Calculus Examples

$y = x + \cos (2 x)$

Step 1

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

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

Step 2

Find the second derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

Differentiate.

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Step 2.1.1.1

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

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

Step 2.1.1.2

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

$f' (x) = 1 + \frac{d}{d x} (\cos (2 x))$

Step 2.1.2

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

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Step 2.1.2.1

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 2.1.2.1.1

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

$f' (x) = 1 + \frac{d}{d u} (\cos (u)) \frac{d}{d x} (2 x)$

Step 2.1.2.1.2

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

$f' (x) = 1 - \sin (u) \frac{d}{d x} (2 x)$

Step 2.1.2.1.3

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

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

Step 2.1.2.2

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

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

Step 2.1.2.3

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

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

Step 2.1.2.4

Multiply $2$ by $1$ .

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

Step 2.1.2.5

Multiply $2$ by $- 1$ .

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

Step 2.2

Find the second derivative.

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Step 2.2.1

Differentiate.

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Step 2.2.1.1

By the Sum Rule, the derivative of $1 - 2 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 2 \sin (2 x)]$ .

$f'' (x) = \frac{d}{d x} (1) + \frac{d}{d x} (- 2 \sin (2 x))$

Step 2.2.1.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 0 + \frac{d}{d x} (- 2 \sin (2 x))$

Step 2.2.2

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

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Step 2.2.2.1

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

$f'' (x) = 0 - 2 \frac{d}{d x} \sin (2 x)$

Step 2.2.2.2

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 2.2.2.2.1

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

$f'' (x) = 0 - 2 (\frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x))$

Step 2.2.2.2.2

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

$f'' (x) = 0 - 2 (\cos (u) \frac{d}{d x} (2 x))$

Step 2.2.2.2.3

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

$f'' (x) = 0 - 2 (\cos (2 x) \frac{d}{d x} (2 x))$

Step 2.2.2.3

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

$f'' (x) = 0 - 2 (\cos (2 x) (2 \frac{d}{d x} (x)))$

Step 2.2.2.4

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

$f'' (x) = 0 - 2 (\cos (2 x) (2 \cdot 1))$

Step 2.2.2.5

Multiply $2$ by $1$ .

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

Step 2.2.2.6

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

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

Step 2.2.2.7

Multiply $2$ by $- 2$ .

$f'' (x) = 0 - 4 \cos (2 x)$

Step 2.2.3

Subtract $4 \cos (2 x)$ from $0$ .

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

Step 2.3

The second derivative of $f (x)$ with respect to $x$ is $- 4 \cos (2 x)$ .

$- 4 \cos (2 x)$

Step 3

Set the second derivative equal to $0$ then solve the equation $- 4 \cos (2 x) = 0$ .

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Step 3.1

Set the second derivative equal to $0$ .

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

Step 3.2

Divide each term in $- 4 \cos (2 x) = 0$ by $- 4$ and simplify.

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Step 3.2.1

Divide each term in $- 4 \cos (2 x) = 0$ by $- 4$ .

$\frac{- 4 \cos (2 x)}{- 4} = \frac{0}{- 4}$

Step 3.2.2

Simplify the left side.

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Step 3.2.2.1

Cancel the common factor of $- 4$ .

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Step 3.2.2.1.1

Cancel the common factor.

$\frac{- 4 \cos (2 x)}{- 4} = \frac{0}{- 4}$

Step 3.2.2.1.2

Divide $\cos (2 x)$ by $1$ .

$\cos (2 x) = \frac{0}{- 4}$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

Divide $0$ by $- 4$ .

$\cos (2 x) = 0$

Step 3.3

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

$2 x = \arccos (0)$

Step 3.4

Simplify the right side.

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Step 3.4.1

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

$2 x = \frac{π}{2}$

Step 3.5

Divide each term in $2 x = \frac{π}{2}$ by $2$ and simplify.

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Step 3.5.1

Divide each term in $2 x = \frac{π}{2}$ by $2$ .

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

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Cancel the common factor of $2$ .

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Step 3.5.2.1.1

Cancel the common factor.

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

Step 3.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{2}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{2}$

Step 3.5.3.2

Multiply $\frac{π}{2} \cdot \frac{1}{2}$ .

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Step 3.5.3.2.1

Multiply $\frac{π}{2}$ by $\frac{1}{2}$ .

$x = \frac{π}{2 \cdot 2}$

Step 3.5.3.2.2

Multiply $2$ by $2$ .

$x = \frac{π}{4}$

Step 3.6

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.

$2 x = 2 π - \frac{π}{2}$

Step 3.7

Solve for $x$ .

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Step 3.7.1

Simplify.

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Step 3.7.1.1

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

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

Step 3.7.1.2

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 3.7.1.3

Combine the numerators over the common denominator.

$2 x = \frac{2 π \cdot 2 - π}{2}$

Step 3.7.1.4

Multiply $2$ by $2$ .

$2 x = \frac{4 π - π}{2}$

Step 3.7.1.5

Subtract $π$ from $4 π$ .

$2 x = \frac{3 π}{2}$

Step 3.7.2

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ and simplify.

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Step 3.7.2.1

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ .

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

Step 3.7.2.2

Simplify the left side.

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Step 3.7.2.2.1

Cancel the common factor of $2$ .

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Step 3.7.2.2.1.1

Cancel the common factor.

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

Step 3.7.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{2}}{2}$

Step 3.7.2.3

Simplify the right side.

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Step 3.7.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{2} \cdot \frac{1}{2}$

Step 3.7.2.3.2

Multiply $\frac{3 π}{2} \cdot \frac{1}{2}$ .

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Step 3.7.2.3.2.1

Multiply $\frac{3 π}{2}$ by $\frac{1}{2}$ .

$x = \frac{3 π}{2 \cdot 2}$

Step 3.7.2.3.2.2

Multiply $2$ by $2$ .

$x = \frac{3 π}{4}$

Step 3.8

Find the period of $\cos (2 x)$ .

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Step 3.8.1

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

$\frac{2 π}{| b |}$

Step 3.8.2

Replace $b$ with $2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 3.8.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{2 π}{2}$

Step 3.8.4

Cancel the common factor of $2$ .

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Step 3.8.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 3.8.4.2

Divide $π$ by $1$ .

$π$

Step 3.9

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

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

Step 3.10

Consolidate the answers.

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

Step 4

Find the points where the second derivative is $0$ .

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Step 4.1

Substitute $\frac{π}{4}$ in $f (x) = x + \cos (2 x)$ to find the value of $y$ .

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Step 4.1.1

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

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

Step 4.1.2

Simplify the result.

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Step 4.1.2.1

Simplify each term.

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Step 4.1.2.1.1

Cancel the common factor of $2$ .

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Step 4.1.2.1.1.1

Factor $2$ out of $4$ .

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

Step 4.1.2.1.1.2

Cancel the common factor.

$f (\frac{π}{4}) = \frac{π}{4} + \cos (2 (\frac{π}{2 \cdot 2}))$

Step 4.1.2.1.1.3

Rewrite the expression.

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

Step 4.1.2.1.2

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

$f (\frac{π}{4}) = \frac{π}{4} + 0$

Step 4.1.2.2

Add $\frac{π}{4}$ and $0$ .

$f (\frac{π}{4}) = \frac{π}{4}$

Step 4.1.2.3

The final answer is $\frac{π}{4}$ .

$\frac{π}{4}$

Step 4.2

The point found by substituting $\frac{π}{4}$ in $f (x) = x + \cos (2 x)$ is $(\frac{π}{4}, \frac{π}{4})$ . This point can be an inflection point.

$(\frac{π}{4}, \frac{π}{4})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, \frac{π}{4}) \cup (\frac{π}{4}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{π}{4})$ into the second derivative to determine if it is increasing or decreasing.

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Step 6.1

Replace the variable $x$ with $0.68539816$ in the expression.

$f'' (0.68539816) = - 4 \cos (2 (0.68539816))$

Step 6.2

Simplify the result.

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Step 6.2.1

Multiply $2$ by $0.68539816$ .

$f'' (0.68539816) = - 4 \cos (1.37079632)$

Step 6.2.2

The final answer is $- 4 \cos (1.37079632)$ .

$- 4 \cos (1.37079632)$

Step 6.3

At $0.68539816$ , the second derivative is $- 0.79467732$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{π}{4})$

Decreasing on $(- \infty, \frac{π}{4})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{π}{4}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $0.88539816$ in the expression.

$f'' (0.88539816) = - 4 \cos (2 (0.88539816))$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $0.88539816$ .

$f'' (0.88539816) = - 4 \cos (1.77079632)$

Step 7.2.2

The final answer is $- 4 \cos (1.77079632)$ .

$- 4 \cos (1.77079632)$

Step 7.3

At $0.88539816$ , the second derivative is $0.79467732$ . Since this is positive, the second derivative is increasing on the interval $(\frac{π}{4}, \infty)$ .

Increasing on $(\frac{π}{4}, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(\frac{π}{4}, \frac{π}{4})$ .

$(\frac{π}{4}, \frac{π}{4})$

Step 9