Calculus Examples

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

Step 1

Write $x + \sin (2 x) + 1$ as a function.

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

Step 2

Find the $x$ values where the second derivative is equal to $0$ .

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

Differentiate.

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Step 2.1.1.1.1

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

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

Step 2.1.1.1.2

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

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

Step 2.1.1.2

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

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Step 2.1.1.2.1

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

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Step 2.1.1.2.1.1

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

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

Step 2.1.1.2.1.2

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

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

Step 2.1.1.2.1.3

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

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

Step 2.1.1.2.2

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

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

Step 2.1.1.2.3

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

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

Step 2.1.1.2.4

Multiply $2$ by $1$ .

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

Step 2.1.1.2.5

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

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

Step 2.1.1.3

Differentiate using the Constant Rule.

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Step 2.1.1.3.1

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

$1 + 2 \cos (2 x) + 0$

Step 2.1.1.3.2

Add $1 + 2 \cos (2 x)$ and $0$ .

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

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

Differentiate.

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Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

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

$0 + \frac{d}{d x} [2 \cos (2 x)]$

Step 2.1.2.2

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

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Step 2.1.2.2.1

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

$0 + 2 \frac{d}{d x} [\cos (2 x)]$

Step 2.1.2.2.2

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

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Step 2.1.2.2.2.1

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

$0 + 2 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x])$

Step 2.1.2.2.2.2

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

$0 + 2 (- \sin (u) \frac{d}{d x} [2 x])$

Step 2.1.2.2.2.3

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

$0 + 2 (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 2.1.2.2.3

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

$0 + 2 (- \sin (2 x) (2 \frac{d}{d x} [x]))$

Step 2.1.2.2.4

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

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

Step 2.1.2.2.5

Multiply $2$ by $1$ .

$0 + 2 (- \sin (2 x) \cdot 2)$

Step 2.1.2.2.6

Multiply $2$ by $- 1$ .

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

Step 2.1.2.2.7

Multiply $- 2$ by $2$ .

$0 - 4 \sin (2 x)$

Step 2.1.2.3

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

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

Step 2.1.3

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

$- 4 \sin (2 x)$

Step 2.2

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

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Step 2.2.1

Set the second derivative equal to $0$ .

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

Step 2.2.2

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

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Step 2.2.2.1

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

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

Step 2.2.2.2

Simplify the left side.

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Step 2.2.2.2.1

Cancel the common factor of $- 4$ .

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Step 2.2.2.2.1.1

Cancel the common factor.

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

Step 2.2.2.2.1.2

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

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

Step 2.2.2.3

Simplify the right side.

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Step 2.2.2.3.1

Divide $0$ by $- 4$ .

$\sin (2 x) = 0$

Step 2.2.3

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

$2 x = \arcsin (0)$

Step 2.2.4

Simplify the right side.

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Step 2.2.4.1

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

$2 x = 0$

Step 2.2.5

Divide each term in $2 x = 0$ by $2$ and simplify.

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Step 2.2.5.1

Divide each term in $2 x = 0$ by $2$ .

$\frac{2 x}{2} = \frac{0}{2}$

Step 2.2.5.2

Simplify the left side.

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Step 2.2.5.2.1

Cancel the common factor of $2$ .

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Step 2.2.5.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 2.2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{2}$

Step 2.2.5.3

Simplify the right side.

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Step 2.2.5.3.1

Divide $0$ by $2$ .

$x = 0$

Step 2.2.6

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.

$2 x = π - 0$

Step 2.2.7

Solve for $x$ .

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Step 2.2.7.1

Simplify.

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Step 2.2.7.1.1

Multiply $- 1$ by $0$ .

$2 x = π + 0$

Step 2.2.7.1.2

Add $π$ and $0$ .

$2 x = π$

Step 2.2.7.2

Divide each term in $2 x = π$ by $2$ and simplify.

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Step 2.2.7.2.1

Divide each term in $2 x = π$ by $2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 2.2.7.2.2

Simplify the left side.

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Step 2.2.7.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.7.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 2.2.7.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{2}$

Step 2.2.8

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

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Step 2.2.8.1

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

$\frac{2 π}{| b |}$

Step 2.2.8.2

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

$\frac{2 π}{| 2 |}$

Step 2.2.8.3

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

$\frac{2 π}{2}$

Step 2.2.8.4

Cancel the common factor of $2$ .

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Step 2.2.8.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 2.2.8.4.2

Divide $π$ by $1$ .

$π$

Step 2.2.9

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

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

Step 2.2.10

Consolidate the answers.

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

Step 3

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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, \infty)$

Step 5

Substitute any number from the interval $(- \infty, \infty)$ into the second derivative and evaluate to determine the concavity.

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Step 5.1

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

$f'' (0) = - 4 \sin (2 (0))$

Step 5.2

Simplify the result.

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Step 5.2.1

Multiply $2$ by $0$ .

$f'' (0) = - 4 \sin (0)$

Step 5.2.2

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

$f'' (0) = - 4 \cdot 0$

Step 5.2.3

Multiply $- 4$ by $0$ .

$f'' (0) = 0$

Step 5.2.4

The final answer is $0$ .

$0$

Step 5.3

The graph is concave up on the interval $(- \infty, \infty)$ because $f'' (0)$ is positive.

Concave up on $(- \infty, \infty)$ since $f'' (x)$ is positive

Step 6