Calculus Examples

$2 \sin (x) + \cos (2 x)$

Step 1

Write $2 \sin (x) + \cos (2 x)$ as a function.

$f (x) = 2 \sin (x) + \cos (2 x)$

Step 2

Find the first derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

$2 \cos (x) + \frac{d}{d x} [\cos (2 x)]$

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$2 \cos (x) + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]$

Step 2.3.1.2

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

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

Step 2.3.1.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 2.3.2

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

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

Step 2.3.3

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

$2 \cos (x) - \sin (2 x) (2 \cdot 1)$

Step 2.3.4

Multiply $2$ by $1$ .

$2 \cos (x) - \sin (2 x) \cdot 2$

Step 2.3.5

Multiply $2$ by $- 1$ .

$2 \cos (x) - 2 \sin (2 x)$

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

Multiply $- 1$ by $2$ .

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

Step 3.3

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

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Step 3.3.1

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

$f'' (x) = - 2 \sin (x) - 2 \frac{d}{d x} \sin (2 x)$

Step 3.3.2

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

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Step 3.3.2.1

To apply the Chain Rule, set $u$ as $2 x$ .

$f'' (x) = - 2 \sin (x) - 2 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x))$

Step 3.3.2.2

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

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

Step 3.3.2.3

Replace all occurrences of $u$ with $2 x$ .

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

Step 3.3.3

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

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

Step 3.3.4

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

$f'' (x) = - 2 \sin (x) - 2 (\cos (2 x) (2 \cdot 1))$

Step 3.3.5

Multiply $2$ by $1$ .

$f'' (x) = - 2 \sin (x) - 2 (\cos (2 x) \cdot 2)$

Step 3.3.6

Move $2$ to the left of $\cos (2 x)$ .

$f'' (x) = - 2 \sin (x) - 2 (2 \cdot \cos (2 x))$

Step 3.3.7

Multiply $2$ by $- 2$ .

$f'' (x) = - 2 \sin (x) - 4 \cos (2 x)$

Step 4

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

$2 \cos (x) - 2 \sin (2 x) = 0$

Step 5

Simplify each term.

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Step 5.1

Apply the sine double-angle identity.

$2 \cos (x) - 2 (2 \sin (x) \cos (x)) = 0$

Step 5.2

Multiply $2$ by $- 2$ .

$2 \cos (x) - 4 \sin (x) \cos (x) = 0$

Step 6

Factor $2 \cos (x)$ out of $2 \cos (x) - 4 \sin (x) \cos (x)$ .

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Step 6.1

Factor $2 \cos (x)$ out of $2 \cos (x)$ .

$2 \cos (x) \cdot 1 - 4 \sin (x) \cos (x) = 0$

Step 6.2

Factor $2 \cos (x)$ out of $- 4 \sin (x) \cos (x)$ .

$2 \cos (x) \cdot 1 + 2 \cos (x) (- 2 \sin (x)) = 0$

Step 6.3

Factor $2 \cos (x)$ out of $2 \cos (x) \cdot 1 + 2 \cos (x) (- 2 \sin (x))$ .

$2 \cos (x) (1 - 2 \sin (x)) = 0$

Step 7

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

$\cos (x) = 0$

$1 - 2 \sin (x) = 0$

Step 8

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

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Step 8.1

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

$\cos (x) = 0$

Step 8.2

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

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Step 8.2.1

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

$x = \arccos (0)$

Step 8.2.2

Simplify the right side.

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Step 8.2.2.1

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

$x = \frac{π}{2}$

Step 8.2.3

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 8.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 8.2.4.1

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 8.2.4.2

Combine fractions.

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Step 8.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 8.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 8.2.4.3

Simplify the numerator.

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Step 8.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 8.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 8.2.5

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 9

Set $1 - 2 \sin (x)$ equal to $0$ and solve for $x$ .

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Step 9.1

Set $1 - 2 \sin (x)$ equal to $0$ .

$1 - 2 \sin (x) = 0$

Step 9.2

Solve $1 - 2 \sin (x) = 0$ for $x$ .

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Step 9.2.1

Subtract $1$ from both sides of the equation.

$- 2 \sin (x) = - 1$

Step 9.2.2

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

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Step 9.2.2.1

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

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

Step 9.2.2.2

Simplify the left side.

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Step 9.2.2.2.1

Cancel the common factor of $- 2$ .

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Step 9.2.2.2.1.1

Cancel the common factor.

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

Step 9.2.2.2.1.2

Divide $\sin (x)$ by $1$ .

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

Step 9.2.2.3

Simplify the right side.

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Step 9.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 9.2.3

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

$x = \arcsin (\frac{1}{2})$

Step 9.2.4

Simplify the right side.

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Step 9.2.4.1

The exact value of $\arcsin (\frac{1}{2})$ is $\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 9.2.5

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.

$x = π - \frac{π}{6}$

Step 9.2.6

Simplify $π - \frac{π}{6}$ .

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Step 9.2.6.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 9.2.6.2

Combine fractions.

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Step 9.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 9.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 9.2.6.3

Simplify the numerator.

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Step 9.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 9.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 9.2.7

The solution to the equation $x = \frac{π}{6}$ .

$x = \frac{π}{6}, \frac{5 π}{6}$

Step 10

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

$x = \frac{π}{2}, \frac{3 π}{2}, \frac{π}{6}, \frac{5 π}{6}$

Step 11

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.

$- 2 \sin (\frac{π}{2}) - 4 \cos (2 (\frac{π}{2}))$

Step 12

Evaluate the second derivative.

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Step 12.1

Simplify each term.

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Step 12.1.1

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

$- 2 \cdot 1 - 4 \cos (2 (\frac{π}{2}))$

Step 12.1.2

Multiply $- 2$ by $1$ .

$- 2 - 4 \cos (2 (\frac{π}{2}))$

Step 12.1.3

Cancel the common factor of $2$ .

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Step 12.1.3.1

Cancel the common factor.

$- 2 - 4 \cos (2 \frac{π}{2})$

Step 12.1.3.2

Rewrite the expression.

$- 2 - 4 \cos (π)$

Step 12.1.4

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

$- 2 - 4 (- \cos (0))$

Step 12.1.5

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

$- 2 - 4 (- 1 \cdot 1)$

Step 12.1.6

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

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Step 12.1.6.1

Multiply $- 1$ by $1$ .

$- 2 - 4 \cdot - 1$

Step 12.1.6.2

Multiply $- 4$ by $- 1$ .

$- 2 + 4$

Step 12.2

Add $- 2$ and $4$ .

$2$

Step 13

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

$x = \frac{π}{2}$ is a local minimum

Step 14

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

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Step 14.1

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

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

Step 14.2

Simplify the result.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

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

$f (\frac{π}{2}) = 2 \cdot 1 + \cos (2 (\frac{π}{2}))$

Step 14.2.1.2

Multiply $2$ by $1$ .

$f (\frac{π}{2}) = 2 + \cos (2 (\frac{π}{2}))$

Step 14.2.1.3

Cancel the common factor of $2$ .

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Step 14.2.1.3.1

Cancel the common factor.

$f (\frac{π}{2}) = 2 + \cos (2 (\frac{π}{2}))$

Step 14.2.1.3.2

Rewrite the expression.

$f (\frac{π}{2}) = 2 + \cos (π)$

Step 14.2.1.4

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

$f (\frac{π}{2}) = 2 - \cos (0)$

Step 14.2.1.5

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

$f (\frac{π}{2}) = 2 - 1 \cdot 1$

Step 14.2.1.6

Multiply $- 1$ by $1$ .

$f (\frac{π}{2}) = 2 - 1$

Step 14.2.2

Subtract $1$ from $2$ .

$f (\frac{π}{2}) = 1$

Step 14.2.3

The final answer is $1$ .

$y = 1$

Step 15

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.

$- 2 \sin (\frac{3 π}{2}) - 4 \cos (2 (\frac{3 π}{2}))$

Step 16

Evaluate the second derivative.

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Step 16.1

Simplify each term.

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Step 16.1.1

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.

$- 2 (- \sin (\frac{π}{2})) - 4 \cos (2 (\frac{3 π}{2}))$

Step 16.1.2

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

$- 2 (- 1 \cdot 1) - 4 \cos (2 (\frac{3 π}{2}))$

Step 16.1.3

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 16.1.3.1

Multiply $- 1$ by $1$ .

$- 2 \cdot - 1 - 4 \cos (2 (\frac{3 π}{2}))$

Step 16.1.3.2

Multiply $- 2$ by $- 1$ .

$2 - 4 \cos (2 (\frac{3 π}{2}))$

Step 16.1.4

Cancel the common factor of $2$ .

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Step 16.1.4.1

Cancel the common factor.

$2 - 4 \cos (2 \frac{3 π}{2})$

Step 16.1.4.2

Rewrite the expression.

$2 - 4 \cos (3 π)$

Step 16.1.5

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$2 - 4 \cos (π)$

Step 16.1.6

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

$2 - 4 (- \cos (0))$

Step 16.1.7

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

$2 - 4 (- 1 \cdot 1)$

Step 16.1.8

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

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Step 16.1.8.1

Multiply $- 1$ by $1$ .

$2 - 4 \cdot - 1$

Step 16.1.8.2

Multiply $- 4$ by $- 1$ .

$2 + 4$

Step 16.2

Add $2$ and $4$ .

$6$

Step 17

$x = \frac{3 π}{2}$ 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 18

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

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Step 18.1

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

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

Step 18.2

Simplify the result.

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Step 18.2.1

Simplify each term.

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Step 18.2.1.1

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.

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

Step 18.2.1.2

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

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

Step 18.2.1.3

Multiply $2 (- 1 \cdot 1)$ .

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Step 18.2.1.3.1

Multiply $- 1$ by $1$ .

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

Step 18.2.1.3.2

Multiply $2$ by $- 1$ .

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

Step 18.2.1.4

Cancel the common factor of $2$ .

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Step 18.2.1.4.1

Cancel the common factor.

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

Step 18.2.1.4.2

Rewrite the expression.

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

Step 18.2.1.5

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$f (\frac{3 π}{2}) = - 2 + \cos (π)$

Step 18.2.1.6

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

$f (\frac{3 π}{2}) = - 2 - \cos (0)$

Step 18.2.1.7

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

$f (\frac{3 π}{2}) = - 2 - 1 \cdot 1$

Step 18.2.1.8

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{2}) = - 2 - 1$

Step 18.2.2

Subtract $1$ from $- 2$ .

$f (\frac{3 π}{2}) = - 3$

Step 18.2.3

The final answer is $- 3$ .

$y = - 3$

Step 19

Evaluate the second derivative at $x = \frac{π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 \sin (\frac{π}{6}) - 4 \cos (2 (\frac{π}{6}))$

Step 20

Evaluate the second derivative.

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Step 20.1

Simplify each term.

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Step 20.1.1

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$- 2 (\frac{1}{2}) - 4 \cos (2 (\frac{π}{6}))$

Step 20.1.2

Cancel the common factor of $2$ .

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Step 20.1.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{1}{2} - 4 \cos (2 (\frac{π}{6}))$

Step 20.1.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{1}{2} - 4 \cos (2 (\frac{π}{6}))$

Step 20.1.2.3

Rewrite the expression.

$- 1 - 4 \cos (2 (\frac{π}{6}))$

Step 20.1.3

Cancel the common factor of $2$ .

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Step 20.1.3.1

Factor $2$ out of $6$ .

$- 1 - 4 \cos (2 \frac{π}{2 (3)})$

Step 20.1.3.2

Cancel the common factor.

$- 1 - 4 \cos (2 \frac{π}{2 \cdot 3})$

Step 20.1.3.3

Rewrite the expression.

$- 1 - 4 \cos (\frac{π}{3})$

Step 20.1.4

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$- 1 - 4 (\frac{1}{2})$

Step 20.1.5

Cancel the common factor of $2$ .

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Step 20.1.5.1

Factor $2$ out of $- 4$ .

$- 1 + 2 (- 2) \frac{1}{2}$

Step 20.1.5.2

Cancel the common factor.

$- 1 + 2 \cdot - 2 \frac{1}{2}$

Step 20.1.5.3

Rewrite the expression.

$- 1 - 2$

Step 20.2

Subtract $2$ from $- 1$ .

$- 3$

Step 21

$x = \frac{π}{6}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{6}$ is a local maximum

Step 22

Find the y-value when $x = \frac{π}{6}$ .

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Step 22.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = 2 \sin (\frac{π}{6}) + \cos (2 (\frac{π}{6}))$

Step 22.2

Simplify the result.

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Step 22.2.1

Simplify each term.

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Step 22.2.1.1

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$f (\frac{π}{6}) = 2 (\frac{1}{2}) + \cos (2 (\frac{π}{6}))$

Step 22.2.1.2

Cancel the common factor of $2$ .

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Step 22.2.1.2.1

Cancel the common factor.

$f (\frac{π}{6}) = 2 (\frac{1}{2}) + \cos (2 (\frac{π}{6}))$

Step 22.2.1.2.2

Rewrite the expression.

$f (\frac{π}{6}) = 1 + \cos (2 (\frac{π}{6}))$

Step 22.2.1.3

Cancel the common factor of $2$ .

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Step 22.2.1.3.1

Factor $2$ out of $6$ .

$f (\frac{π}{6}) = 1 + \cos (2 (\frac{π}{2 (3)}))$

Step 22.2.1.3.2

Cancel the common factor.

$f (\frac{π}{6}) = 1 + \cos (2 (\frac{π}{2 \cdot 3}))$

Step 22.2.1.3.3

Rewrite the expression.

$f (\frac{π}{6}) = 1 + \cos (\frac{π}{3})$

Step 22.2.1.4

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$f (\frac{π}{6}) = 1 + \frac{1}{2}$

Step 22.2.2

Simplify the expression.

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Step 22.2.2.1

Write $1$ as a fraction with a common denominator.

$f (\frac{π}{6}) = \frac{2}{2} + \frac{1}{2}$

Step 22.2.2.2

Combine the numerators over the common denominator.

$f (\frac{π}{6}) = \frac{2 + 1}{2}$

Step 22.2.2.3

Add $2$ and $1$ .

$f (\frac{π}{6}) = \frac{3}{2}$

Step 22.2.3

The final answer is $\frac{3}{2}$ .

$y = \frac{3}{2}$

Step 23

Evaluate the second derivative at $x = \frac{5 π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 \sin (\frac{5 π}{6}) - 4 \cos (2 (\frac{5 π}{6}))$

Step 24

Evaluate the second derivative.

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Step 24.1

Simplify each term.

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Step 24.1.1

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

$- 2 \sin (\frac{π}{6}) - 4 \cos (2 (\frac{5 π}{6}))$

Step 24.1.2

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$- 2 (\frac{1}{2}) - 4 \cos (2 (\frac{5 π}{6}))$

Step 24.1.3

Cancel the common factor of $2$ .

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Step 24.1.3.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{1}{2} - 4 \cos (2 (\frac{5 π}{6}))$

Step 24.1.3.2

Cancel the common factor.

$2 \cdot - 1 \frac{1}{2} - 4 \cos (2 (\frac{5 π}{6}))$

Step 24.1.3.3

Rewrite the expression.

$- 1 - 4 \cos (2 (\frac{5 π}{6}))$

Step 24.1.4

Cancel the common factor of $2$ .

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Step 24.1.4.1

Factor $2$ out of $6$ .

$- 1 - 4 \cos (2 \frac{5 π}{2 (3)})$

Step 24.1.4.2

Cancel the common factor.

$- 1 - 4 \cos (2 \frac{5 π}{2 \cdot 3})$

Step 24.1.4.3

Rewrite the expression.

$- 1 - 4 \cos (\frac{5 π}{3})$

Step 24.1.5

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

$- 1 - 4 \cos (\frac{π}{3})$

Step 24.1.6

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$- 1 - 4 (\frac{1}{2})$

Step 24.1.7

Cancel the common factor of $2$ .

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Step 24.1.7.1

Factor $2$ out of $- 4$ .

$- 1 + 2 (- 2) \frac{1}{2}$

Step 24.1.7.2

Cancel the common factor.

$- 1 + 2 \cdot - 2 \frac{1}{2}$

Step 24.1.7.3

Rewrite the expression.

$- 1 - 2$

Step 24.2

Subtract $2$ from $- 1$ .

$- 3$

Step 25

$x = \frac{5 π}{6}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{5 π}{6}$ is a local maximum

Step 26

Find the y-value when $x = \frac{5 π}{6}$ .

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Step 26.1

Replace the variable $x$ with $\frac{5 π}{6}$ in the expression.

$f (\frac{5 π}{6}) = 2 \sin (\frac{5 π}{6}) + \cos (2 (\frac{5 π}{6}))$

Step 26.2

Simplify the result.

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Step 26.2.1

Simplify each term.

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Step 26.2.1.1

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

$f (\frac{5 π}{6}) = 2 \sin (\frac{π}{6}) + \cos (2 (\frac{5 π}{6}))$

Step 26.2.1.2

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$f (\frac{5 π}{6}) = 2 (\frac{1}{2}) + \cos (2 (\frac{5 π}{6}))$

Step 26.2.1.3

Cancel the common factor of $2$ .

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Step 26.2.1.3.1

Cancel the common factor.

$f (\frac{5 π}{6}) = 2 (\frac{1}{2}) + \cos (2 (\frac{5 π}{6}))$

Step 26.2.1.3.2

Rewrite the expression.

$f (\frac{5 π}{6}) = 1 + \cos (2 (\frac{5 π}{6}))$

Step 26.2.1.4

Cancel the common factor of $2$ .

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Step 26.2.1.4.1

Factor $2$ out of $6$ .

$f (\frac{5 π}{6}) = 1 + \cos (2 (\frac{5 π}{2 (3)}))$

Step 26.2.1.4.2

Cancel the common factor.

$f (\frac{5 π}{6}) = 1 + \cos (2 (\frac{5 π}{2 \cdot 3}))$

Step 26.2.1.4.3

Rewrite the expression.

$f (\frac{5 π}{6}) = 1 + \cos (\frac{5 π}{3})$

Step 26.2.1.5

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

$f (\frac{5 π}{6}) = 1 + \cos (\frac{π}{3})$

Step 26.2.1.6

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$f (\frac{5 π}{6}) = 1 + \frac{1}{2}$

Step 26.2.2

Simplify the expression.

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Step 26.2.2.1

Write $1$ as a fraction with a common denominator.

$f (\frac{5 π}{6}) = \frac{2}{2} + \frac{1}{2}$

Step 26.2.2.2

Combine the numerators over the common denominator.

$f (\frac{5 π}{6}) = \frac{2 + 1}{2}$

Step 26.2.2.3

Add $2$ and $1$ .

$f (\frac{5 π}{6}) = \frac{3}{2}$

Step 26.2.3

The final answer is $\frac{3}{2}$ .

$y = \frac{3}{2}$

Step 27

These are the local extrema for $f (x) = 2 \sin (x) + \cos (2 x)$ .

$(\frac{π}{2}, 1)$ is a local minima

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

$(\frac{π}{6}, \frac{3}{2})$ is a local maxima

$(\frac{5 π}{6}, \frac{3}{2})$ is a local maxima

Step 28