Calculus Examples

$6 x^{2} + 6 \sin (2 x)$

Step 1

Write $6 x^{2} + 6 \sin (2 x)$ as a function.

$f (x) = 6 x^{2} + 6 \sin (2 x)$

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

By the Sum Rule, the derivative of $6 x^{2} + 6 \sin (2 x)$ with respect to $x$ is $\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [6 \sin (2 x)]$ .

$\frac{d}{d x} [6 x^{2}] + \frac{d}{d x} [6 \sin (2 x)]$

Step 2.1.1.2

Evaluate $\frac{d}{d x} [6 x^{2}]$ .

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Step 2.1.1.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{2}$ with respect to $x$ is $6 \frac{d}{d x} [x^{2}]$ .

$6 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [6 \sin (2 x)]$

Step 2.1.1.2.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$6 (2 x) + \frac{d}{d x} [6 \sin (2 x)]$

Step 2.1.1.2.3

Multiply $2$ by $6$ .

$12 x + \frac{d}{d x} [6 \sin (2 x)]$

Step 2.1.1.3

Evaluate $\frac{d}{d x} [6 \sin (2 x)]$ .

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Step 2.1.1.3.1

Since $6$ is constant with respect to $x$ , the derivative of $6 \sin (2 x)$ with respect to $x$ is $6 \frac{d}{d x} [\sin (2 x)]$ .

$12 x + 6 \frac{d}{d x} [\sin (2 x)]$

Step 2.1.1.3.2

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 2.1.1.3.2.1

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

$12 x + 6 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x])$

Step 2.1.1.3.2.2

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

$12 x + 6 (\cos (u) \frac{d}{d x} [2 x])$

Step 2.1.1.3.2.3

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

$12 x + 6 (\cos (2 x) \frac{d}{d x} [2 x])$

Step 2.1.1.3.3

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

$12 x + 6 (\cos (2 x) (2 \frac{d}{d x} [x]))$

Step 2.1.1.3.4

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

$12 x + 6 (\cos (2 x) (2 \cdot 1))$

Step 2.1.1.3.5

Multiply $2$ by $1$ .

$12 x + 6 (\cos (2 x) \cdot 2)$

Step 2.1.1.3.6

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

$12 x + 6 (2 \cdot \cos (2 x))$

Step 2.1.1.3.7

Multiply $2$ by $6$ .

$f' (x) = 12 x + 12 \cos (2 x)$

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

By the Sum Rule, the derivative of $12 x + 12 \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [12 x] + \frac{d}{d x} [12 \cos (2 x)]$ .

$\frac{d}{d x} [12 x] + \frac{d}{d x} [12 \cos (2 x)]$

Step 2.1.2.2

Evaluate $\frac{d}{d x} [12 x]$ .

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Step 2.1.2.2.1

Since $12$ is constant with respect to $x$ , the derivative of $12 x$ with respect to $x$ is $12 \frac{d}{d x} [x]$ .

$12 \frac{d}{d x} [x] + \frac{d}{d x} [12 \cos (2 x)]$

Step 2.1.2.2.2

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

$12 \cdot 1 + \frac{d}{d x} [12 \cos (2 x)]$

Step 2.1.2.2.3

Multiply $12$ by $1$ .

$12 + \frac{d}{d x} [12 \cos (2 x)]$

Step 2.1.2.3

Evaluate $\frac{d}{d x} [12 \cos (2 x)]$ .

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Step 2.1.2.3.1

Since $12$ is constant with respect to $x$ , the derivative of $12 \cos (2 x)$ with respect to $x$ is $12 \frac{d}{d x} [\cos (2 x)]$ .

$12 + 12 \frac{d}{d x} [\cos (2 x)]$

Step 2.1.2.3.2

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 2.1.2.3.2.1

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

$12 + 12 (\frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x])$

Step 2.1.2.3.2.2

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

$12 + 12 (- \sin (u) \frac{d}{d x} [2 x])$

Step 2.1.2.3.2.3

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

$12 + 12 (- \sin (2 x) \frac{d}{d x} [2 x])$

Step 2.1.2.3.3

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

$12 + 12 (- \sin (2 x) (2 \frac{d}{d x} [x]))$

Step 2.1.2.3.4

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

$12 + 12 (- \sin (2 x) (2 \cdot 1))$

Step 2.1.2.3.5

Multiply $2$ by $1$ .

$12 + 12 (- \sin (2 x) \cdot 2)$

Step 2.1.2.3.6

Multiply $2$ by $- 1$ .

$12 + 12 (- 2 \sin (2 x))$

Step 2.1.2.3.7

Multiply $- 2$ by $12$ .

$f'' (x) = 12 - 24 \sin (2 x)$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $12 - 24 \sin (2 x)$ .

$12 - 24 \sin (2 x)$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $12 - 24 \sin (2 x) = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$12 - 24 \sin (2 x) = 0$

Step 2.2.2

Subtract $12$ from both sides of the equation.

$- 24 \sin (2 x) = - 12$

Step 2.2.3

Divide each term in $- 24 \sin (2 x) = - 12$ by $- 24$ and simplify.

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Step 2.2.3.1

Divide each term in $- 24 \sin (2 x) = - 12$ by $- 24$ .

$\frac{- 24 \sin (2 x)}{- 24} = \frac{- 12}{- 24}$

Step 2.2.3.2

Simplify the left side.

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Step 2.2.3.2.1

Cancel the common factor of $- 24$ .

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Step 2.2.3.2.1.1

Cancel the common factor.

$\frac{- 24 \sin (2 x)}{- 24} = \frac{- 12}{- 24}$

Step 2.2.3.2.1.2

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

$\sin (2 x) = \frac{- 12}{- 24}$

Step 2.2.3.3

Simplify the right side.

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Step 2.2.3.3.1

Cancel the common factor of $- 12$ and $- 24$ .

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Step 2.2.3.3.1.1

Factor $- 12$ out of $- 12$ .

$\sin (2 x) = \frac{- 12 (1)}{- 24}$

Step 2.2.3.3.1.2

Cancel the common factors.

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Step 2.2.3.3.1.2.1

Factor $- 12$ out of $- 24$ .

$\sin (2 x) = \frac{- 12 \cdot 1}{- 12 \cdot 2}$

Step 2.2.3.3.1.2.2

Cancel the common factor.

$\sin (2 x) = \frac{- 12 \cdot 1}{- 12 \cdot 2}$

Step 2.2.3.3.1.2.3

Rewrite the expression.

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

Step 2.2.4

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

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

Step 2.2.5

Simplify the right side.

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Step 2.2.5.1

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

$2 x = \frac{π}{6}$

Step 2.2.6

Divide each term in $2 x = \frac{π}{6}$ by $2$ and simplify.

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Step 2.2.6.1

Divide each term in $2 x = \frac{π}{6}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{6}}{2}$

Step 2.2.6.2

Simplify the left side.

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Step 2.2.6.2.1

Cancel the common factor of $2$ .

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Step 2.2.6.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{6}}{2}$

Step 2.2.6.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{6}}{2}$

Step 2.2.6.3

Simplify the right side.

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Step 2.2.6.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{6} \cdot \frac{1}{2}$

Step 2.2.6.3.2

Multiply $\frac{π}{6} \cdot \frac{1}{2}$ .

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Step 2.2.6.3.2.1

Multiply $\frac{π}{6}$ by $\frac{1}{2}$ .

$x = \frac{π}{6 \cdot 2}$

Step 2.2.6.3.2.2

Multiply $6$ by $2$ .

$x = \frac{π}{12}$

Step 2.2.7

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.

$2 x = π - \frac{π}{6}$

Step 2.2.8

Solve for $x$ .

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Step 2.2.8.1

Simplify.

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Step 2.2.8.1.1

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

$2 x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 2.2.8.1.2

Combine $π$ and $\frac{6}{6}$ .

$2 x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 2.2.8.1.3

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 6 - π}{6}$

Step 2.2.8.1.4

Subtract $π$ from $π \cdot 6$ .

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Step 2.2.8.1.4.1

Reorder $π$ and $6$ .

$2 x = \frac{6 \cdot π - π}{6}$

Step 2.2.8.1.4.2

Subtract $π$ from $6 \cdot π$ .

$2 x = \frac{5 \cdot π}{6}$

Step 2.2.8.2

Divide each term in $2 x = \frac{5 \cdot π}{6}$ by $2$ and simplify.

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Step 2.2.8.2.1

Divide each term in $2 x = \frac{5 \cdot π}{6}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{6}}{2}$

Step 2.2.8.2.2

Simplify the left side.

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Step 2.2.8.2.2.1

Cancel the common factor of $2$ .

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Step 2.2.8.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{6}}{2}$

Step 2.2.8.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 \cdot π}{6}}{2}$

Step 2.2.8.2.3

Simplify the right side.

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Step 2.2.8.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 \cdot π}{6} \cdot \frac{1}{2}$

Step 2.2.8.2.3.2

Multiply $\frac{5 π}{6} \cdot \frac{1}{2}$ .

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Step 2.2.8.2.3.2.1

Multiply $\frac{5 π}{6}$ by $\frac{1}{2}$ .

$x = \frac{5 π}{6 \cdot 2}$

Step 2.2.8.2.3.2.2

Multiply $6$ by $2$ .

$x = \frac{5 π}{12}$

Step 2.2.9

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

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Step 2.2.9.1

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

$\frac{2 π}{| b |}$

Step 2.2.9.2

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

$\frac{2 π}{| 2 |}$

Step 2.2.9.3

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

$\frac{2 π}{2}$

Step 2.2.9.4

Cancel the common factor of $2$ .

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Step 2.2.9.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 2.2.9.4.2

Divide $π$ by $1$ .

$π$

Step 2.2.10

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

$x = \frac{π}{12} + π n, \frac{5 π}{12} + π n$ , 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.

$f'' (0) = 12 - 24 \sin (2 (0))$

Step 5.2

Simplify the result.

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $2$ by $0$ .

$f'' (0) = 12 - 24 \sin (0)$

Step 5.2.1.2

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

$f'' (0) = 12 - 24 \cdot 0$

Step 5.2.1.3

Multiply $- 24$ by $0$ .

$f'' (0) = 12 + 0$

Step 5.2.2

Add $12$ and $0$ .

$f'' (0) = 12$

Step 5.2.3

The final answer is $12$ .

$12$

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