Calculus Examples

$f (x) = 4 x - 4 \sin (x)$

Step 1

Find the derivative.

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Step 1.1

By the Sum Rule, the derivative of $4 x - 4 \sin (x)$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 4 \sin (x)]$ .

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 4 \sin (x)]$

Step 1.2

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 4 \sin (x)]$

Step 1.2.2

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

$4 \cdot 1 + \frac{d}{d x} [- 4 \sin (x)]$

Step 1.2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 4 \sin (x)]$

Step 1.3

Evaluate $\frac{d}{d x} [- 4 \sin (x)]$ .

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Step 1.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 \sin (x)$ with respect to $x$ is $- 4 \frac{d}{d x} [\sin (x)]$ .

$4 - 4 \frac{d}{d x} [\sin (x)]$

Step 1.3.2

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

$4 - 4 \cos (x)$

Step 2

Set the derivative equal to $0$ then solve the equation $4 - 4 \cos (x) = 0$ .

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Step 2.1

Subtract $4$ from both sides of the equation.

$- 4 \cos (x) = - 4$

Step 2.2

Divide each term in $- 4 \cos (x) = - 4$ by $- 4$ and simplify.

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Step 2.2.1

Divide each term in $- 4 \cos (x) = - 4$ by $- 4$ .

$\frac{- 4 \cos (x)}{- 4} = \frac{- 4}{- 4}$

Step 2.2.2

Simplify the left side.

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Step 2.2.2.1

Cancel the common factor of $- 4$ .

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Step 2.2.2.1.1

Cancel the common factor.

$\frac{- 4 \cos (x)}{- 4} = \frac{- 4}{- 4}$

Step 2.2.2.1.2

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

$\cos (x) = \frac{- 4}{- 4}$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

Divide $- 4$ by $- 4$ .

$\cos (x) = 1$

Step 2.3

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

$x = \arccos (1)$

Step 2.4

Simplify the right side.

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Step 2.4.1

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

$x = 0$

Step 2.5

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.

$x = 2 π - 0$

Step 2.6

Subtract $0$ from $2 π$ .

$x = 2 π$

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.

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

Step 2.9

Consolidate the answers.

$x = 2 π n$ , for any integer $n$

Step 3

Solve the original function $f (x) = 4 x - 4 \sin (x)$ at $x = 2 π$ .

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Step 3.1

$f (2 π) = 4 (2 π) - 4 \sin (2 π)$

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

Multiply $2$ by $4$ .

$f (2 π) = 8 π - 4 \sin (2 π)$

Step 3.2.1.2

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

$f (2 π) = 8 π - 4 \sin (0)$

Step 3.2.1.3

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

$f (2 π) = 8 π - 4 \cdot 0$

Step 3.2.1.4

Multiply $- 4$ by $0$ .

$f (2 π) = 8 π + 0$

Step 3.2.2

Add $8 π$ and $0$ .

$f (2 π) = 8 π$

Step 3.2.3

The final answer is $8 π$ .

$8 π$

Step 4

The horizontal tangent line on function $f (x) = 4 x - 4 \sin (x)$ is $y = x = 2 π n, y = 8 π$ .

$y = x = 2 π n, y = 8 π$

Step 5