Calculus Examples

$f (x) = 2 \sin (x) + \sin (2 x)$ , $[0, π]$

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 $f (x) = 2 \sin (x) + \sin (2 x)$ 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 $[0, π]$ .

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

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

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

Step 3.1.2

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

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Step 3.1.2.1

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

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

Step 3.1.2.2

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

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

Step 3.1.3

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

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Step 3.1.3.1

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 3.1.3.1.1

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

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

Step 3.1.3.1.2

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

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

Step 3.1.3.1.3

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

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

Step 3.1.3.2

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

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

Step 3.1.3.3

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

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

Step 3.1.3.4

Multiply $2$ by $1$ .

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

Step 3.1.3.5

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

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

Step 3.2

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

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

Step 4

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

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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 $(0, π)$ .

The function is continuous.

Step 5

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

The function is differentiable.

Step 6

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

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

Step 7

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

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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify each term.

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Step 7.2.1.1

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

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

Step 7.2.1.2

Multiply $2$ by $0$ .

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

Step 7.2.1.3

Multiply $2$ by $0$ .

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

Step 7.2.1.4

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

$f (0) = 0 + 0$

Step 7.2.2

Add $0$ and $0$ .

$f (0) = 0$

Step 7.2.3

The final answer is $0$ .

$0$

Step 8

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

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Step 8.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

$2 \cos (x) + 2 (1 - 2 \sin^{2} (x)) = \frac{(0) - (0)}{(π) - (0)}$

Step 8.2

Simplify the left side.

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Step 8.2.1

Simplify each term.

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Step 8.2.1.1

Apply the distributive property.

$2 \cos (x) + 2 \cdot 1 + 2 (- 2 \sin^{2} (x)) = \frac{(0) - (0)}{(π) - (0)}$

Step 8.2.1.2

Multiply $2$ by $1$ .

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

Step 8.2.1.3

Multiply $- 2$ by $2$ .

$2 \cos (x) + 2 - 4 \sin^{2} (x) = \frac{(0) - (0)}{(π) - (0)}$

Step 8.3

Simplify the right side.

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Step 8.3.1

Simplify $\frac{(0) - (0)}{(π) - (0)}$ .

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Step 8.3.1.1

Simplify the numerator.

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Step 8.3.1.1.1

Multiply $- 1$ by $0$ .

$2 \cos (x) + 2 - 4 \sin^{2} (x) = \frac{0 + 0}{π - (0)}$

Step 8.3.1.1.2

Add $0$ and $0$ .

$2 \cos (x) + 2 - 4 \sin^{2} (x) = \frac{0}{π - (0)}$

Step 8.3.1.2

Simplify the denominator.

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Step 8.3.1.2.1

Multiply $- 1$ by $0$ .

$2 \cos (x) + 2 - 4 \sin^{2} (x) = \frac{0}{π + 0}$

Step 8.3.1.2.2

Add $π$ and $0$ .

$2 \cos (x) + 2 - 4 \sin^{2} (x) = \frac{0}{π}$

Step 8.3.1.3

Divide $0$ by $π$ .

$2 \cos (x) + 2 - 4 \sin^{2} (x) = 0$

Step 8.4

Solve the equation for $x$ .

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Step 8.4.1

Replace the $- 4 \sin^{2} (x)$ with $- 4 (1 - \cos^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

$2 \cos (x) + 2 - 4 (1 - \cos^{2} (x)) = 0$

Step 8.4.2

Simplify each term.

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Step 8.4.2.1

Apply the distributive property.

$2 \cos (x) + 2 - 4 \cdot 1 - 4 (- \cos^{2} (x)) = 0$

Step 8.4.2.2

Multiply $- 4$ by $1$ .

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

Step 8.4.2.3

Multiply $- 1$ by $- 4$ .

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

Step 8.4.3

Subtract $4$ from $2$ .

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

Step 8.4.4

Reorder the polynomial.

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

Step 8.4.5

Substitute $u$ for $\cos (x)$ .

$4 {(u)}^{2} + 2 (u) - 2 = 0$

Step 8.4.6

Factor the left side of the equation.

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Step 8.4.6.1

Factor $2$ out of $4 u^{2} + 2 u - 2$ .

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Step 8.4.6.1.1

Factor $2$ out of $4 u^{2}$ .

$2 (2 u^{2}) + 2 u - 2 = 0$

Step 8.4.6.1.2

Factor $2$ out of $2 u$ .

$2 (2 u^{2}) + 2 u - 2 = 0$

Step 8.4.6.1.3

Factor $2$ out of $- 2$ .

$2 (2 u^{2}) + 2 u + 2 (- 1) = 0$

Step 8.4.6.1.4

Factor $2$ out of $2 (2 u^{2}) + 2 u$ .

$2 (2 u^{2} + u) + 2 (- 1) = 0$

Step 8.4.6.1.5

Factor $2$ out of $2 (2 u^{2} + u) + 2 (- 1)$ .

$2 (2 u^{2} + u - 1) = 0$

Step 8.4.6.2

Factor.

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Step 8.4.6.2.1

Factor by grouping.

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Step 8.4.6.2.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = 1$ .

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Step 8.4.6.2.1.1.1

Multiply by $1$ .

$2 (2 u^{2} + 1 u - 1) = 0$

Step 8.4.6.2.1.1.2

Rewrite $1$ as $- 1$ plus $2$

$2 (2 u^{2} + (- 1 + 2) u - 1) = 0$

Step 8.4.6.2.1.1.3

Apply the distributive property.

$2 (2 u^{2} - 1 u + 2 u - 1) = 0$

Step 8.4.6.2.1.2

Factor out the greatest common factor from each group.

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Step 8.4.6.2.1.2.1

Group the first two terms and the last two terms.

$2 (2 u^{2} - 1 u + 2 u - 1) = 0$

Step 8.4.6.2.1.2.2

Factor out the greatest common factor (GCF) from each group.

$2 (u (2 u - 1) + 1 (2 u - 1)) = 0$

Step 8.4.6.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

$2 ((2 u - 1) (u + 1)) = 0$

Step 8.4.6.2.2

Remove unnecessary parentheses.

$2 (2 u - 1) (u + 1) = 0$

Step 8.4.7

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$2 u - 1 = 0$

$u + 1 = 0$

Step 8.4.8

Set $2 u - 1$ equal to $0$ and solve for $u$ .

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Step 8.4.8.1

Set $2 u - 1$ equal to $0$ .

$2 u - 1 = 0$

Step 8.4.8.2

Solve $2 u - 1 = 0$ for $u$ .

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Step 8.4.8.2.1

Add $1$ to both sides of the equation.

$2 u = 1$

Step 8.4.8.2.2

Divide each term in $2 u = 1$ by $2$ and simplify.

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Step 8.4.8.2.2.1

Divide each term in $2 u = 1$ by $2$ .

$\frac{2 u}{2} = \frac{1}{2}$

Step 8.4.8.2.2.2

Simplify the left side.

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Step 8.4.8.2.2.2.1

Cancel the common factor of $2$ .

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Step 8.4.8.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{1}{2}$

Step 8.4.8.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{2}$

Step 8.4.9

Set $u + 1$ equal to $0$ and solve for $u$ .

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Step 8.4.9.1

Set $u + 1$ equal to $0$ .

$u + 1 = 0$

Step 8.4.9.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 8.4.10

The final solution is all the values that make $2 (2 u - 1) (u + 1) = 0$ true.

$u = \frac{1}{2}, - 1$

Step 8.4.11

Substitute $\cos (x)$ for $u$ .

$\cos (x) = \frac{1}{2}, - 1$

Step 8.4.12

Set up each of the solutions to solve for $x$ .

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

$\cos (x) = - 1$

Step 8.4.13

Solve for $x$ in $\cos (x) = \frac{1}{2}$ .

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Step 8.4.13.1

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

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

Step 8.4.13.2

Simplify the right side.

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Step 8.4.13.2.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$x = \frac{π}{3}$

Step 8.4.13.3

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.

$x = 2 π - \frac{π}{3}$

Step 8.4.13.4

Simplify $2 π - \frac{π}{3}$ .

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Step 8.4.13.4.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Step 8.4.13.4.2

Combine fractions.

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Step 8.4.13.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 8.4.13.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 8.4.13.4.3

Simplify the numerator.

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Step 8.4.13.4.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 8.4.13.4.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 8.4.13.5

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

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Step 8.4.13.5.1

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

$\frac{2 π}{| b |}$

Step 8.4.13.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 8.4.13.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 8.4.13.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 8.4.13.6

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

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 8.4.14

Solve for $x$ in $\cos (x) = - 1$ .

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Step 8.4.14.1

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

$x = \arccos (- 1)$

Step 8.4.14.2

Simplify the right side.

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Step 8.4.14.2.1

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

$x = π$

Step 8.4.14.3

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$x = 2 π - π$

Step 8.4.14.4

Subtract $π$ from $2 π$ .

$x = π$

Step 8.4.14.5

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

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Step 8.4.14.5.1

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

$\frac{2 π}{| b |}$

Step 8.4.14.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 8.4.14.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 8.4.14.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 8.4.14.6

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

$x = π + 2 π n$ , for any integer $n$

Step 8.4.15

List all of the solutions.

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n, π + 2 π n$ , for any integer $n$

Step 8.4.16

Consolidate the answers.

$x = \frac{π}{3} + \frac{2 π n}{3}$ , for any integer $n$

Step 9

There is a tangent line found at $x = \frac{π}{3} + \frac{2 π n}{3}$ parallel to the line that passes through the end points $a = 0$ and $b = π$ .

There is a tangent line at $x = \frac{π}{3} + \frac{2 π n}{3}$ parallel to the line that passes through the end points $a = 0$ and $b = π$

Step 10