Calculus Examples

$f (x) = \sin (2 x)$ , $[0, 2 π]$

Step 1

If $f$ is continuous on the interval $[a, b]$ and differentiable on $(a, b)$ , then at least one real number $c$ exists in the interval $(a, b)$ such that $f' (c) = \frac{f (b) - f a}{b - a}$ . The mean value theorem expresses the relationship between the slope of the tangent to the curve at $x = c$ and the slope of the line through the points $(a, f (a))$ and $(b, f (b))$ .

If $f (x)$ is continuous on $[a, b]$

and if $f (x)$ differentiable on $(a, b)$ ,

then there exists at least one point, $c$ in $[a, b]$ : $f' (c) = \frac{f (b) - f a}{b - a}$ .

Step 2

Check if $f (x) = \sin (2 x)$ is continuous.

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Step 2.1

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 2.2

$f (x)$ is continuous on $[0, 2 π]$ .

The function is continuous.

Step 3

Find the derivative.

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Step 3.1

Find the first derivative.

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Step 3.1.1

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 3.1.1.1

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

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

Step 3.1.1.2

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

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

Step 3.1.1.3

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

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

Step 3.1.2

Differentiate.

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Step 3.1.2.1

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

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

Step 3.1.2.2

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

$\cos (2 x) (2 \cdot 1)$

Step 3.1.2.3

Simplify the expression.

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Step 3.1.2.3.1

Multiply $2$ by $1$ .

$\cos (2 x) \cdot 2$

Step 3.1.2.3.2

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

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

Step 3.2

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

$2 \cos (2 x)$

Step 4

Find if the derivative is continuous on $(0, 2 π)$ .

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Step 4.1

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 4.2

$f' (x)$ is continuous on $(0, 2 π)$ .

The function is continuous.

Step 5

The function is differentiable on $(0, 2 π)$ because the derivative is continuous on $(0, 2 π)$ .

The function is differentiable.

Step 6

$f (x)$ satisfies the two conditions for the mean value theorem. It is continuous on $[0, 2 π]$ and differentiable on $(0, 2 π)$ .

$f (x)$ is continuous on $[0, 2 π]$ and differentiable on $(0, 2 π)$ .

Step 7

Evaluate $f (a)$ from the interval $[0, 2 π]$ .

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Step 7.1

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

$f (0) = \sin (2 (0))$

Step 7.2

Simplify the result.

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Step 7.2.1

Multiply $2$ by $0$ .

$f (0) = \sin (0)$

Step 7.2.2

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

$f (0) = 0$

Step 7.2.3

The final answer is $0$ .

$0$

Step 8

Solve $2 \cos (2 x) = \frac{- (f (b) + f (a))}{- (b + a)}$ for $x$ . $2 \cos (2 x) = \frac{- (f (2 π) + f (0))}{- (2 π + 0)}$ .

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Step 8.1

Simplify.

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Step 8.1.1

Multiply $- 1$ by $0$ .

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

Step 8.1.2

Multiply $- 1$ by $0$ .

$2 \cos (2 x) = \frac{0 + 0}{2 π + 0}$

Step 8.1.3

Reduce the expression $\frac{0 + 0}{2 π + 0}$ by cancelling the common factors.

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Step 8.1.3.1

Factor $2$ out of $0$ .

$2 \cos (2 x) = \frac{2 (0) + 0}{2 π + 0}$

Step 8.1.3.2

Factor $2$ out of $0$ .

$2 \cos (2 x) = \frac{2 \cdot 0 + 2 \cdot 0}{2 π + 0}$

Step 8.1.3.3

Factor $2$ out of $2 \cdot 0 + 2 \cdot 0$ .

$2 \cos (2 x) = \frac{2 \cdot (0 + 0)}{2 π + 0}$

Step 8.1.3.4

Factor $2$ out of $2 π$ .

$2 \cos (2 x) = \frac{2 \cdot (0 + 0)}{2 (π) + 0}$

Step 8.1.3.5

Factor $2$ out of $0$ .

$2 \cos (2 x) = \frac{2 \cdot (0 + 0)}{2 (π) + 2 (0)}$

Step 8.1.3.6

Factor $2$ out of $2 (π) + 2 (0)$ .

$2 \cos (2 x) = \frac{2 \cdot (0 + 0)}{2 (π + 0)}$

Step 8.1.3.7

Cancel the common factor.

$2 \cos (2 x) = \frac{2 \cdot (0 + 0)}{2 (π + 0)}$

Step 8.1.3.8

Rewrite the expression.

$2 \cos (2 x) = \frac{0 + 0}{π + 0}$

Step 8.1.4

Add $0$ and $0$ .

$2 \cos (2 x) = \frac{0}{π + 0}$

Step 8.1.5

Add $π$ and $0$ .

$2 \cos (2 x) = \frac{0}{π}$

Step 8.1.6

Divide $0$ by $π$ .

$2 \cos (2 x) = 0$

Step 8.2

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

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Step 8.2.1

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

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

Step 8.2.2

Simplify the left side.

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Step 8.2.2.1

Cancel the common factor of $2$ .

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Step 8.2.2.1.1

Cancel the common factor.

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

Step 8.2.2.1.2

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

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

Step 8.2.3

Simplify the right side.

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Step 8.2.3.1

Divide $0$ by $2$ .

$\cos (2 x) = 0$

Step 8.3

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

$2 x = \arccos (0)$

Step 8.4

Simplify the right side.

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Step 8.4.1

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

$2 x = \frac{π}{2}$

Step 8.5

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

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Step 8.5.1

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

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

Step 8.5.2

Simplify the left side.

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Step 8.5.2.1

Cancel the common factor of $2$ .

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Step 8.5.2.1.1

Cancel the common factor.

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

Step 8.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{2}$

Step 8.5.3

Simplify the right side.

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Step 8.5.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{2}$

Step 8.5.3.2

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

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Step 8.5.3.2.1

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

$x = \frac{π}{2 \cdot 2}$

Step 8.5.3.2.2

Multiply $2$ by $2$ .

$x = \frac{π}{4}$

Step 8.6

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 8.7

Solve for $x$ .

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Step 8.7.1

Simplify.

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Step 8.7.1.1

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 8.7.1.2

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 8.7.1.3

Combine the numerators over the common denominator.

$2 x = \frac{2 π \cdot 2 - π}{2}$

Step 8.7.1.4

Multiply $2$ by $2$ .

$2 x = \frac{4 π - π}{2}$

Step 8.7.1.5

Subtract $π$ from $4 π$ .

$2 x = \frac{3 π}{2}$

Step 8.7.2

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

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Step 8.7.2.1

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

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

Step 8.7.2.2

Simplify the left side.

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Step 8.7.2.2.1

Cancel the common factor of $2$ .

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Step 8.7.2.2.1.1

Cancel the common factor.

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

Step 8.7.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{2}}{2}$

Step 8.7.2.3

Simplify the right side.

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Step 8.7.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{2} \cdot \frac{1}{2}$

Step 8.7.2.3.2

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

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Step 8.7.2.3.2.1

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

$x = \frac{3 π}{2 \cdot 2}$

Step 8.7.2.3.2.2

Multiply $2$ by $2$ .

$x = \frac{3 π}{4}$

Step 8.8

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

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Step 8.8.1

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

$\frac{2 π}{| b |}$

Step 8.8.2

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

$\frac{2 π}{| 2 |}$

Step 8.8.3

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

$\frac{2 π}{2}$

Step 8.8.4

Cancel the common factor of $2$ .

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Step 8.8.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 8.8.4.2

Divide $π$ by $1$ .

$π$

Step 8.9

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 8.10

Consolidate the answers.

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

Step 9

There is a tangent line found at $x = \frac{π}{4} + \frac{π n}{2}$ parallel to the line that passes through the end points $a = 0$ and $b = 2 π$ .

There is a tangent line at $x = \frac{π}{4} + \frac{π n}{2}$ parallel to the line that passes through the end points $a = 0$ and $b = 2 π$

Step 10