Calculus Examples

f (x) = 5 {cos}^{2} (x)

$f (x) = 5 \cos^{2} (x)$ on

[0, π]

$[0, π]$

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

Since

5

$5$ is constant with respect to

x

$x$ , the derivative of

5 {cos}^{2} (x)

$5 \cos^{2} (x)$ with respect to

x

$x$ is

5 \frac{d}{d x} [{cos}^{2} (x)]

$5 \frac{d}{d x} [\cos^{2} (x)]$ .

5 \frac{d}{d x} [{cos}^{2} (x)]

$5 \frac{d}{d x} [\cos^{2} (x)]$

Step 1.1.1.2

Differentiate using the chain rule, which states that

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ is

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ where

f (x) = x^{2}

$f (x) = x^{2}$ and

g (x) = cos (x)

$g (x) = \cos (x)$ .

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Step 1.1.1.2.1

To apply the Chain Rule, set

u

$u$ as

cos (x)

$\cos (x)$ .

5 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [cos (x)])

$5 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\cos (x)])$

Step 1.1.1.2.2

Differentiate using the Power Rule which states that

\frac{d}{d u} [u^{n}]

$\frac{d}{d u} [u^{n}]$ is

n u^{n - 1}

$n u^{n - 1}$ where

n = 2

$n = 2$ .

5 (2 u \frac{d}{d x} [cos (x)])

$5 (2 u \frac{d}{d x} [\cos (x)])$

Step 1.1.1.2.3

Replace all occurrences of

u

$u$ with

cos (x)

$\cos (x)$ .

5 (2 cos (x) \frac{d}{d x} [cos (x)])

$5 (2 \cos (x) \frac{d}{d x} [\cos (x)])$

5 (2 cos (x) \frac{d}{d x} [cos (x)])

$5 (2 \cos (x) \frac{d}{d x} [\cos (x)])$

Step 1.1.1.3

Multiply

2

$2$ by

5

$5$ .

10 (cos (x) \frac{d}{d x} [cos (x)])

$10 (\cos (x) \frac{d}{d x} [\cos (x)])$

Step 1.1.1.4

The derivative of

cos (x)

$\cos (x)$ with respect to

x

$x$ is

- sin (x)

$- \sin (x)$ .

10 cos (x) (- sin (x))

$10 \cos (x) (- \sin (x))$

Step 1.1.1.5

Multiply

- 1

$- 1$ by

10

$10$ .

f' (x) = - 10 cos (x) sin (x)

$f' (x) = - 10 \cos (x) \sin (x)$

f' (x) = - 10 cos (x) sin (x)

$f' (x) = - 10 \cos (x) \sin (x)$

Step 1.1.2

The first derivative of

f (x)

$f (x)$ with respect to

x

$x$ is

- 10 cos (x) sin (x)

$- 10 \cos (x) \sin (x)$ .

- 10 cos (x) sin (x)

$- 10 \cos (x) \sin (x)$

- 10 cos (x) sin (x)

$- 10 \cos (x) \sin (x)$

Step 1.2

Set the first derivative equal to

0

$0$ then solve the equation

- 10 cos (x) sin (x) = 0

$- 10 \cos (x) \sin (x) = 0$ .

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Step 1.2.1

Set the first derivative equal to

0

$0$ .

- 10 cos (x) sin (x) = 0

$- 10 \cos (x) \sin (x) = 0$

Step 1.2.2

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

cos (x) = 0

$\cos (x) = 0$

sin (x) = 0

$\sin (x) = 0$

Step 1.2.3

Set

cos (x)

$\cos (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.3.1

Set

cos (x)

$\cos (x)$ equal to

0

$0$ .

cos (x) = 0

$\cos (x) = 0$

Step 1.2.3.2

Solve

cos (x) = 0

$\cos (x) = 0$ for

x

$x$ .

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Step 1.2.3.2.1

Take the inverse cosine of both sides of the equation to extract

x

$x$ from inside the cosine.

x = arccos (0)

$x = \arccos (0)$

Step 1.2.3.2.2

Simplify the right side.

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Step 1.2.3.2.2.1

The exact value of

arccos (0)

$\arccos (0)$ is

\frac{π}{2}

$\frac{π}{2}$ .

x = \frac{π}{2}

$x = \frac{π}{2}$

x = \frac{π}{2}

$x = \frac{π}{2}$

Step 1.2.3.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from

2 π

$2 π$ to find the solution in the fourth quadrant.

x = 2 π - \frac{π}{2}

$x = 2 π - \frac{π}{2}$

Step 1.2.3.2.4

Simplify

2 π - \frac{π}{2}

$2 π - \frac{π}{2}$ .

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Step 1.2.3.2.4.1

To write

2 π

$2 π$ as a fraction with a common denominator, multiply by

\frac{2}{2}

$\frac{2}{2}$ .

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

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

Step 1.2.3.2.4.2

Combine fractions.

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Step 1.2.3.2.4.2.1

Combine

2 π

$2 π$ and

\frac{2}{2}

$\frac{2}{2}$ .

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

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

Step 1.2.3.2.4.2.2

Combine the numerators over the common denominator.

x = \frac{2 π \cdot 2 - π}{2}

$x = \frac{2 π \cdot 2 - π}{2}$

x = \frac{2 π \cdot 2 - π}{2}

$x = \frac{2 π \cdot 2 - π}{2}$

Step 1.2.3.2.4.3

Simplify the numerator.

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Step 1.2.3.2.4.3.1

Multiply

2

$2$ by

2

$2$ .

x = \frac{4 π - π}{2}

$x = \frac{4 π - π}{2}$

Step 1.2.3.2.4.3.2

Subtract

π

$π$ from

4 π

$4 π$ .

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

x = \frac{3 π}{2}

$x = \frac{3 π}{2}$

Step 1.2.3.2.5

Find the period of

cos (x)

$\cos (x)$ .

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Step 1.2.3.2.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 1.2.3.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 1.2.3.2.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 1.2.3.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.2.3.2.6

The period of the

cos (x)

$\cos (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n

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

n

$n$

Step 1.2.4

Set

sin (x)

$\sin (x)$ equal to

0

$0$ and solve for

x

$x$ .

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Step 1.2.4.1

Set

sin (x)

$\sin (x)$ equal to

0

$0$ .

sin (x) = 0

$\sin (x) = 0$

Step 1.2.4.2

Solve

sin (x) = 0

$\sin (x) = 0$ for

x

$x$ .

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Step 1.2.4.2.1

Take the inverse sine of both sides of the equation to extract

x

$x$ from inside the sine.

x = arcsin (0)

$x = \arcsin (0)$

Step 1.2.4.2.2

Simplify the right side.

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Step 1.2.4.2.2.1

The exact value of

arcsin (0)

$\arcsin (0)$ is

0

$0$ .

x = 0

$x = 0$

x = 0

$x = 0$

Step 1.2.4.2.3

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from

π

$π$ to find the solution in the second quadrant.

x = π - 0

$x = π - 0$

Step 1.2.4.2.4

Subtract

0

$0$ from

π

$π$ .

x = π

$x = π$

Step 1.2.4.2.5

Find the period of

sin (x)

$\sin (x)$ .

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Step 1.2.4.2.5.1

The period of the function can be calculated using

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$ .

\frac{2 π}{| b |}

$\frac{2 π}{| b |}$

Step 1.2.4.2.5.2

Replace

b

$b$ with

1

$1$ in the formula for period.

\frac{2 π}{| 1 |}

$\frac{2 π}{| 1 |}$

Step 1.2.4.2.5.3

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

0

$0$ and

1

$1$ is

1

$1$ .

\frac{2 π}{1}

$\frac{2 π}{1}$

Step 1.2.4.2.5.4

Divide

2 π

$2 π$ by

1

$1$ .

2 π

$2 π$

2 π

$2 π$

Step 1.2.4.2.6

The period of the

sin (x)

$\sin (x)$ function is

2 π

$2 π$ so values will repeat every

2 π

$2 π$ radians in both directions.

x = 2 π n, π + 2 π n

$x = 2 π n, π + 2 π n$ , for any integer

n

$n$

x = 2 π n, π + 2 π n

$x = 2 π n, π + 2 π n$ , for any integer

n

$n$

x = 2 π n, π + 2 π n

$x = 2 π n, π + 2 π n$ , for any integer

n

$n$

Step 1.2.5

The final solution is all the values that make

- 10 cos (x) sin (x) = 0

$- 10 \cos (x) \sin (x) = 0$ true.

x = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, 2 π n, π + 2 π n

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

n

$n$

Step 1.2.6

Consolidate the answers.

x = \frac{π n}{2}

$x = \frac{π n}{2}$ , for any integer

n

$n$

x = \frac{π n}{2}

$x = \frac{π n}{2}$ , for any integer

n

$n$

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

5 {cos}^{2} (x)

$5 \cos^{2} (x)$ at each

x

$x$ value where the derivative is

0

$0$ or undefined.

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Step 1.4.1

Evaluate at

x = 0

$x = 0$ .

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Step 1.4.1.1

Substitute

0

$0$ for

x

$x$ .

5 {cos}^{2} (0)

$5 \cos^{2} (0)$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

The exact value of

cos (0)

$\cos (0)$ is

1

$1$ .

5 \cdot 1^{2}

$5 \cdot 1^{2}$

Step 1.4.1.2.2

One to any power is one.

$5 \cdot 1$

Step 1.4.1.2.3

Multiply

$5$ by

$1$ .

$5$

Step 1.4.2

Evaluate at

$x = \frac{π}{2}$ .

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Step 1.4.2.1

Substitute

$\frac{π}{2}$ for

$x$ .

$5 \cos^{2} (\frac{π}{2})$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

The exact value of

$\cos (\frac{π}{2})$ is

$0$ .

$5 \cdot 0^{2}$

Step 1.4.2.2.2

Raising

$0$ to any positive power yields

$0$ .

$5 \cdot 0$

Step 1.4.2.2.3

Multiply

$5$ by

$0$ .

$0$

Step 1.4.3

List all of the points.

$(0 + π n, 5), (\frac{π}{2} + π n, 0)$ , for any integer

$n$

$(0 + π n, 5), (\frac{π}{2} + π n, 0)$ , for any integer

$n$

$(0 + π n, 5), (\frac{π}{2} + π n, 0)$ , for any integer

$n$

Step 2

Exclude the points that are not on the interval.

$(\frac{π}{2}, 0)$

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.

$(- \infty, 0) \cup (0, \frac{π}{2}) \cup (\frac{π}{2}, π) \cup (π, \infty)$

Step 3.2

Substitute any number, such as

$- 2$ , from the interval

$(- \infty, 0)$ in the first derivative

$- 10 \cos (x) \sin (x)$ to check if the result is negative or positive.

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Step 3.2.1

Replace the variable

$x$ with

$- 2$ in the expression.

$f' (- 2) = - 10 \cos (- 2) \sin (- 2)$

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Evaluate

$\cos (- 2)$ .

$f' (- 2) = - 10 \cdot (- 0.41614683 \sin (- 2))$

Step 3.2.2.2

Multiply

$- 10$ by

$- 0.41614683$ .

$f' (- 2) = 4.16146836 \sin (- 2)$

Step 3.2.2.3

Evaluate

$\sin (- 2)$ .

$f' (- 2) = 4.16146836 \cdot - 0.90929742$

Step 3.2.2.4

Multiply

$4.16146836$ by

$- 0.90929742$ .

$f' (- 2) = - 3.78401247$

Step 3.2.2.5

The final answer is

$- 3.78401247$ .

$- 3.78401247$

Step 3.3

Substitute any number, such as

$1$ , from the interval

$(0, \frac{π}{2})$ in the first derivative

$- 10 \cos (x) \sin (x)$ to check if the result is negative or positive.

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Step 3.3.1

Replace the variable

$x$ with

$1$ in the expression.

$f' (1) = - 10 \cos (1) \sin (1)$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Evaluate

$\cos (1)$ .

$f' (1) = - 10 \cdot (0.5403023 \sin (1))$

Step 3.3.2.2

Multiply

$- 10$ by

$0.5403023$ .

$f' (1) = - 5.40302305 \sin (1)$

Step 3.3.2.3

Evaluate

$\sin (1)$ .

$f' (1) = - 5.40302305 \cdot 0.84147098$

Step 3.3.2.4

Multiply

$- 5.40302305$ by

$0.84147098$ .

$f' (1) = - 4.54648713$

Step 3.3.2.5

The final answer is

$- 4.54648713$ .

$- 4.54648713$

Step 3.4

Substitute any number, such as

$2$ , from the interval

$(\frac{π}{2}, π)$ in the first derivative

$- 10 \cos (x) \sin (x)$ to check if the result is negative or positive.

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Step 3.4.1

Replace the variable

$x$ with

$2$ in the expression.

$f' (2) = - 10 \cos (2) \sin (2)$

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Evaluate

$\cos (2)$ .

$f' (2) = - 10 \cdot (- 0.41614683 \sin (2))$

Step 3.4.2.2

Multiply

$- 10$ by

$- 0.41614683$ .

$f' (2) = 4.16146836 \sin (2)$

Step 3.4.2.3

Evaluate

$\sin (2)$ .

$f' (2) = 4.16146836 \cdot 0.90929742$

Step 3.4.2.4

Multiply

$4.16146836$ by

$0.90929742$ .

$f' (2) = 3.78401247$

Step 3.4.2.5

The final answer is

$3.78401247$ .

$3.78401247$

Step 3.5

Substitute any number, such as

$6$ , from the interval

$(π, \infty)$ in the first derivative

$- 10 \cos (x) \sin (x)$ to check if the result is negative or positive.

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Step 3.5.1

Replace the variable

$x$ with

$6$ in the expression.

$f' (6) = - 10 \cos (6) \sin (6)$

Step 3.5.2

Simplify the result.

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Step 3.5.2.1

Evaluate

$\cos (6)$ .

$f' (6) = - 10 \cdot (0.96017028 \sin (6))$

Step 3.5.2.2

Multiply

$- 10$ by

$0.96017028$ .

$f' (6) = - 9.60170286 \sin (6)$

Step 3.5.2.3

Evaluate

$\sin (6)$ .

$f' (6) = - 9.60170286 \cdot - 0.27941549$

Step 3.5.2.4

Multiply

$- 9.60170286$ by

$- 0.27941549$ .

$f' (6) = 2.68286459$

Step 3.5.2.5

The final answer is

$2.68286459$ .

$2.68286459$

Step 3.6

Since the first derivative did not change signs around

$x = 0$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 3.7

Since the first derivative changed signs from negative to positive around

$x = \frac{π}{2}$ , then

$x = \frac{π}{2}$ is a local minimum.

$x = \frac{π}{2}$ is a local minimum

Step 3.8

Since the first derivative did not change signs around

$x = π$ , this is not a local maximum or minimum.

Not a local maximum or minimum

Step 3.9

These are the local extrema for

$f (x) = 5 \cos^{2} (x)$ .

$x = \frac{π}{2}$ 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.

No absolute maximum

Absolute Minimum:

$(\frac{π}{2}, 0)$

Step 5