Calculus Examples

$f (x) = \sin (x) + 8$ , $0 < x < 2 π$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

By the Sum Rule, the derivative of $\sin (x) + 8$ with respect to $x$ is $\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [8]$ .

$\frac{d}{d x} [\sin (x)] + \frac{d}{d x} [8]$

Step 1.1.1.2

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

$\cos (x) + \frac{d}{d x} [8]$

Step 1.1.1.3

Differentiate using the Constant Rule.

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Step 1.1.1.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\cos (x) + 0$

Step 1.1.1.3.2

Add $\cos (x)$ and $0$ .

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

Step 1.1.2

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

$\cos (x)$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $\cos (x) = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$\cos (x) = 0$

Step 1.2.2

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

$x = \arccos (0)$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

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

$x = \frac{π}{2}$

Step 1.2.4

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.

$x = 2 π - \frac{π}{2}$

Step 1.2.5

Simplify $2 π - \frac{π}{2}$ .

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Step 1.2.5.1

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

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

Step 1.2.5.2

Combine fractions.

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Step 1.2.5.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 1.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 1.2.5.3

Simplify the numerator.

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Step 1.2.5.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 1.2.5.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 1.2.6

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

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Step 1.2.6.1

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

$\frac{2 π}{| b |}$

Step 1.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 1.2.6.3

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

$\frac{2 π}{1}$

Step 1.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.7

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

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

Step 1.2.8

Consolidate the answers.

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

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

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.

Step 1.4

Evaluate $\sin (x) + 8$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = \frac{π}{2}$ .

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Step 1.4.1.1

Substitute $\frac{π}{2}$ for $x$ .

$\sin (\frac{π}{2}) + 8$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

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

$1 + 8$

Step 1.4.1.2.2

Add $1$ and $8$ .

$9$

Step 1.4.2

Evaluate at $x = \frac{3 π}{2}$ .

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Step 1.4.2.1

Substitute $\frac{3 π}{2}$ for $x$ .

$\sin (\frac{3 π}{2}) + 8$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

Simplify each term.

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Step 1.4.2.2.1.1

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 fourth quadrant.

$- \sin (\frac{π}{2}) + 8$

Step 1.4.2.2.1.2

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

$- 1 \cdot 1 + 8$

Step 1.4.2.2.1.3

Multiply $- 1$ by $1$ .

$- 1 + 8$

Step 1.4.2.2.2

Add $- 1$ and $8$ .

$7$

Step 1.4.3

List all of the points.

$(\frac{π}{2} + 2 π n, 9), (\frac{3 π}{2} + 2 π n, 7)$ , for any integer $n$

Step 2

Exclude the points that are not on the interval.

$(\frac{π}{2}, 9), (\frac{3 π}{2}, 7)$

Step 3

Use the first derivative test to determine which points can be maxima or minima.

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Step 3.1

Split $(- \infty, \infty)$ into separate intervals around the $x$ values that make the first derivative $0$ or undefined.

$(- \infty, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{2}) \cup (\frac{3 π}{2}, \infty)$

Step 3.2

Substitute any number, such as $0$ , from the interval $(- \infty, \frac{π}{2})$ in the first derivative $\cos (x)$ to check if the result is negative or positive.

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Step 3.2.1

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

$f' (0) = \cos (0)$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

The exact value of $\cos (0)$ is $1$ .

$f' (0) = 1$

Step 3.2.2.2

The final answer is $1$ .

$1$

Step 3.3

Substitute any number, such as $3$ , from the interval $(\frac{π}{2}, \frac{3 π}{2})$ in the first derivative $\cos (x)$ to check if the result is negative or positive.

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Step 3.3.1

Replace the variable $x$ with $3$ in the expression.

$f' (3) = \cos (3)$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Evaluate $\cos (3)$ .

$f' (3) = - 0.98999249$

Step 3.3.2.2

The final answer is $- 0.98999249$ .

$- 0.98999249$

Step 3.4

Substitute any number, such as $7$ , from the interval $(\frac{3 π}{2}, \infty)$ in the first derivative $\cos (x)$ to check if the result is negative or positive.

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Step 3.4.1

Replace the variable $x$ with $7$ in the expression.

$f' (7) = \cos (7)$

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Evaluate $\cos (7)$ .

$f' (7) = 0.75390225$

Step 3.4.2.2

The final answer is $0.75390225$ .

$0.75390225$

Step 3.5

Since the first derivative changed signs from positive to negative around $x = \frac{π}{2}$ , then $x = \frac{π}{2}$ is a local maximum.

$x = \frac{π}{2}$ is a local maximum

Step 3.6

Since the first derivative changed signs from negative to positive around $x = \frac{3 π}{2}$ , then $x = \frac{3 π}{2}$ is a local minimum.

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

Step 3.7

These are the local extrema for $f (x) = \sin (x) + 8$ .

$x = \frac{π}{2}$ is a local maximum

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

$x = \frac{π}{2}$ is a local maximum

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

Step 4

Compare the $f (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $f (x)$ value and the minimum will occur at the lowest $f (x)$ value.

Absolute Maximum: $(\frac{π}{2}, 9)$

Absolute Minimum: $(\frac{3 π}{2}, 7)$

Step 5