Calculus Examples

$f (x) = x - 2 \cos (x)$ , $0 < x < 2 π$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Differentiate.

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Step 1.1.1.1.1

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

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

Step 1.1.1.1.2

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

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

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

$1 - 2 \frac{d}{d x} [\cos (x)]$

Step 1.1.1.2.2

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

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

Step 1.1.1.2.3

Multiply $- 1$ by $- 2$ .

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

Step 1.1.2

The first derivative of $f (x)$ with respect to $x$ is $1 + 2 \sin (x)$ .

$1 + 2 \sin (x)$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $1 + 2 \sin (x) = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

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

Step 1.2.2

Subtract $1$ from both sides of the equation.

$2 \sin (x) = - 1$

Step 1.2.3

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

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Step 1.2.3.1

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

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

Step 1.2.3.2

Simplify the left side.

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Step 1.2.3.2.1

Cancel the common factor of $2$ .

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Step 1.2.3.2.1.1

Cancel the common factor.

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

Step 1.2.3.2.1.2

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

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

Step 1.2.3.3

Simplify the right side.

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Step 1.2.3.3.1

Move the negative in front of the fraction.

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

Step 1.2.4

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

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

Step 1.2.5

Simplify the right side.

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Step 1.2.5.1

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

$x = - \frac{π}{6}$

Step 1.2.6

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.

$x = 2 π + \frac{π}{6} + π$

Step 1.2.7

Simplify the expression to find the second solution.

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Step 1.2.7.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$x = 2 π + \frac{π}{6} + π - 2 π$

Step 1.2.7.2

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$x = \frac{7 π}{6}$

Step 1.2.8

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

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Step 1.2.8.1

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

$\frac{2 π}{| b |}$

Step 1.2.8.2

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

$\frac{2 π}{| 1 |}$

Step 1.2.8.3

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

$\frac{2 π}{1}$

Step 1.2.8.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.9

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

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Step 1.2.9.1

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

Step 1.2.9.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 1.2.9.3

Combine fractions.

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Step 1.2.9.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 1.2.9.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 1.2.9.4

Simplify the numerator.

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Step 1.2.9.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 1.2.9.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 1.2.9.5

List the new angles.

$x = \frac{11 π}{6}$

Step 1.2.10

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

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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.

Step 1.4

Evaluate $x - 2 \cos (x)$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = \frac{7 π}{6}$ .

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Step 1.4.1.1

Substitute $\frac{7 π}{6}$ for $x$ .

$(\frac{7 π}{6}) - 2 \cos (\frac{7 π}{6})$

Step 1.4.1.2

Simplify each term.

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Step 1.4.1.2.1

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 third quadrant.

$\frac{7 π}{6} - 2 (- \cos (\frac{π}{6}))$

Step 1.4.1.2.2

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

$\frac{7 π}{6} - 2 (- \frac{\sqrt{3}}{2})$

Step 1.4.1.2.3

Cancel the common factor of $2$ .

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Step 1.4.1.2.3.1

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

$\frac{7 π}{6} - 2 \frac{- \sqrt{3}}{2}$

Step 1.4.1.2.3.2

Factor $2$ out of $- 2$ .

$\frac{7 π}{6} + 2 (- 1) \frac{- \sqrt{3}}{2}$

Step 1.4.1.2.3.3

Cancel the common factor.

$\frac{7 π}{6} + 2 \cdot - 1 \frac{- \sqrt{3}}{2}$

Step 1.4.1.2.3.4

Rewrite the expression.

$\frac{7 π}{6} - 1 (- \sqrt{3})$

Step 1.4.1.2.4

Multiply $- 1$ by $- 1$ .

$\frac{7 π}{6} + 1 \sqrt{3}$

Step 1.4.1.2.5

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

$\frac{7 π}{6} + \sqrt{3}$

Step 1.4.2

Evaluate at $x = \frac{19 π}{6}$ .

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Step 1.4.2.1

Substitute $\frac{19 π}{6}$ for $x$ .

$(\frac{19 π}{6}) - 2 \cos (\frac{19 π}{6})$

Step 1.4.2.2

Simplify each term.

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Step 1.4.2.2.1

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

$\frac{19 π}{6} - 2 \cos (\frac{7 π}{6})$

Step 1.4.2.2.2

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 third quadrant.

$\frac{19 π}{6} - 2 (- \cos (\frac{π}{6}))$

Step 1.4.2.2.3

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

$\frac{19 π}{6} - 2 (- \frac{\sqrt{3}}{2})$

Step 1.4.2.2.4

Cancel the common factor of $2$ .

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Step 1.4.2.2.4.1

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

$\frac{19 π}{6} - 2 \frac{- \sqrt{3}}{2}$

Step 1.4.2.2.4.2

Factor $2$ out of $- 2$ .

$\frac{19 π}{6} + 2 (- 1) \frac{- \sqrt{3}}{2}$

Step 1.4.2.2.4.3

Cancel the common factor.

$\frac{19 π}{6} + 2 \cdot - 1 \frac{- \sqrt{3}}{2}$

Step 1.4.2.2.4.4

Rewrite the expression.

$\frac{19 π}{6} - 1 (- \sqrt{3})$

Step 1.4.2.2.5

Multiply $- 1$ by $- 1$ .

$\frac{19 π}{6} + 1 \sqrt{3}$

Step 1.4.2.2.6

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

$\frac{19 π}{6} + \sqrt{3}$

Step 1.4.3

Evaluate at $x = \frac{31 π}{6}$ .

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Step 1.4.3.1

Substitute $\frac{31 π}{6}$ for $x$ .

$(\frac{31 π}{6}) - 2 \cos (\frac{31 π}{6})$

Step 1.4.3.2

Simplify each term.

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Step 1.4.3.2.1

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

$\frac{31 π}{6} - 2 \cos (\frac{7 π}{6})$

Step 1.4.3.2.2

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 third quadrant.

$\frac{31 π}{6} - 2 (- \cos (\frac{π}{6}))$

Step 1.4.3.2.3

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

$\frac{31 π}{6} - 2 (- \frac{\sqrt{3}}{2})$

Step 1.4.3.2.4

Cancel the common factor of $2$ .

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Step 1.4.3.2.4.1

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

$\frac{31 π}{6} - 2 \frac{- \sqrt{3}}{2}$

Step 1.4.3.2.4.2

Factor $2$ out of $- 2$ .

$\frac{31 π}{6} + 2 (- 1) \frac{- \sqrt{3}}{2}$

Step 1.4.3.2.4.3

Cancel the common factor.

$\frac{31 π}{6} + 2 \cdot - 1 \frac{- \sqrt{3}}{2}$

Step 1.4.3.2.4.4

Rewrite the expression.

$\frac{31 π}{6} - 1 (- \sqrt{3})$

Step 1.4.3.2.5

Multiply $- 1$ by $- 1$ .

$\frac{31 π}{6} + 1 \sqrt{3}$

Step 1.4.3.2.6

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

$\frac{31 π}{6} + \sqrt{3}$

Step 1.4.4

Evaluate at $x = \frac{43 π}{6}$ .

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Step 1.4.4.1

Substitute $\frac{43 π}{6}$ for $x$ .

$(\frac{43 π}{6}) - 2 \cos (\frac{43 π}{6})$

Step 1.4.4.2

Simplify each term.

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Step 1.4.4.2.1

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

$\frac{43 π}{6} - 2 \cos (\frac{7 π}{6})$

Step 1.4.4.2.2

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 third quadrant.

$\frac{43 π}{6} - 2 (- \cos (\frac{π}{6}))$

Step 1.4.4.2.3

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

$\frac{43 π}{6} - 2 (- \frac{\sqrt{3}}{2})$

Step 1.4.4.2.4

Cancel the common factor of $2$ .

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Step 1.4.4.2.4.1

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

$\frac{43 π}{6} - 2 \frac{- \sqrt{3}}{2}$

Step 1.4.4.2.4.2

Factor $2$ out of $- 2$ .

$\frac{43 π}{6} + 2 (- 1) \frac{- \sqrt{3}}{2}$

Step 1.4.4.2.4.3

Cancel the common factor.

$\frac{43 π}{6} + 2 \cdot - 1 \frac{- \sqrt{3}}{2}$

Step 1.4.4.2.4.4

Rewrite the expression.

$\frac{43 π}{6} - 1 (- \sqrt{3})$

Step 1.4.4.2.5

Multiply $- 1$ by $- 1$ .

$\frac{43 π}{6} + 1 \sqrt{3}$

Step 1.4.4.2.6

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

$\frac{43 π}{6} + \sqrt{3}$

Step 1.4.5

Evaluate at $x = \frac{55 π}{6}$ .

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Step 1.4.5.1

Substitute $\frac{55 π}{6}$ for $x$ .

$(\frac{55 π}{6}) - 2 \cos (\frac{55 π}{6})$

Step 1.4.5.2

Simplify each term.

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Step 1.4.5.2.1

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

$\frac{55 π}{6} - 2 \cos (\frac{7 π}{6})$

Step 1.4.5.2.2

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 third quadrant.

$\frac{55 π}{6} - 2 (- \cos (\frac{π}{6}))$

Step 1.4.5.2.3

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

$\frac{55 π}{6} - 2 (- \frac{\sqrt{3}}{2})$

Step 1.4.5.2.4

Cancel the common factor of $2$ .

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Step 1.4.5.2.4.1

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

$\frac{55 π}{6} - 2 \frac{- \sqrt{3}}{2}$

Step 1.4.5.2.4.2

Factor $2$ out of $- 2$ .

$\frac{55 π}{6} + 2 (- 1) \frac{- \sqrt{3}}{2}$

Step 1.4.5.2.4.3

Cancel the common factor.

$\frac{55 π}{6} + 2 \cdot - 1 \frac{- \sqrt{3}}{2}$

Step 1.4.5.2.4.4

Rewrite the expression.

$\frac{55 π}{6} - 1 (- \sqrt{3})$

Step 1.4.5.2.5

Multiply $- 1$ by $- 1$ .

$\frac{55 π}{6} + 1 \sqrt{3}$

Step 1.4.5.2.6

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

$\frac{55 π}{6} + \sqrt{3}$

Step 1.4.6

Evaluate at $x = \frac{11 π}{6}$ .

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Step 1.4.6.1

Substitute $\frac{11 π}{6}$ for $x$ .

$(\frac{11 π}{6}) - 2 \cos (\frac{11 π}{6})$

Step 1.4.6.2

Simplify each term.

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Step 1.4.6.2.1

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

$\frac{11 π}{6} - 2 \cos (\frac{π}{6})$

Step 1.4.6.2.2

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

$\frac{11 π}{6} - 2 \frac{\sqrt{3}}{2}$

Step 1.4.6.2.3

Cancel the common factor of $2$ .

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Step 1.4.6.2.3.1

Factor $2$ out of $- 2$ .

$\frac{11 π}{6} + 2 (- 1) \frac{\sqrt{3}}{2}$

Step 1.4.6.2.3.2

Cancel the common factor.

$\frac{11 π}{6} + 2 \cdot - 1 \frac{\sqrt{3}}{2}$

Step 1.4.6.2.3.3

Rewrite the expression.

$\frac{11 π}{6} - 1 \sqrt{3}$

Step 1.4.6.2.4

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

$\frac{11 π}{6} - \sqrt{3}$

Step 1.4.7

Evaluate at $x = \frac{23 π}{6}$ .

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Step 1.4.7.1

Substitute $\frac{23 π}{6}$ for $x$ .

$(\frac{23 π}{6}) - 2 \cos (\frac{23 π}{6})$

Step 1.4.7.2

Simplify each term.

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Step 1.4.7.2.1

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

$\frac{23 π}{6} - 2 \cos (\frac{11 π}{6})$

Step 1.4.7.2.2

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

$\frac{23 π}{6} - 2 \cos (\frac{π}{6})$

Step 1.4.7.2.3

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

$\frac{23 π}{6} - 2 \frac{\sqrt{3}}{2}$

Step 1.4.7.2.4

Cancel the common factor of $2$ .

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Step 1.4.7.2.4.1

Factor $2$ out of $- 2$ .

$\frac{23 π}{6} + 2 (- 1) \frac{\sqrt{3}}{2}$

Step 1.4.7.2.4.2

Cancel the common factor.

$\frac{23 π}{6} + 2 \cdot - 1 \frac{\sqrt{3}}{2}$

Step 1.4.7.2.4.3

Rewrite the expression.

$\frac{23 π}{6} - 1 \sqrt{3}$

Step 1.4.7.2.5

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

$\frac{23 π}{6} - \sqrt{3}$

Step 1.4.8

Evaluate at $x = \frac{35 π}{6}$ .

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Step 1.4.8.1

Substitute $\frac{35 π}{6}$ for $x$ .

$(\frac{35 π}{6}) - 2 \cos (\frac{35 π}{6})$

Step 1.4.8.2

Simplify each term.

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Step 1.4.8.2.1

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

$\frac{35 π}{6} - 2 \cos (\frac{11 π}{6})$

Step 1.4.8.2.2

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

$\frac{35 π}{6} - 2 \cos (\frac{π}{6})$

Step 1.4.8.2.3

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

$\frac{35 π}{6} - 2 \frac{\sqrt{3}}{2}$

Step 1.4.8.2.4

Cancel the common factor of $2$ .

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Step 1.4.8.2.4.1

Factor $2$ out of $- 2$ .

$\frac{35 π}{6} + 2 (- 1) \frac{\sqrt{3}}{2}$

Step 1.4.8.2.4.2

Cancel the common factor.

$\frac{35 π}{6} + 2 \cdot - 1 \frac{\sqrt{3}}{2}$

Step 1.4.8.2.4.3

Rewrite the expression.

$\frac{35 π}{6} - 1 \sqrt{3}$

Step 1.4.8.2.5

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

$\frac{35 π}{6} - \sqrt{3}$

Step 1.4.9

Evaluate at $x = \frac{47 π}{6}$ .

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Step 1.4.9.1

Substitute $\frac{47 π}{6}$ for $x$ .

$(\frac{47 π}{6}) - 2 \cos (\frac{47 π}{6})$

Step 1.4.9.2

Simplify each term.

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Step 1.4.9.2.1

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

$\frac{47 π}{6} - 2 \cos (\frac{11 π}{6})$

Step 1.4.9.2.2

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

$\frac{47 π}{6} - 2 \cos (\frac{π}{6})$

Step 1.4.9.2.3

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

$\frac{47 π}{6} - 2 \frac{\sqrt{3}}{2}$

Step 1.4.9.2.4

Cancel the common factor of $2$ .

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Step 1.4.9.2.4.1

Factor $2$ out of $- 2$ .

$\frac{47 π}{6} + 2 (- 1) \frac{\sqrt{3}}{2}$

Step 1.4.9.2.4.2

Cancel the common factor.

$\frac{47 π}{6} + 2 \cdot - 1 \frac{\sqrt{3}}{2}$

Step 1.4.9.2.4.3

Rewrite the expression.

$\frac{47 π}{6} - 1 \sqrt{3}$

Step 1.4.9.2.5

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

$\frac{47 π}{6} - \sqrt{3}$

Step 1.4.10

Evaluate at $x = \frac{59 π}{6}$ .

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Step 1.4.10.1

Substitute $\frac{59 π}{6}$ for $x$ .

$(\frac{59 π}{6}) - 2 \cos (\frac{59 π}{6})$

Step 1.4.10.2

Simplify each term.

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Step 1.4.10.2.1

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

$\frac{59 π}{6} - 2 \cos (\frac{11 π}{6})$

Step 1.4.10.2.2

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

$\frac{59 π}{6} - 2 \cos (\frac{π}{6})$

Step 1.4.10.2.3

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

$\frac{59 π}{6} - 2 \frac{\sqrt{3}}{2}$

Step 1.4.10.2.4

Cancel the common factor of $2$ .

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Step 1.4.10.2.4.1

Factor $2$ out of $- 2$ .

$\frac{59 π}{6} + 2 (- 1) \frac{\sqrt{3}}{2}$

Step 1.4.10.2.4.2

Cancel the common factor.

$\frac{59 π}{6} + 2 \cdot - 1 \frac{\sqrt{3}}{2}$

Step 1.4.10.2.4.3

Rewrite the expression.

$\frac{59 π}{6} - 1 \sqrt{3}$

Step 1.4.10.2.5

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

$\frac{59 π}{6} - \sqrt{3}$

Step 1.4.11

List all of the points.

$(\frac{7 π}{6}, \frac{7 π}{6} + \sqrt{3}), (\frac{19 π}{6}, \frac{19 π}{6} + \sqrt{3}), (\frac{31 π}{6}, \frac{31 π}{6} + \sqrt{3}), (\frac{43 π}{6}, \frac{43 π}{6} + \sqrt{3}), (\frac{55 π}{6}, \frac{55 π}{6} + \sqrt{3}), (\frac{11 π}{6}, \frac{11 π}{6} - \sqrt{3}), (\frac{23 π}{6}, \frac{23 π}{6} - \sqrt{3}), (\frac{35 π}{6}, \frac{35 π}{6} - \sqrt{3}), (\frac{47 π}{6}, \frac{47 π}{6} - \sqrt{3}), (\frac{59 π}{6}, \frac{59 π}{6} - \sqrt{3})$

Step 2

Exclude the points that are not on the interval.

$(\frac{7 π}{6}, \frac{7 π}{6} + \sqrt{3}), (\frac{11 π}{6}, \frac{11 π}{6} - \sqrt{3})$

Step 3

Use the first derivative test to determine which points can be maxima or minima.

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Step 3.1

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

$(- \infty, \frac{7 π}{6}) \cup (\frac{7 π}{6}, \frac{11 π}{6}) \cup (\frac{11 π}{6}, \frac{19 π}{6}) \cup (\frac{19 π}{6}, \frac{23 π}{6}) \cup (\frac{23 π}{6}, \frac{31 π}{6}) \cup (\frac{31 π}{6}, \frac{35 π}{6}) \cup (\frac{35 π}{6}, \frac{43 π}{6}) \cup (\frac{43 π}{6}, \frac{47 π}{6}) \cup (\frac{47 π}{6}, \frac{55 π}{6}) \cup (\frac{55 π}{6}, \frac{59 π}{6}) \cup (\frac{59 π}{6}, \infty)$

Step 3.2

Substitute any number, such as $0$ , from the interval $(- \infty, \frac{7 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.2.1

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

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

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

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

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

Step 3.2.2.1.2

Multiply $2$ by $0$ .

$f' (0) = 1 + 0$

Step 3.2.2.2

Add $1$ and $0$ .

$f' (0) = 1$

Step 3.2.2.3

The final answer is $1$ .

$1$

Step 3.3

Substitute any number, such as $5$ , from the interval $(\frac{7 π}{6}, \frac{11 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.3.1

Replace the variable $x$ with $5$ in the expression.

$f' (5) = 1 + 2 \sin (5)$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Evaluate $\sin (5)$ .

$f' (5) = 1 + 2 \cdot - 0.95892427$

Step 3.3.2.1.2

Multiply $2$ by $- 0.95892427$ .

$f' (5) = 1 - 1.91784854$

Step 3.3.2.2

Subtract $1.91784854$ from $1$ .

$f' (5) = - 0.91784854$

Step 3.3.2.3

The final answer is $- 0.91784854$ .

$- 0.91784854$

Step 3.4

Substitute any number, such as $8$ , from the interval $(\frac{11 π}{6}, \frac{19 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.4.1

Replace the variable $x$ with $8$ in the expression.

$f' (8) = 1 + 2 \sin (8)$

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Evaluate $\sin (8)$ .

$f' (8) = 1 + 2 \cdot 0.98935824$

Step 3.4.2.1.2

Multiply $2$ by $0.98935824$ .

$f' (8) = 1 + 1.97871649$

Step 3.4.2.2

Add $1$ and $1.97871649$ .

$f' (8) = 2.97871649$

Step 3.4.2.3

The final answer is $2.97871649$ .

$2.97871649$

Step 3.5

Substitute any number, such as $11$ , from the interval $(\frac{19 π}{6}, \frac{23 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.5.1

Replace the variable $x$ with $11$ in the expression.

$f' (11) = 1 + 2 \sin (11)$

Step 3.5.2

Simplify the result.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

Evaluate $\sin (11)$ .

$f' (11) = 1 + 2 \cdot - 0.9999902$

Step 3.5.2.1.2

Multiply $2$ by $- 0.9999902$ .

$f' (11) = 1 - 1.99998041$

Step 3.5.2.2

Subtract $1.99998041$ from $1$ .

$f' (11) = - 0.99998041$

Step 3.5.2.3

The final answer is $- 0.99998041$ .

$- 0.99998041$

Step 3.6

Substitute any number, such as $14$ , from the interval $(\frac{23 π}{6}, \frac{31 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.6.1

Replace the variable $x$ with $14$ in the expression.

$f' (14) = 1 + 2 \sin (14)$

Step 3.6.2

Simplify the result.

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Step 3.6.2.1

Simplify each term.

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Step 3.6.2.1.1

Evaluate $\sin (14)$ .

$f' (14) = 1 + 2 \cdot 0.99060735$

Step 3.6.2.1.2

Multiply $2$ by $0.99060735$ .

$f' (14) = 1 + 1.98121471$

Step 3.6.2.2

Add $1$ and $1.98121471$ .

$f' (14) = 2.98121471$

Step 3.6.2.3

The final answer is $2.98121471$ .

$2.98121471$

Step 3.7

Substitute any number, such as $17$ , from the interval $(\frac{31 π}{6}, \frac{35 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.7.1

Replace the variable $x$ with $17$ in the expression.

$f' (17) = 1 + 2 \sin (17)$

Step 3.7.2

Simplify the result.

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Step 3.7.2.1

Simplify each term.

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Step 3.7.2.1.1

Evaluate $\sin (17)$ .

$f' (17) = 1 + 2 \cdot - 0.96139749$

Step 3.7.2.1.2

Multiply $2$ by $- 0.96139749$ .

$f' (17) = 1 - 1.92279498$

Step 3.7.2.2

Subtract $1.92279498$ from $1$ .

$f' (17) = - 0.92279498$

Step 3.7.2.3

The final answer is $- 0.92279498$ .

$- 0.92279498$

Step 3.8

Substitute any number, such as $20$ , from the interval $(\frac{35 π}{6}, \frac{43 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.8.1

Replace the variable $x$ with $20$ in the expression.

$f' (20) = 1 + 2 \sin (20)$

Step 3.8.2

Simplify the result.

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Step 3.8.2.1

Simplify each term.

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Step 3.8.2.1.1

Evaluate $\sin (20)$ .

$f' (20) = 1 + 2 \cdot 0.91294525$

Step 3.8.2.1.2

Multiply $2$ by $0.91294525$ .

$f' (20) = 1 + 1.8258905$

Step 3.8.2.2

Add $1$ and $1.8258905$ .

$f' (20) = 2.8258905$

Step 3.8.2.3

The final answer is $2.8258905$ .

$2.8258905$

Step 3.9

Substitute any number, such as $24$ , from the interval $(\frac{43 π}{6}, \frac{47 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.9.1

Replace the variable $x$ with $24$ in the expression.

$f' (24) = 1 + 2 \sin (24)$

Step 3.9.2

Simplify the result.

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Step 3.9.2.1

Simplify each term.

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Step 3.9.2.1.1

Evaluate $\sin (24)$ .

$f' (24) = 1 + 2 \cdot - 0.90557836$

Step 3.9.2.1.2

Multiply $2$ by $- 0.90557836$ .

$f' (24) = 1 - 1.81115672$

Step 3.9.2.2

Subtract $1.81115672$ from $1$ .

$f' (24) = - 0.81115672$

Step 3.9.2.3

The final answer is $- 0.81115672$ .

$- 0.81115672$

Step 3.10

Substitute any number, such as $27$ , from the interval $(\frac{47 π}{6}, \frac{55 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.10.1

Replace the variable $x$ with $27$ in the expression.

$f' (27) = 1 + 2 \sin (27)$

Step 3.10.2

Simplify the result.

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Step 3.10.2.1

Simplify each term.

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Step 3.10.2.1.1

Evaluate $\sin (27)$ .

$f' (27) = 1 + 2 \cdot 0.95637592$

Step 3.10.2.1.2

Multiply $2$ by $0.95637592$ .

$f' (27) = 1 + 1.91275185$

Step 3.10.2.2

Add $1$ and $1.91275185$ .

$f' (27) = 2.91275185$

Step 3.10.2.3

The final answer is $2.91275185$ .

$2.91275185$

Step 3.11

Substitute any number, such as $30$ , from the interval $(\frac{55 π}{6}, \frac{59 π}{6})$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.11.1

Replace the variable $x$ with $30$ in the expression.

$f' (30) = 1 + 2 \sin (30)$

Step 3.11.2

Simplify the result.

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Step 3.11.2.1

Simplify each term.

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Step 3.11.2.1.1

The exact value of $\sin (30)$ is $\frac{1}{2}$ .

$f' (30) = 1 + 2 (\frac{1}{2})$

Step 3.11.2.1.2

Cancel the common factor of $2$ .

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Step 3.11.2.1.2.1

Cancel the common factor.

$f' (30) = 1 + 2 (\frac{1}{2})$

Step 3.11.2.1.2.2

Rewrite the expression.

$f' (30) = 1 + 1$

Step 3.11.2.2

Add $1$ and $1$ .

$f' (30) = 2$

Step 3.11.2.3

The final answer is $2$ .

$2$

Step 3.12

Substitute any number, such as $33$ , from the interval $(\frac{59 π}{6}, \infty)$ in the first derivative $1 + 2 \sin (x)$ to check if the result is negative or positive.

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Step 3.12.1

Replace the variable $x$ with $33$ in the expression.

$f' (33) = 1 + 2 \sin (33)$

Step 3.12.2

Simplify the result.

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Step 3.12.2.1

Simplify each term.

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Step 3.12.2.1.1

Evaluate $\sin (33)$ .

$f' (33) = 1 + 2 \cdot 0.99991186$

Step 3.12.2.1.2

Multiply $2$ by $0.99991186$ .

$f' (33) = 1 + 1.99982372$

Step 3.12.2.2

Add $1$ and $1.99982372$ .

$f' (33) = 2.99982372$

Step 3.12.2.3

The final answer is $2.99982372$ .

$2.99982372$

Step 3.13

Since the first derivative changed signs from positive to negative around $x = \frac{7 π}{6}$ , then $x = \frac{7 π}{6}$ is a local maximum.

$x = \frac{7 π}{6}$ is a local maximum

Step 3.14

Since the first derivative changed signs from negative to positive around $x = \frac{11 π}{6}$ , then $x = \frac{11 π}{6}$ is a local minimum.

$x = \frac{11 π}{6}$ is a local minimum

Step 3.15

Since the first derivative changed signs from positive to negative around $x = \frac{19 π}{6}$ , then $x = \frac{19 π}{6}$ is a local maximum.

$x = \frac{19 π}{6}$ is a local maximum

Step 3.16

Since the first derivative changed signs from negative to positive around $x = \frac{23 π}{6}$ , then $x = \frac{23 π}{6}$ is a local minimum.

$x = \frac{23 π}{6}$ is a local minimum

Step 3.17

Since the first derivative changed signs from positive to negative around $x = \frac{31 π}{6}$ , then $x = \frac{31 π}{6}$ is a local maximum.

$x = \frac{31 π}{6}$ is a local maximum

Step 3.18

Since the first derivative changed signs from negative to positive around $x = \frac{35 π}{6}$ , then $x = \frac{35 π}{6}$ is a local minimum.

$x = \frac{35 π}{6}$ is a local minimum

Step 3.19

Since the first derivative changed signs from positive to negative around $x = \frac{43 π}{6}$ , then $x = \frac{43 π}{6}$ is a local maximum.

$x = \frac{43 π}{6}$ is a local maximum

Step 3.20

Since the first derivative changed signs from negative to positive around $x = \frac{47 π}{6}$ , then $x = \frac{47 π}{6}$ is a local minimum.

$x = \frac{47 π}{6}$ is a local minimum

Step 3.21

Since the first derivative did not change signs around $x = \frac{55 π}{6}$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 3.22

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

$x = \frac{7 π}{6}$ is a local maximum

$x = \frac{11 π}{6}$ is a local minimum

$x = \frac{19 π}{6}$ is a local maximum

$x = \frac{23 π}{6}$ is a local minimum

$x = \frac{31 π}{6}$ is a local maximum

$x = \frac{35 π}{6}$ is a local minimum

$x = \frac{43 π}{6}$ is a local maximum

$x = \frac{47 π}{6}$ is a local minimum

$x = \frac{7 π}{6}$ is a local maximum

$x = \frac{11 π}{6}$ is a local minimum

$x = \frac{19 π}{6}$ is a local maximum

$x = \frac{23 π}{6}$ is a local minimum

$x = \frac{31 π}{6}$ is a local maximum

$x = \frac{35 π}{6}$ is a local minimum

$x = \frac{43 π}{6}$ is a local maximum

$x = \frac{47 π}{6}$ is a local minimum

Step 4

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

Absolute Maximum: $(\frac{7 π}{6}, \frac{7 π}{6} + \sqrt{3})$

Absolute Minimum: $(\frac{11 π}{6}, \frac{11 π}{6} - \sqrt{3})$

Step 5