Calculus Examples

$f (x) = \sqrt{3} x - 2 \sin (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $\sqrt{3} x - 2 \sin (x)$ with respect to $x$ is $\frac{d}{d x} [\sqrt{3} x] + \frac{d}{d x} [- 2 \sin (x)]$ .

$\frac{d}{d x} [\sqrt{3} x] + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2

Evaluate $\frac{d}{d x} [\sqrt{3} x]$ .

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Step 1.2.1

Since $\sqrt{3}$ is constant with respect to $x$ , the derivative of $\sqrt{3} x$ with respect to $x$ is $\sqrt{3} \frac{d}{d x} [x]$ .

$\sqrt{3} \frac{d}{d x} [x] + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.2

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

$\sqrt{3} \cdot 1 + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.2.3

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

$\sqrt{3} + \frac{d}{d x} [- 2 \sin (x)]$

Step 1.3

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

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Step 1.3.1

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

$\sqrt{3} - 2 \frac{d}{d x} [\sin (x)]$

Step 1.3.2

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

$\sqrt{3} - 2 \cos (x)$

Step 1.4

Reorder terms.

$- 2 \cos (x) + \sqrt{3}$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- 2 \cos (x) + \sqrt{3}$ with respect to $x$ is $\frac{d}{d x} [- 2 \cos (x)] + \frac{d}{d x} [\sqrt{3}]$ .

$f'' (x) = \frac{d}{d x} (- 2 \cos (x)) + \frac{d}{d x} (\sqrt{3})$

Step 2.2

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

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Step 2.2.1

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

$f'' (x) = - 2 \frac{d}{d x} \cos (x) + \frac{d}{d x} (\sqrt{3})$

Step 2.2.2

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

$f'' (x) = - 2 (- \sin (x)) + \frac{d}{d x} (\sqrt{3})$

Step 2.2.3

Multiply $- 1$ by $- 2$ .

$f'' (x) = 2 \sin (x) + \frac{d}{d x} (\sqrt{3})$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $\sqrt{3}$ is constant with respect to $x$ , the derivative of $\sqrt{3}$ with respect to $x$ is $0$ .

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

Step 2.3.2

Add $2 \sin (x)$ and $0$ .

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

Step 3

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

$- 2 \cos (x) + \sqrt{3} = 0$

Step 4

Subtract $\sqrt{3}$ from both sides of the equation.

$- 2 \cos (x) = - \sqrt{3}$

Step 5

Divide each term in $- 2 \cos (x) = - \sqrt{3}$ by $- 2$ and simplify.

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Step 5.1

Divide each term in $- 2 \cos (x) = - \sqrt{3}$ by $- 2$ .

$\frac{- 2 \cos (x)}{- 2} = \frac{- \sqrt{3}}{- 2}$

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.

$\frac{- 2 \cos (x)}{- 2} = \frac{- \sqrt{3}}{- 2}$

Step 5.2.1.2

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

$\cos (x) = \frac{- \sqrt{3}}{- 2}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Dividing two negative values results in a positive value.

$\cos (x) = \frac{\sqrt{3}}{2}$

Step 6

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

$x = \arccos (\frac{\sqrt{3}}{2})$

Step 7

Simplify the right side.

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Step 7.1

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

$x = \frac{π}{6}$

Step 8

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.

$x = 2 π - \frac{π}{6}$

Step 9

Simplify $2 π - \frac{π}{6}$ .

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Step 9.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

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

Step 9.2

Combine fractions.

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Step 9.2.1

Combine $2 π$ and $\frac{6}{6}$ .

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

Step 9.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 6 - π}{6}$

Step 9.3

Simplify the numerator.

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Step 9.3.1

Multiply $6$ by $2$ .

$x = \frac{12 π - π}{6}$

Step 9.3.2

Subtract $π$ from $12 π$ .

$x = \frac{11 π}{6}$

Step 10

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

$x = \frac{π}{6}, \frac{11 π}{6}$

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.

$2 \sin (\frac{π}{6})$

Step 12

Evaluate the second derivative.

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Step 12.1

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

$2 (\frac{1}{2})$

Step 12.2

Cancel the common factor of $2$ .

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Step 12.2.1

Cancel the common factor.

$2 (\frac{1}{2})$

Step 12.2.2

Rewrite the expression.

$1$

Step 13

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

$x = \frac{π}{6}$ is a local minimum

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.

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

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

Combine $\sqrt{3}$ and $\frac{π}{6}$ .

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

Step 14.2.1.2

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

$f (\frac{π}{6}) = \frac{\sqrt{3} π}{6} - 2 (\frac{1}{2})$

Step 14.2.1.3

Cancel the common factor of $2$ .

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Step 14.2.1.3.1

Factor $2$ out of $- 2$ .

$f (\frac{π}{6}) = \frac{\sqrt{3} π}{6} + 2 (- 1) (\frac{1}{2})$

Step 14.2.1.3.2

Cancel the common factor.

$f (\frac{π}{6}) = \frac{\sqrt{3} π}{6} + 2 \cdot (- 1 (\frac{1}{2}))$

Step 14.2.1.3.3

Rewrite the expression.

$f (\frac{π}{6}) = \frac{\sqrt{3} π}{6} - 1$

Step 14.2.2

The final answer is $\frac{\sqrt{3} π}{6} - 1$ .

$y = \frac{\sqrt{3} π}{6} - 1$

Step 15

Evaluate the second derivative at $x = \frac{11 π}{6}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \sin (\frac{11 π}{6})$

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 sine is negative in the fourth quadrant.

$2 (- \sin (\frac{π}{6}))$

Step 16.2

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

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

Step 16.3

Cancel the common factor of $2$ .

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Step 16.3.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 16.3.2

Cancel the common factor.

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

Step 16.3.3

Rewrite the expression.

$- 1$

Step 17

$x = \frac{11 π}{6}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{11 π}{6}$ is a local maximum

Step 18

Find the y-value when $x = \frac{11 π}{6}$ .

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Step 18.1

Replace the variable $x$ with $\frac{11 π}{6}$ in the expression.

$f (\frac{11 π}{6}) = \sqrt{3} (\frac{11 π}{6}) - 2 \sin (\frac{11 π}{6})$

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

Combine $\sqrt{3}$ and $\frac{11 π}{6}$ .

$f (\frac{11 π}{6}) = \frac{\sqrt{3} (11 π)}{6} - 2 \sin (\frac{11 π}{6})$

Step 18.2.1.2

Move $11$ to the left of $\sqrt{3}$ .

$f (\frac{11 π}{6}) = \frac{11 \cdot (\sqrt{3} π)}{6} - 2 \sin (\frac{11 π}{6})$

Step 18.2.1.3

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

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} - 2 (- \sin (\frac{π}{6}))$

Step 18.2.1.4

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

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} - 2 (- \frac{1}{2})$

Step 18.2.1.5

Cancel the common factor of $2$ .

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Step 18.2.1.5.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} - 2 (\frac{- 1}{2})$

Step 18.2.1.5.2

Factor $2$ out of $- 2$ .

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} + 2 (- 1) (\frac{- 1}{2})$

Step 18.2.1.5.3

Cancel the common factor.

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} + 2 \cdot (- 1 (\frac{- 1}{2}))$

Step 18.2.1.5.4

Rewrite the expression.

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} - 1 \cdot - 1$

Step 18.2.1.6

Multiply $- 1$ by $- 1$ .

$f (\frac{11 π}{6}) = \frac{11 \sqrt{3} π}{6} + 1$

Step 18.2.2

The final answer is $\frac{11 \sqrt{3} π}{6} + 1$ .

$y = \frac{11 \sqrt{3} π}{6} + 1$

Step 19

These are the local extrema for $f (x) = \sqrt{3} x - 2 \sin (x)$ .

$(\frac{π}{6}, \frac{\sqrt{3} π}{6} - 1)$ is a local minima

$(\frac{11 π}{6}, \frac{11 \sqrt{3} π}{6} + 1)$ is a local maxima

Step 20