Calculus Examples

$h (x) = \sin (x) + \frac{x}{2}$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

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

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Step 1.3.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{x}{2}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [x]$ .

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

Step 1.3.2

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

$\cos (x) + \frac{1}{2} \cdot 1$

Step 1.3.3

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

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

Step 1.4

Reorder terms.

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

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 $\frac{1}{2} + \cos (x)$ with respect to $x$ is $\frac{d}{d x} [\frac{1}{2}] + \frac{d}{d x} [\cos (x)]$ .

$h'' (x) = \frac{d}{d x} (\frac{1}{2}) + \frac{d}{d x} (\cos (x))$

Step 2.1.2

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2}$ with respect to $x$ is $0$ .

$h'' (x) = 0 + \frac{d}{d x} (\cos (x))$

Step 2.2

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

$h'' (x) = 0 - \sin (x)$

Step 2.3

Subtract $\sin (x)$ from $0$ .

$h'' (x) = - \sin (x)$

Step 3

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

$\frac{1}{2} + \cos (x) = 0$

Step 4

Subtract $\frac{1}{2}$ from both sides of the equation.

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

Step 5

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

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

Step 6

Simplify the right side.

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Step 6.1

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

$x = \frac{2 π}{3}$

Step 7

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.

$x = 2 π - \frac{2 π}{3}$

Step 8

Simplify $2 π - \frac{2 π}{3}$ .

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Step 8.1

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

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

Step 8.2

Combine fractions.

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Step 8.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 8.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - 2 π}{3}$

Step 8.3

Simplify the numerator.

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Step 8.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - 2 π}{3}$

Step 8.3.2

Subtract $2 π$ from $6 π$ .

$x = \frac{4 π}{3}$

Step 9

The solution to the equation $x = \frac{2 π}{3}$ .

$x = \frac{2 π}{3}, \frac{4 π}{3}$

Step 10

Evaluate the second derivative at $x = \frac{2 π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin (\frac{2 π}{3})$

Step 11

Evaluate the second derivative.

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Step 11.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$- \sin (\frac{π}{3})$

Step 11.2

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

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

Step 12

$x = \frac{2 π}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{2 π}{3}$ is a local maximum

Step 13

Find the y-value when $x = \frac{2 π}{3}$ .

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Step 13.1

Replace the variable $x$ with $\frac{2 π}{3}$ in the expression.

$h (\frac{2 π}{3}) = \sin (\frac{2 π}{3}) + \frac{\frac{2 π}{3}}{2}$

Step 13.2

Simplify the result.

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Step 13.2.1

Simplify each term.

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Step 13.2.1.1

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

$h (\frac{2 π}{3}) = \sin (\frac{π}{3}) + \frac{\frac{2 π}{3}}{2}$

Step 13.2.1.2

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

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

Step 13.2.1.3

Multiply the numerator by the reciprocal of the denominator.

$h (\frac{2 π}{3}) = \frac{\sqrt{3}}{2} + \frac{2 π}{3} \cdot \frac{1}{2}$

Step 13.2.1.4

Cancel the common factor of $2$ .

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Step 13.2.1.4.1

Factor $2$ out of $2 π$ .

$h (\frac{2 π}{3}) = \frac{\sqrt{3}}{2} + \frac{2 (π)}{3} \cdot \frac{1}{2}$

Step 13.2.1.4.2

Cancel the common factor.

$h (\frac{2 π}{3}) = \frac{\sqrt{3}}{2} + \frac{2 π}{3} \cdot \frac{1}{2}$

Step 13.2.1.4.3

Rewrite the expression.

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

Step 13.2.2

The final answer is $\frac{\sqrt{3}}{2} + \frac{π}{3}$ .

$y = \frac{\sqrt{3}}{2} + \frac{π}{3}$

Step 14

Evaluate the second derivative at $x = \frac{4 π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- \sin (\frac{4 π}{3})$

Step 15

Evaluate the second derivative.

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Step 15.1

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 third quadrant.

$- - \sin (\frac{π}{3})$

Step 15.2

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

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

Step 15.3

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

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Step 15.3.1

Multiply $- 1$ by $- 1$ .

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

Step 15.3.2

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

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

Step 16

$x = \frac{4 π}{3}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{4 π}{3}$ is a local minimum

Step 17

Find the y-value when $x = \frac{4 π}{3}$ .

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Step 17.1

Replace the variable $x$ with $\frac{4 π}{3}$ in the expression.

$h (\frac{4 π}{3}) = \sin (\frac{4 π}{3}) + \frac{\frac{4 π}{3}}{2}$

Step 17.2

Simplify the result.

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Step 17.2.1

Simplify each term.

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Step 17.2.1.1

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 third quadrant.

$h (\frac{4 π}{3}) = - \sin (\frac{π}{3}) + \frac{\frac{4 π}{3}}{2}$

Step 17.2.1.2

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

$h (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2} + \frac{\frac{4 π}{3}}{2}$

Step 17.2.1.3

Multiply the numerator by the reciprocal of the denominator.

$h (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2} + \frac{4 π}{3} \cdot \frac{1}{2}$

Step 17.2.1.4

Cancel the common factor of $2$ .

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Step 17.2.1.4.1

Factor $2$ out of $4 π$ .

$h (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2} + \frac{2 (2 π)}{3} \cdot \frac{1}{2}$

Step 17.2.1.4.2

Cancel the common factor.

$h (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2} + \frac{2 (2 π)}{3} \cdot \frac{1}{2}$

Step 17.2.1.4.3

Rewrite the expression.

$h (\frac{4 π}{3}) = - \frac{\sqrt{3}}{2} + \frac{2 π}{3}$

Step 17.2.2

The final answer is $- \frac{\sqrt{3}}{2} + \frac{2 π}{3}$ .

$y = - \frac{\sqrt{3}}{2} + \frac{2 π}{3}$

Step 18

These are the local extrema for $h (x) = \sin (x) + \frac{x}{2}$ .

$(\frac{2 π}{3}, \frac{\sqrt{3}}{2} + \frac{π}{3})$ is a local maxima

$(\frac{4 π}{3}, - \frac{\sqrt{3}}{2} + \frac{2 π}{3})$ is a local minima

Step 19