Calculus Examples

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

Step 1

Find the derivative.

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Step 1.1

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

$1 + 2 \cos (x)$

Step 2

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

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Step 2.1

Subtract $1$ from both sides of the equation.

$2 \cos (x) = - 1$

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.1.2

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

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

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Move the negative in front of the fraction.

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

Step 2.3

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

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

Step 2.4

Simplify the right side.

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Step 2.4.1

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

$x = \frac{2 π}{3}$

Step 2.5

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

$x = 2 π - \frac{2 π}{3}$

Step 2.6

Simplify $2 π - \frac{2 π}{3}$ .

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Step 2.6.1

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

$x = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Step 2.6.2

Combine fractions.

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Step 2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

$x = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Step 2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 2.6.3

Simplify the numerator.

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Step 2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 2.6.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 2.7

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

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Step 2.7.1

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

$\frac{2 π}{| b |}$

Step 2.7.2

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

$\frac{2 π}{| 1 |}$

Step 2.7.3

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

$\frac{2 π}{1}$

Step 2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.8

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

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

Step 3

Solve the original function $f (x) = x + 2 \sin (x)$ at $x = \frac{2 π}{3}$ .

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Step 3.1

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

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

Step 3.2

Simplify the result.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

Cancel the common factor of $2$ .

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Step 3.2.1.3.1

Cancel the common factor.

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

Step 3.2.1.3.2

Rewrite the expression.

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

Step 3.2.2

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

$\frac{2 π}{3} + \sqrt{3}$

Step 4

Solve the original function $f (x) = x + 2 \sin (x)$ at $x = \frac{4 π}{3}$ .

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Step 4.1

Replace the variable $x$ with $\frac{4 π}{3}$ in the expression.

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

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

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 third quadrant.

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

Step 4.2.1.2

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

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

Step 4.2.1.3

Cancel the common factor of $2$ .

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Step 4.2.1.3.1

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

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

Step 4.2.1.3.2

Cancel the common factor.

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

Step 4.2.1.3.3

Rewrite the expression.

$f (\frac{4 π}{3}) = \frac{4 π}{3} - \sqrt{3}$

Step 4.2.2

The final answer is $\frac{4 π}{3} - \sqrt{3}$ .

$\frac{4 π}{3} - \sqrt{3}$

Step 5

The horizontal tangent lines on function $f (x) = x + 2 \sin (x)$ are $y = \frac{2 π}{3} + \sqrt{3}, y = \frac{4 π}{3} - \sqrt{3}$ .

$y = \frac{2 π}{3} + \sqrt{3}, y = \frac{4 π}{3} - \sqrt{3}$

Step 6