Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

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

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

Step 1.1.2

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

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

Step 1.2

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

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

Multiply $- 1$ by $2$ .

$1 - 2 \sin (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$1 - 2 \sin (x) = 0$

Step 4

Subtract $1$ from both sides of the equation.

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

Step 5

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ and simplify.

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Step 5.1

Divide each term in $- 2 \sin (x) = - 1$ by $- 2$ .

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $- 2$ .

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Step 5.2.1.1

Cancel the common factor.

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

Step 5.2.1.2

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

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Dividing two negative values results in a positive value.

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

Step 6

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

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

Step 7

Simplify the right side.

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Step 7.1

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

$x = \frac{π}{6}$

Step 8

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.

$x = π - \frac{π}{6}$

Step 9

Simplify $π - \frac{π}{6}$ .

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Step 9.1

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

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

Step 9.2

Combine fractions.

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Step 9.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 9.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 9.3

Simplify the numerator.

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Step 9.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 9.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 10

The solution to the equation $x = \frac{π}{6}$ .

$x = \frac{π}{6}, \frac{5 π}{6}$

Step 11

Evaluate the second derivative at $x = \frac{π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 \cos (\frac{π}{6})$

Step 12

Evaluate the second derivative.

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Step 12.1

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- 2 \frac{\sqrt{3}}{2}$

Step 12.2

Cancel the common factor of $2$ .

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Step 12.2.1

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{\sqrt{3}}{2}$

Step 12.2.2

Cancel the common factor.

$2 \cdot - 1 \frac{\sqrt{3}}{2}$

Step 12.2.3

Rewrite the expression.

$- 1 \sqrt{3}$

Step 12.3

Rewrite $- 1 \sqrt{3}$ as $- \sqrt{3}$ .

$- \sqrt{3}$

Step 13

$x = \frac{π}{6}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{6}$ is a local maximum

Step 14

Find the y-value when $x = \frac{π}{6}$ .

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Step 14.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = (\frac{π}{6}) + 2 \cos (\frac{π}{6})$

Step 14.2

Simplify the result.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{π}{6}) = \frac{π}{6} + 2 (\frac{\sqrt{3}}{2})$

Step 14.2.1.2

Cancel the common factor of $2$ .

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Step 14.2.1.2.1

Cancel the common factor.

$f (\frac{π}{6}) = \frac{π}{6} + 2 (\frac{\sqrt{3}}{2})$

Step 14.2.1.2.2

Rewrite the expression.

$f (\frac{π}{6}) = \frac{π}{6} + \sqrt{3}$

Step 14.2.2

The final answer is $\frac{π}{6} + \sqrt{3}$ .

$y = \frac{π}{6} + \sqrt{3}$

Step 15

Evaluate the second derivative at $x = \frac{5 π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 2 \cos (\frac{5 π}{6})$

Step 16

Evaluate the second derivative.

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Step 16.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$- 2 (- \cos (\frac{π}{6}))$

Step 16.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$- 2 (- \frac{\sqrt{3}}{2})$

Step 16.3

Cancel the common factor of $2$ .

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Step 16.3.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$- 2 \frac{- \sqrt{3}}{2}$

Step 16.3.2

Factor $2$ out of $- 2$ .

$2 (- 1) \frac{- \sqrt{3}}{2}$

Step 16.3.3

Cancel the common factor.

$2 \cdot - 1 \frac{- \sqrt{3}}{2}$

Step 16.3.4

Rewrite the expression.

$- 1 (- \sqrt{3})$

Step 16.4

Multiply.

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Step 16.4.1

Multiply $- 1$ by $- 1$ .

$1 \sqrt{3}$

Step 16.4.2

Multiply $\sqrt{3}$ by $1$ .

$\sqrt{3}$

Step 17

$x = \frac{5 π}{6}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5 π}{6}$ is a local minimum

Step 18

Find the y-value when $x = \frac{5 π}{6}$ .

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Step 18.1

Replace the variable $x$ with $\frac{5 π}{6}$ in the expression.

$f (\frac{5 π}{6}) = (\frac{5 π}{6}) + 2 \cos (\frac{5 π}{6})$

Step 18.2

Simplify the result.

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Step 18.2.1

Simplify each term.

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Step 18.2.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$f (\frac{5 π}{6}) = \frac{5 π}{6} + 2 (- \cos (\frac{π}{6}))$

Step 18.2.1.2

The exact value of $\cos (\frac{π}{6})$ is $\frac{\sqrt{3}}{2}$ .

$f (\frac{5 π}{6}) = \frac{5 π}{6} + 2 (- \frac{\sqrt{3}}{2})$

Step 18.2.1.3

Cancel the common factor of $2$ .

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Step 18.2.1.3.1

Move the leading negative in $- \frac{\sqrt{3}}{2}$ into the numerator.

$f (\frac{5 π}{6}) = \frac{5 π}{6} + 2 (\frac{- \sqrt{3}}{2})$

Step 18.2.1.3.2

Cancel the common factor.

$f (\frac{5 π}{6}) = \frac{5 π}{6} + 2 (\frac{- \sqrt{3}}{2})$

Step 18.2.1.3.3

Rewrite the expression.

$f (\frac{5 π}{6}) = \frac{5 π}{6} - \sqrt{3}$

Step 18.2.2

The final answer is $\frac{5 π}{6} - \sqrt{3}$ .

$y = \frac{5 π}{6} - \sqrt{3}$

Step 19

These are the local extrema for $f (x) = x + 2 \cos (x)$ .

$(\frac{π}{6}, \frac{π}{6} + \sqrt{3})$ is a local maxima

$(\frac{5 π}{6}, \frac{5 π}{6} - \sqrt{3})$ is a local minima

Step 20