Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

Multiply $- 1$ by $2$ .

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

Step 1.3

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

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Step 1.3.1

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

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Step 1.3.1.1

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

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

Step 1.3.1.2

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

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

Step 1.3.1.3

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

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Multiply $2$ by $1$ .

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

Step 1.3.5

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

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

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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Step 2.3.2.1

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

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $2$ by $1$ .

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

Step 2.3.6

Multiply $2$ by $- 1$ .

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

Step 2.3.7

Multiply $- 2$ by $2$ .

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

Step 3

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

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

Step 4

Simplify each term.

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Step 4.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 4.2

Apply the distributive property.

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

Step 4.3

Multiply $2$ by $1$ .

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

Step 4.4

Multiply $- 2$ by $2$ .

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

Step 5

Factor $- 2 \sin (x) + 2 - 4 \sin^{2} (x)$ .

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Step 5.1

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

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Step 5.1.1

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

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

Step 5.1.2

Factor $2$ out of $2$ .

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

Step 5.1.3

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

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

Step 5.1.4

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

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

Step 5.1.5

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

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

Step 5.2

Factor.

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Step 5.2.1

Factor by grouping.

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Step 5.2.1.1

Reorder terms.

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

Step 5.2.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 2 \cdot 1 = - 2$ and whose sum is $b = - 1$ .

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Step 5.2.1.2.1

Factor $- 1$ out of $- \sin (x)$ .

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

Step 5.2.1.2.2

Rewrite $- 1$ as $1$ plus $- 2$

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

Step 5.2.1.2.3

Apply the distributive property.

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

Step 5.2.1.2.4

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

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

Step 5.2.1.3

Factor out the greatest common factor from each group.

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Step 5.2.1.3.1

Group the first two terms and the last two terms.

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

Step 5.2.1.3.2

Factor out the greatest common factor (GCF) from each group.

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

Step 5.2.1.4

Factor the polynomial by factoring out the greatest common factor, $- 2 \sin (x) + 1$ .

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

Step 5.2.2

Remove unnecessary parentheses.

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

Step 6

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

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

$\sin (x) + 1 = 0$

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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Step 7.2.1

Subtract $1$ from both sides of the equation.

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

Step 7.2.2

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

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Step 7.2.2.1

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

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

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.

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

Step 7.2.2.2.1.2

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

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

Step 7.2.2.3

Simplify the right side.

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Step 7.2.2.3.1

Dividing two negative values results in a positive value.

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

Step 7.2.3

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

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

Step 7.2.4

Simplify the right side.

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Step 7.2.4.1

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

$x = \frac{π}{6}$

Step 7.2.5

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 7.2.6

Simplify $π - \frac{π}{6}$ .

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Step 7.2.6.1

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 7.2.6.2

Combine fractions.

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Step 7.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 7.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 7.2.6.3

Simplify the numerator.

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Step 7.2.6.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 7.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 7.2.7

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

$x = \frac{π}{6}, \frac{5 π}{6}$

Step 8

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

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Step 8.1

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

$\sin (x) + 1 = 0$

Step 8.2

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

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Step 8.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 8.2.2

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

$x = \arcsin (- 1)$

Step 8.2.3

Simplify the right side.

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Step 8.2.3.1

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

$x = - \frac{π}{2}$

Step 8.2.4

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.

$x = 2 π + \frac{π}{2} + π$

Step 8.2.5

Simplify the expression to find the second solution.

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Step 8.2.5.1

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

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 8.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 8.2.6

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

$x = - \frac{π}{2}, \frac{3 π}{2}$

Step 9

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

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

Step 10

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.

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

Step 11

Evaluate the second derivative.

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Step 11.1

Simplify each term.

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Step 11.1.1

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- 2 \frac{\sqrt{3}}{2} - 4 \sin (2 (\frac{π}{6}))$

Step 11.1.2

Cancel the common factor of $2$ .

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Step 11.1.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{\sqrt{3}}{2} - 4 \sin (2 (\frac{π}{6}))$

Step 11.1.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{\sqrt{3}}{2} - 4 \sin (2 (\frac{π}{6}))$

Step 11.1.2.3

Rewrite the expression.

$- 1 \sqrt{3} - 4 \sin (2 (\frac{π}{6}))$

Step 11.1.3

Rewrite $- 1 \sqrt{3}$ as $- \sqrt{3}$ .

$- \sqrt{3} - 4 \sin (2 (\frac{π}{6}))$

Step 11.1.4

Cancel the common factor of $2$ .

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Step 11.1.4.1

Factor $2$ out of $6$ .

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

Step 11.1.4.2

Cancel the common factor.

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

Step 11.1.4.3

Rewrite the expression.

$- \sqrt{3} - 4 \sin (\frac{π}{3})$

Step 11.1.5

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

$- \sqrt{3} - 4 \frac{\sqrt{3}}{2}$

Step 11.1.6

Cancel the common factor of $2$ .

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Step 11.1.6.1

Factor $2$ out of $- 4$ .

$- \sqrt{3} + 2 (- 2) \frac{\sqrt{3}}{2}$

Step 11.1.6.2

Cancel the common factor.

$- \sqrt{3} + 2 \cdot - 2 \frac{\sqrt{3}}{2}$

Step 11.1.6.3

Rewrite the expression.

$- \sqrt{3} - 2 \sqrt{3}$

Step 11.2

Subtract $2 \sqrt{3}$ from $- \sqrt{3}$ .

$- 3 \sqrt{3}$

Step 12

$x = \frac{π}{6}$ 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 13

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

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Step 13.1

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

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

Step 13.2

Simplify the result.

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Step 13.2.1

Simplify each term.

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Step 13.2.1.1

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = 2 (\frac{\sqrt{3}}{2}) + \sin (2 (\frac{π}{6}))$

Step 13.2.1.2

Cancel the common factor of $2$ .

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Step 13.2.1.2.1

Cancel the common factor.

$f (\frac{π}{6}) = 2 (\frac{\sqrt{3}}{2}) + \sin (2 (\frac{π}{6}))$

Step 13.2.1.2.2

Rewrite the expression.

$f (\frac{π}{6}) = \sqrt{3} + \sin (2 (\frac{π}{6}))$

Step 13.2.1.3

Cancel the common factor of $2$ .

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Step 13.2.1.3.1

Factor $2$ out of $6$ .

$f (\frac{π}{6}) = \sqrt{3} + \sin (2 (\frac{π}{2 (3)}))$

Step 13.2.1.3.2

Cancel the common factor.

$f (\frac{π}{6}) = \sqrt{3} + \sin (2 (\frac{π}{2 \cdot 3}))$

Step 13.2.1.3.3

Rewrite the expression.

$f (\frac{π}{6}) = \sqrt{3} + \sin (\frac{π}{3})$

Step 13.2.1.4

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

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

Step 13.2.2

To write $\sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 13.2.3

Combine fractions.

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Step 13.2.3.1

Combine $\sqrt{3}$ and $\frac{2}{2}$ .

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

Step 13.2.3.2

Combine the numerators over the common denominator.

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

Step 13.2.4

Simplify the numerator.

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Step 13.2.4.1

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

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

Step 13.2.4.2

Add $2 \sqrt{3}$ and $\sqrt{3}$ .

$f (\frac{π}{6}) = \frac{3 \sqrt{3}}{2}$

Step 13.2.5

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

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

Step 14

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.

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

Step 15

Evaluate the second derivative.

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Step 15.1

Simplify each term.

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Step 15.1.1

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.

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

Step 15.1.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- 2 (- \frac{\sqrt{3}}{2}) - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.3

Cancel the common factor of $2$ .

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Step 15.1.3.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$- 2 \frac{- \sqrt{3}}{2} - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.3.2

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{- \sqrt{3}}{2} - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.3.3

Cancel the common factor.

$2 \cdot - 1 \frac{- \sqrt{3}}{2} - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.3.4

Rewrite the expression.

$- 1 (- \sqrt{3}) - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.4

Multiply $- 1$ by $- 1$ .

$1 \sqrt{3} - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.5

Multiply $\sqrt{3}$ by $1$ .

$\sqrt{3} - 4 \sin (2 (\frac{5 π}{6}))$

Step 15.1.6

Cancel the common factor of $2$ .

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Step 15.1.6.1

Factor $2$ out of $6$ .

$\sqrt{3} - 4 \sin (2 \frac{5 π}{2 (3)})$

Step 15.1.6.2

Cancel the common factor.

$\sqrt{3} - 4 \sin (2 \frac{5 π}{2 \cdot 3})$

Step 15.1.6.3

Rewrite the expression.

$\sqrt{3} - 4 \sin (\frac{5 π}{3})$

Step 15.1.7

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.

$\sqrt{3} - 4 (- \sin (\frac{π}{3}))$

Step 15.1.8

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

$\sqrt{3} - 4 (- \frac{\sqrt{3}}{2})$

Step 15.1.9

Cancel the common factor of $2$ .

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Step 15.1.9.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$\sqrt{3} - 4 \frac{- \sqrt{3}}{2}$

Step 15.1.9.2

Factor $2$ out of $- 4$ .

$\sqrt{3} + 2 (- 2) \frac{- \sqrt{3}}{2}$

Step 15.1.9.3

Cancel the common factor.

$\sqrt{3} + 2 \cdot - 2 \frac{- \sqrt{3}}{2}$

Step 15.1.9.4

Rewrite the expression.

$\sqrt{3} - 2 (- \sqrt{3})$

Step 15.1.10

Multiply $- 1$ by $- 2$ .

$\sqrt{3} + 2 \sqrt{3}$

Step 15.2

Add $\sqrt{3}$ and $2 \sqrt{3}$ .

$3 \sqrt{3}$

Step 16

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

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

Step 17

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

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Step 17.1

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

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

Step 17.2

Simplify the result.

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Step 17.2.1

Simplify each term.

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Step 17.2.1.1

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.

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

Step 17.2.1.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{5 π}{6}) = 2 (- \frac{\sqrt{3}}{2}) + \sin (2 (\frac{5 π}{6}))$

Step 17.2.1.3

Cancel the common factor of $2$ .

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Step 17.2.1.3.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$f (\frac{5 π}{6}) = 2 (\frac{- \sqrt{3}}{2}) + \sin (2 (\frac{5 π}{6}))$

Step 17.2.1.3.2

Cancel the common factor.

$f (\frac{5 π}{6}) = 2 (\frac{- \sqrt{3}}{2}) + \sin (2 (\frac{5 π}{6}))$

Step 17.2.1.3.3

Rewrite the expression.

$f (\frac{5 π}{6}) = - \sqrt{3} + \sin (2 (\frac{5 π}{6}))$

Step 17.2.1.4

Cancel the common factor of $2$ .

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Step 17.2.1.4.1

Factor $2$ out of $6$ .

$f (\frac{5 π}{6}) = - \sqrt{3} + \sin (2 (\frac{5 π}{2 (3)}))$

Step 17.2.1.4.2

Cancel the common factor.

$f (\frac{5 π}{6}) = - \sqrt{3} + \sin (2 (\frac{5 π}{2 \cdot 3}))$

Step 17.2.1.4.3

Rewrite the expression.

$f (\frac{5 π}{6}) = - \sqrt{3} + \sin (\frac{5 π}{3})$

Step 17.2.1.5

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.

$f (\frac{5 π}{6}) = - \sqrt{3} - \sin (\frac{π}{3})$

Step 17.2.1.6

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

$f (\frac{5 π}{6}) = - \sqrt{3} - \frac{\sqrt{3}}{2}$

Step 17.2.2

To write $- \sqrt{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$f (\frac{5 π}{6}) = - \sqrt{3} \cdot \frac{2}{2} - \frac{\sqrt{3}}{2}$

Step 17.2.3

Combine fractions.

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Step 17.2.3.1

Combine $- \sqrt{3}$ and $\frac{2}{2}$ .

$f (\frac{5 π}{6}) = \frac{- \sqrt{3} \cdot 2}{2} - \frac{\sqrt{3}}{2}$

Step 17.2.3.2

Combine the numerators over the common denominator.

$f (\frac{5 π}{6}) = \frac{- \sqrt{3} \cdot 2 - \sqrt{3}}{2}$

Step 17.2.4

Simplify the numerator.

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Step 17.2.4.1

Multiply $2$ by $- 1$ .

$f (\frac{5 π}{6}) = \frac{- 2 \sqrt{3} - \sqrt{3}}{2}$

Step 17.2.4.2

Subtract $\sqrt{3}$ from $- 2 \sqrt{3}$ .

$f (\frac{5 π}{6}) = \frac{- 3 \sqrt{3}}{2}$

Step 17.2.5

Move the negative in front of the fraction.

$f (\frac{5 π}{6}) = - \frac{3 \sqrt{3}}{2}$

Step 17.2.6

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

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

Step 18

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.

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

Step 19

Evaluate the second derivative.

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Step 19.1

Simplify each term.

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Step 19.1.1

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

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

Step 19.1.2

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

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

Step 19.1.3

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

$- 2 \cdot 0 - 4 \sin (2 (- \frac{π}{2}))$

Step 19.1.4

Multiply $- 2$ by $0$ .

$0 - 4 \sin (2 (- \frac{π}{2}))$

Step 19.1.5

Cancel the common factor of $2$ .

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Step 19.1.5.1

Move the leading negative in $- \frac{π}{2}$ into the numerator.

$0 - 4 \sin (2 \frac{- π}{2})$

Step 19.1.5.2

Cancel the common factor.

$0 - 4 \sin (2 \frac{- π}{2})$

Step 19.1.5.3

Rewrite the expression.

$0 - 4 \sin (- π)$

Step 19.1.6

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

$0 - 4 \sin (0)$

Step 19.1.7

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

$0 - 4 \cdot 0$

Step 19.1.8

Multiply $- 4$ by $0$ .

$0 + 0$

Step 19.2

Add $0$ and $0$ .

$0$

Step 20

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 20.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, - \frac{π}{2}) \cup (- \frac{π}{2}, \frac{π}{6}) \cup (\frac{π}{6}, \frac{5 π}{6}) \cup (\frac{5 π}{6}, \frac{3 π}{2}) \cup (\frac{3 π}{2}, \infty)$

Step 20.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - \frac{π}{2})$ in the first derivative $- 2 \sin (x) + 2 \cos (2 x)$ to check if the result is negative or positive.

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Step 20.2.1

Replace the variable $x$ with $- 4$ in the expression.

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

Step 20.2.2

Simplify the result.

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Step 20.2.2.1

Simplify each term.

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Step 20.2.2.1.1

Evaluate $\sin (- 4)$ .

$f' (- 4) = - 2 \cdot 0.75680249 + 2 \cos (2 (- 4))$

Step 20.2.2.1.2

Multiply $- 2$ by $0.75680249$ .

$f' (- 4) = - 1.51360499 + 2 \cos (2 (- 4))$

Step 20.2.2.1.3

Multiply $2$ by $- 4$ .

$f' (- 4) = - 1.51360499 + 2 \cos (- 8)$

Step 20.2.2.1.4

Evaluate $\cos (- 8)$ .

$f' (- 4) = - 1.51360499 + 2 \cdot - 0.14550003$

Step 20.2.2.1.5

Multiply $2$ by $- 0.14550003$ .

$f' (- 4) = - 1.51360499 - 0.29100006$

Step 20.2.2.2

Subtract $0.29100006$ from $- 1.51360499$ .

$f' (- 4) = - 1.80460505$

Step 20.2.2.3

The final answer is $- 1.80460505$ .

$- 1.80460505$

Step 20.3

Substitute any number, such as $0$ , from the interval $(- \frac{π}{2}, \frac{π}{6})$ in the first derivative $- 2 \sin (x) + 2 \cos (2 x)$ to check if the result is negative or positive.

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Step 20.3.1

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

$f' (0) = - 2 \sin (0) + 2 \cos (2 (0))$

Step 20.3.2

Simplify the result.

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Step 20.3.2.1

Simplify each term.

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Step 20.3.2.1.1

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

$f' (0) = - 2 \cdot 0 + 2 \cos (2 (0))$

Step 20.3.2.1.2

Multiply $- 2$ by $0$ .

$f' (0) = 0 + 2 \cos (2 (0))$

Step 20.3.2.1.3

Multiply $2$ by $0$ .

$f' (0) = 0 + 2 \cos (0)$

Step 20.3.2.1.4

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

$f' (0) = 0 + 2 \cdot 1$

Step 20.3.2.1.5

Multiply $2$ by $1$ .

$f' (0) = 0 + 2$

Step 20.3.2.2

Add $0$ and $2$ .

$f' (0) = 2$

Step 20.3.2.3

The final answer is $2$ .

$2$

Step 20.4

Substitute any number, such as $2$ , from the interval $(\frac{π}{6}, \frac{5 π}{6})$ in the first derivative $- 2 \sin (x) + 2 \cos (2 x)$ to check if the result is negative or positive.

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Step 20.4.1

Replace the variable $x$ with $2$ in the expression.

$f' (2) = - 2 \sin (2) + 2 \cos (2 (2))$

Step 20.4.2

Simplify the result.

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Step 20.4.2.1

Simplify each term.

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Step 20.4.2.1.1

Evaluate $\sin (2)$ .

$f' (2) = - 2 \cdot 0.90929742 + 2 \cos (2 (2))$

Step 20.4.2.1.2

Multiply $- 2$ by $0.90929742$ .

$f' (2) = - 1.81859485 + 2 \cos (2 (2))$

Step 20.4.2.1.3

Multiply $2$ by $2$ .

$f' (2) = - 1.81859485 + 2 \cos (4)$

Step 20.4.2.1.4

Evaluate $\cos (4)$ .

$f' (2) = - 1.81859485 + 2 \cdot - 0.65364362$

Step 20.4.2.1.5

Multiply $2$ by $- 0.65364362$ .

$f' (2) = - 1.81859485 - 1.30728724$

Step 20.4.2.2

Subtract $1.30728724$ from $- 1.81859485$ .

$f' (2) = - 3.12588209$

Step 20.4.2.3

The final answer is $- 3.12588209$ .

$- 3.12588209$

Step 20.5

Substitute any number, such as $4$ , from the interval $(\frac{5 π}{6}, \frac{3 π}{2})$ in the first derivative $- 2 \sin (x) + 2 \cos (2 x)$ to check if the result is negative or positive.

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Step 20.5.1

Replace the variable $x$ with $4$ in the expression.

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

Step 20.5.2

Simplify the result.

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Step 20.5.2.1

Simplify each term.

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Step 20.5.2.1.1

Evaluate $\sin (4)$ .

$f' (4) = - 2 \cdot - 0.75680249 + 2 \cos (2 (4))$

Step 20.5.2.1.2

Multiply $- 2$ by $- 0.75680249$ .

$f' (4) = 1.51360499 + 2 \cos (2 (4))$

Step 20.5.2.1.3

Multiply $2$ by $4$ .

$f' (4) = 1.51360499 + 2 \cos (8)$

Step 20.5.2.1.4

Evaluate $\cos (8)$ .

$f' (4) = 1.51360499 + 2 \cdot - 0.14550003$

Step 20.5.2.1.5

Multiply $2$ by $- 0.14550003$ .

$f' (4) = 1.51360499 - 0.29100006$

Step 20.5.2.2

Subtract $0.29100006$ from $1.51360499$ .

$f' (4) = 1.22260492$

Step 20.5.2.3

The final answer is $1.22260492$ .

$1.22260492$

Step 20.6

Substitute any number, such as $7$ , from the interval $(\frac{3 π}{2}, \infty)$ in the first derivative $- 2 \sin (x) + 2 \cos (2 x)$ to check if the result is negative or positive.

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Step 20.6.1

Replace the variable $x$ with $7$ in the expression.

$f' (7) = - 2 \sin (7) + 2 \cos (2 (7))$

Step 20.6.2

Simplify the result.

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Step 20.6.2.1

Simplify each term.

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Step 20.6.2.1.1

Evaluate $\sin (7)$ .

$f' (7) = - 2 \cdot 0.65698659 + 2 \cos (2 (7))$

Step 20.6.2.1.2

Multiply $- 2$ by $0.65698659$ .

$f' (7) = - 1.31397319 + 2 \cos (2 (7))$

Step 20.6.2.1.3

Multiply $2$ by $7$ .

$f' (7) = - 1.31397319 + 2 \cos (14)$

Step 20.6.2.1.4

Evaluate $\cos (14)$ .

$f' (7) = - 1.31397319 + 2 \cdot 0.13673721$

Step 20.6.2.1.5

Multiply $2$ by $0.13673721$ .

$f' (7) = - 1.31397319 + 0.27347443$

Step 20.6.2.2

Add $- 1.31397319$ and $0.27347443$ .

$f' (7) = - 1.04049876$

Step 20.6.2.3

The final answer is $- 1.04049876$ .

$- 1.04049876$

Step 20.7

Since the first derivative changed signs from negative to positive around $x = - \frac{π}{2}$ , then $x = - \frac{π}{2}$ is a local minimum.

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

Step 20.8

Since the first derivative changed signs from positive to negative around $x = \frac{π}{6}$ , then $x = \frac{π}{6}$ is a local maximum.

$x = \frac{π}{6}$ is a local maximum

Step 20.9

Since the first derivative changed signs from negative to positive around $x = \frac{5 π}{6}$ , then $x = \frac{5 π}{6}$ is a local minimum.

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

Step 20.10

Since the first derivative changed signs from positive to negative around $x = \frac{3 π}{2}$ , then $x = \frac{3 π}{2}$ is a local maximum.

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

Step 20.11

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

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

$x = \frac{π}{6}$ is a local maximum

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

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

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

$x = \frac{π}{6}$ is a local maximum

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

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

Step 21