Calculus Examples

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

Step 1

Find the $x$ values where the second derivative is equal to $0$ .

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Step 1.1

Find the second derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

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

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

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 1.1.1.2.2.1

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

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

Step 1.1.1.2.2.2

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

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

Step 1.1.1.2.2.3

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

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

Step 1.1.1.2.3

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

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

Step 1.1.1.2.4

Multiply $3$ by $2$ .

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

Step 1.1.1.3

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

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Step 1.1.1.3.1

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

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

Step 1.1.1.3.2

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

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

Step 1.1.1.4

Differentiate using the Constant Rule.

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Step 1.1.1.4.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

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

Step 1.1.1.4.2

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

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

Step 1.1.2

Find the second derivative.

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Step 1.1.2.1

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

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

Step 1.1.2.2

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

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Step 1.1.2.2.1

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

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

Step 1.1.2.2.2

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 $g (x) = \cos (x)$ .

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

Step 1.1.2.2.3

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

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

Step 1.1.2.2.4

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 1.1.2.2.4.1

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

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

Step 1.1.2.2.4.2

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

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

Step 1.1.2.2.4.3

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

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

Step 1.1.2.2.5

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

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

Step 1.1.2.2.6

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

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Step 1.1.2.2.6.1

Move $\sin (x)$ .

$6 (\sin (x) \sin^{2} (x) \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.6.2

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

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Step 1.1.2.2.6.2.1

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

$6 (\sin^{1} (x) \sin^{2} (x) \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.6.2.2

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

$6 ({\sin (x)}^{1 + 2} \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.6.3

Add $1$ and $2$ .

$6 (\sin^{3} (x) \cdot - 1 + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.7

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

$6 (- 1 \cdot \sin^{3} (x) + \cos (x) (2 \sin (x) \cos (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.8

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

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

Step 1.1.2.2.9

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

$6 (- \sin^{3} (x) + 2 \sin (x) (\cos^{1} (x) \cos (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.10

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

$6 (- \sin^{3} (x) + 2 \sin (x) (\cos^{1} (x) \cos^{1} (x))) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.11

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

$6 (- \sin^{3} (x) + 2 \sin (x) {\cos (x)}^{1 + 1}) + \frac{d}{d x} [3 \cos (x)]$

Step 1.1.2.2.12

Add $1$ and $1$ .

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

Step 1.1.2.3

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

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Step 1.1.2.3.1

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

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

Step 1.1.2.3.2

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

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

Step 1.1.2.3.3

Multiply $- 1$ by $3$ .

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

Step 1.1.2.4

Simplify.

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Step 1.1.2.4.1

Apply the distributive property.

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

Step 1.1.2.4.2

Combine terms.

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Step 1.1.2.4.2.1

Multiply $- 1$ by $6$ .

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

Step 1.1.2.4.2.2

Multiply $2$ by $6$ .

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

Step 1.1.2.4.3

Reorder terms.

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

Step 1.1.3

The second derivative of $f (x)$ with respect to $x$ is $12 \cos^{2} (x) \sin (x) - 6 \sin^{3} (x) - 3 \sin (x)$ .

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

Step 1.2

Set the second derivative equal to $0$ then solve the equation $12 \cos^{2} (x) \sin (x) - 6 \sin^{3} (x) - 3 \sin (x) = 0$ .

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Step 1.2.1

Set the second derivative equal to $0$ .

$12 \cos^{2} (x) \sin (x) - 6 \sin^{3} (x) - 3 \sin (x) = 0$

Step 1.2.2

Factor $3 \sin (x)$ out of $12 \cos^{2} (x) \sin (x) - 6 \sin^{3} (x) - 3 \sin (x)$ .

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Step 1.2.2.1

Factor $3 \sin (x)$ out of $12 \cos^{2} (x) \sin (x)$ .

$3 \sin (x) (4 \cos^{2} (x)) - 6 \sin^{3} (x) - 3 \sin (x) = 0$

Step 1.2.2.2

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

$3 \sin (x) (4 \cos^{2} (x)) + 3 \sin (x) (- 2 \sin^{2} (x)) - 3 \sin (x) = 0$

Step 1.2.2.3

Factor $3 \sin (x)$ out of $- 3 \sin (x)$ .

$3 \sin (x) (4 \cos^{2} (x)) + 3 \sin (x) (- 2 \sin^{2} (x)) + 3 \sin (x) \cdot - 1 = 0$

Step 1.2.2.4

Factor $3 \sin (x)$ out of $3 \sin (x) (4 \cos^{2} (x)) + 3 \sin (x) (- 2 \sin^{2} (x))$ .

$3 \sin (x) (4 \cos^{2} (x) - 2 \sin^{2} (x)) + 3 \sin (x) \cdot - 1 = 0$

Step 1.2.2.5

Factor $3 \sin (x)$ out of $3 \sin (x) (4 \cos^{2} (x) - 2 \sin^{2} (x)) + 3 \sin (x) \cdot - 1$ .

$3 \sin (x) (4 \cos^{2} (x) - 2 \sin^{2} (x) - 1) = 0$

Step 1.2.3

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

$\sin (x) = 0$

$4 \cos^{2} (x) - 2 \sin^{2} (x) - 1 = 0$

Step 1.2.4

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

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Step 1.2.4.1

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

$\sin (x) = 0$

Step 1.2.4.2

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

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Step 1.2.4.2.1

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

$x = \arcsin (0)$

Step 1.2.4.2.2

Simplify the right side.

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Step 1.2.4.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 1.2.4.2.3

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

$x = π - 0$

Step 1.2.4.2.4

Subtract $0$ from $π$ .

$x = π$

Step 1.2.4.2.5

Find the period of $\sin (x)$ .

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Step 1.2.4.2.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.4.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.4.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.2.4.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.4.2.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 2 π n, π + 2 π n$ , for any integer $n$

Step 1.2.5

Set $4 \cos^{2} (x) - 2 \sin^{2} (x) - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $4 \cos^{2} (x) - 2 \sin^{2} (x) - 1$ equal to $0$ .

$4 \cos^{2} (x) - 2 \sin^{2} (x) - 1 = 0$

Step 1.2.5.2

Solve $4 \cos^{2} (x) - 2 \sin^{2} (x) - 1 = 0$ for $x$ .

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Step 1.2.5.2.1

Replace the $4 \cos^{2} (x)$ with $4 (1 - \sin^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$4 (1 - \sin^{2} (x)) - 2 \sin^{2} (x) - 1 = 0$

Step 1.2.5.2.2

Simplify each term.

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Step 1.2.5.2.2.1

Apply the distributive property.

$4 \cdot 1 + 4 (- \sin^{2} (x)) - 2 \sin^{2} (x) - 1 = 0$

Step 1.2.5.2.2.2

Multiply $4$ by $1$ .

$4 + 4 (- \sin^{2} (x)) - 2 \sin^{2} (x) - 1 = 0$

Step 1.2.5.2.2.3

Multiply $- 1$ by $4$ .

$4 - 4 \sin^{2} (x) - 2 \sin^{2} (x) - 1 = 0$

Step 1.2.5.2.3

Simplify by adding terms.

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Step 1.2.5.2.3.1

Subtract $1$ from $4$ .

$- 4 \sin^{2} (x) - 2 \sin^{2} (x) + 3 = 0$

Step 1.2.5.2.3.2

Subtract $2 \sin^{2} (x)$ from $- 4 \sin^{2} (x)$ .

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

Step 1.2.5.2.4

Subtract $3$ from both sides of the equation.

$- 6 \sin^{2} (x) = - 3$

Step 1.2.5.2.5

Divide each term in $- 6 \sin^{2} (x) = - 3$ by $- 6$ and simplify.

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Step 1.2.5.2.5.1

Divide each term in $- 6 \sin^{2} (x) = - 3$ by $- 6$ .

$\frac{- 6 \sin^{2} (x)}{- 6} = \frac{- 3}{- 6}$

Step 1.2.5.2.5.2

Simplify the left side.

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Step 1.2.5.2.5.2.1

Cancel the common factor of $- 6$ .

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Step 1.2.5.2.5.2.1.1

Cancel the common factor.

$\frac{- 6 \sin^{2} (x)}{- 6} = \frac{- 3}{- 6}$

Step 1.2.5.2.5.2.1.2

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

$\sin^{2} (x) = \frac{- 3}{- 6}$

Step 1.2.5.2.5.3

Simplify the right side.

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Step 1.2.5.2.5.3.1

Cancel the common factor of $- 3$ and $- 6$ .

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Step 1.2.5.2.5.3.1.1

Factor $- 3$ out of $- 3$ .

$\sin^{2} (x) = \frac{- 3 (1)}{- 6}$

Step 1.2.5.2.5.3.1.2

Cancel the common factors.

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Step 1.2.5.2.5.3.1.2.1

Factor $- 3$ out of $- 6$ .

$\sin^{2} (x) = \frac{- 3 \cdot 1}{- 3 \cdot 2}$

Step 1.2.5.2.5.3.1.2.2

Cancel the common factor.

$\sin^{2} (x) = \frac{- 3 \cdot 1}{- 3 \cdot 2}$

Step 1.2.5.2.5.3.1.2.3

Rewrite the expression.

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

Step 1.2.5.2.6

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 1.2.5.2.7

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

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Step 1.2.5.2.7.1

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

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

Step 1.2.5.2.7.2

Any root of $1$ is $1$ .

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

Step 1.2.5.2.7.3

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

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

Step 1.2.5.2.7.4

Combine and simplify the denominator.

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Step 1.2.5.2.7.4.1

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

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

Step 1.2.5.2.7.4.2

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

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

Step 1.2.5.2.7.4.3

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

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

Step 1.2.5.2.7.4.4

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

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

Step 1.2.5.2.7.4.5

Add $1$ and $1$ .

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

Step 1.2.5.2.7.4.6

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

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Step 1.2.5.2.7.4.6.1

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

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

Step 1.2.5.2.7.4.6.2

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

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

Step 1.2.5.2.7.4.6.3

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

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

Step 1.2.5.2.7.4.6.4

Cancel the common factor of $2$ .

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Step 1.2.5.2.7.4.6.4.1

Cancel the common factor.

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

Step 1.2.5.2.7.4.6.4.2

Rewrite the expression.

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

Step 1.2.5.2.7.4.6.5

Evaluate the exponent.

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

Step 1.2.5.2.8

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

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Step 1.2.5.2.8.1

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

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

Step 1.2.5.2.8.2

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

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

Step 1.2.5.2.8.3

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

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

Step 1.2.5.2.9

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

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

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

Step 1.2.5.2.10

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

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Step 1.2.5.2.10.1

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

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

Step 1.2.5.2.10.2

Simplify the right side.

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Step 1.2.5.2.10.2.1

The exact value of $\arcsin (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$x = \frac{π}{4}$

Step 1.2.5.2.10.3

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

$x = π - \frac{π}{4}$

Step 1.2.5.2.10.4

Simplify $π - \frac{π}{4}$ .

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Step 1.2.5.2.10.4.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x = π \cdot \frac{4}{4} - \frac{π}{4}$

Step 1.2.5.2.10.4.2

Combine fractions.

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Step 1.2.5.2.10.4.2.1

Combine $π$ and $\frac{4}{4}$ .

$x = \frac{π \cdot 4}{4} - \frac{π}{4}$

Step 1.2.5.2.10.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 4 - π}{4}$

Step 1.2.5.2.10.4.3

Simplify the numerator.

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Step 1.2.5.2.10.4.3.1

Move $4$ to the left of $π$ .

$x = \frac{4 \cdot π - π}{4}$

Step 1.2.5.2.10.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{4}$

Step 1.2.5.2.10.5

Find the period of $\sin (x)$ .

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Step 1.2.5.2.10.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.5.2.10.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.5.2.10.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.2.5.2.10.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.5.2.10.6

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{π}{4} + 2 π n, \frac{3 π}{4} + 2 π n$ , for any integer $n$

Step 1.2.5.2.11

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

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Step 1.2.5.2.11.1

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

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

Step 1.2.5.2.11.2

Simplify the right side.

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Step 1.2.5.2.11.2.1

The exact value of $\arcsin (- \frac{\sqrt{2}}{2})$ is $- \frac{π}{4}$ .

$x = - \frac{π}{4}$

Step 1.2.5.2.11.3

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{4} + π$

Step 1.2.5.2.11.4

Simplify the expression to find the second solution.

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Step 1.2.5.2.11.4.1

Subtract $2 π$ from $2 π + \frac{π}{4} + π$ .

$x = 2 π + \frac{π}{4} + π - 2 π$

Step 1.2.5.2.11.4.2

The resulting angle of $\frac{5 π}{4}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{4} + π$ .

$x = \frac{5 π}{4}$

Step 1.2.5.2.11.5

Find the period of $\sin (x)$ .

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Step 1.2.5.2.11.5.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.5.2.11.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 1.2.5.2.11.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 1.2.5.2.11.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.5.2.11.6

Add $2 π$ to every negative angle to get positive angles.

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Step 1.2.5.2.11.6.1

Add $2 π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + 2 π$

Step 1.2.5.2.11.6.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 π \cdot \frac{4}{4} - \frac{π}{4}$

Step 1.2.5.2.11.6.3

Combine fractions.

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Step 1.2.5.2.11.6.3.1

Combine $2 π$ and $\frac{4}{4}$ .

$\frac{2 π \cdot 4}{4} - \frac{π}{4}$

Step 1.2.5.2.11.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 4 - π}{4}$

Step 1.2.5.2.11.6.4

Simplify the numerator.

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Step 1.2.5.2.11.6.4.1

Multiply $4$ by $2$ .

$\frac{8 π - π}{4}$

Step 1.2.5.2.11.6.4.2

Subtract $π$ from $8 π$ .

$\frac{7 π}{4}$

Step 1.2.5.2.11.6.5

List the new angles.

$x = \frac{7 π}{4}$

Step 1.2.5.2.11.7

The period of the $\sin (x)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Step 1.2.5.2.12

List all of the solutions.

$x = \frac{π}{4} + 2 π n, \frac{3 π}{4} + 2 π n, \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Step 1.2.5.2.13

Consolidate the answers.

$x = \frac{π}{4} + \frac{π n}{2}$ , for any integer $n$

Step 1.2.6

The final solution is all the values that make $3 \sin (x) (4 \cos^{2} (x) - 2 \sin^{2} (x) - 1) = 0$ true.

$x = 2 π n, π + 2 π n, \frac{π}{4} + \frac{π n}{2}$ , for any integer $n$

Step 1.2.7

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

$x = π n, \frac{π}{4} + \frac{π n}{2}$ , for any integer $n$

Step 2

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 3

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, \infty)$

Step 4

Substitute any number from the interval $(- \infty, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 4.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = 12 \cos^{2} (0) \sin (0) - 6 \sin^{3} (0) - 3 \sin (0)$

Step 4.2

Simplify the result.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

The exact value of $\cos (0)$ is $1$ .

$f'' (0) = 12 \cdot (1^{2} \sin (0)) - 6 \sin^{3} (0) - 3 \sin (0)$

Step 4.2.1.2

One to any power is one.

$f'' (0) = 12 \cdot (1 \sin (0)) - 6 \sin^{3} (0) - 3 \sin (0)$

Step 4.2.1.3

Multiply $12$ by $1$ .

$f'' (0) = 12 \sin (0) - 6 \sin^{3} (0) - 3 \sin (0)$

Step 4.2.1.4

The exact value of $\sin (0)$ is $0$ .

$f'' (0) = 12 \cdot 0 - 6 \sin^{3} (0) - 3 \sin (0)$

Step 4.2.1.5

Multiply $12$ by $0$ .

$f'' (0) = 0 - 6 \sin^{3} (0) - 3 \sin (0)$

Step 4.2.1.6

The exact value of $\sin (0)$ is $0$ .

$f'' (0) = 0 - 6 \cdot 0^{3} - 3 \sin (0)$

Step 4.2.1.7

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

$f'' (0) = 0 - 6 \cdot 0 - 3 \sin (0)$

Step 4.2.1.8

Multiply $- 6$ by $0$ .

$f'' (0) = 0 + 0 - 3 \sin (0)$

Step 4.2.1.9

The exact value of $\sin (0)$ is $0$ .

$f'' (0) = 0 + 0 - 3 \cdot 0$

Step 4.2.1.10

Multiply $- 3$ by $0$ .

$f'' (0) = 0 + 0 + 0$

Step 4.2.2

Simplify by adding numbers.

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Step 4.2.2.1

Add $0$ and $0$ .

$f'' (0) = 0 + 0$

Step 4.2.2.2

Add $0$ and $0$ .

$f'' (0) = 0$

Step 4.2.3

The final answer is $0$ .

$0$

Step 4.3

The graph is concave up on the interval $(- \infty, \infty)$ because $f'' (0)$ is positive.

Concave up on $(- \infty, \infty)$ since $f'' (x)$ is positive

Step 5