}) + 4$

Step 16.2.1.5

Multiply $2$ by $- 1$ .

$f (\frac{3 π}{2}) = - 2 + 3 \sin (\frac{3 π}{2}) + 4$

Step 16.2.1.6

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because sine is negative in the fourth quadrant.

$f (\frac{3 π}{2}) = - 2 + 3 (- \sin (\frac{π}{2})) + 4$

Step 16.2.1.7

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$f (\frac{3 π}{2}) = - 2 + 3 (- 1 \cdot 1) + 4$

Step 16.2.1.8

Multiply $3 (- 1 \cdot 1)$ .

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Step 16.2.1.8.1

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = - 2 + 3 \cdot - 1 + 4$

Step 16.2.1.8.2

Multiply $3$ by $- 1$ .

$f (\frac{3 π}{2}) = - 2 - 3 + 4$

Step 16.2.2

Simplify by adding and subtracting.

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Step 16.2.2.1

Subtract $3$ from $- 2$ .

$f (\frac{3 π}{2}) = - 5 + 4$

Step 16.2.2.2

Add $- 5$ and $4$ .

$f (\frac{3 π}{2}) = - 1$

Step 16.2.3

The final answer is $- 1$ .

$y = - 1$

Step 17

These are the local extrema for $f (x) = 2 \sin^{3} (x) + 3 \sin (x) + 4$ .

$(\frac{π}{2}, 9)$ is a local maxima

$(\frac{3 π}{2}, - 1)$ is a local minima

Step 18