Calculus Examples

f (x) = {sin}^{2} (x)

$f (x) = \sin^{2} (x)$ 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

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) = sin (x)

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

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Step 1.1.1.1.1

To apply the Chain Rule, set

u

$u$ as

sin (x)

$\sin (x)$ .

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

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

Step 1.1.1.1.2

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

$n = 2$ .

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

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

Step 1.1.1.1.3

Replace all occurrences of

u

$u$ with

sin (x)

$\sin (x)$ .

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

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

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

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

Step 1.1.1.2

The derivative of

sin (x)

$\sin (x)$ with respect to

x

$x$ is

cos (x)

$\cos (x)$ .

2 sin (x) cos (x)

$2 \sin (x) \cos (x)$

Step 1.1.1.3

Simplify.

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Step 1.1.1.3.1

Reorder the factors of

2 sin (x) cos (x)

$2 \sin (x) \cos (x)$ .

2 cos (x) sin (x)

$2 \cos (x) \sin (x)$

Step 1.1.1.3.2

Reorder

2 cos (x)

$2 \cos (x)$ and

sin (x)

$\sin (x)$ .

sin (x) (2 cos (x))

$\sin (x) (2 \cos (x))$

Step 1.1.1.3.3

Reorder

sin (x)

$\sin (x)$ and

2

$2$ .

2 \cdot sin (x) cos (x)

$2 \cdot \sin (x) \cos (x)$

Step 1.1.1.3.4

Apply the sine double-angle identity.

f' (x) = sin (2 x)

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

f' (x) = sin (2 x)

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

f' (x) = sin (2 x)

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

Step 1.1.2

The first derivative of

f (x)

$f (x)$ with respect to

x

$x$ is

sin (2 x)

$\sin (2 x)$ .

sin (2 x)

$\sin (2 x)$

sin (2 x)

$\sin (2 x)$

Step 1.2

Set the first derivative equal to

0

$0$ then solve the equation

sin (2 x) = 0

$\sin (2 x) = 0$ .

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Step 1.2.1

Set the first derivative equal to

0

$0$ .

sin (2 x) = 0

$\sin (2 x) = 0$

Step 1.2.2

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

x

$x$ from inside the sine.

2 x = arcsin (0)

$2 x = \arcsin (0)$

Step 1.2.3

Simplify the right side.

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Step 1.2.3.1

The exact value of

arcsin (0)

$\arcsin (0)$ is

0

$0$ .

2 x = 0

$2 x = 0$

2 x = 0

$2 x = 0$

Step 1.2.4

Divide each term in

2 x = 0

$2 x = 0$ by

2

$2$ and simplify.

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Step 1.2.4.1

Divide each term in

2 x = 0

$2 x = 0$ by

2

$2$ .

\frac{2 x}{2} = \frac{0}{2}

$\frac{2 x}{2} = \frac{0}{2}$

Step 1.2.4.2

Simplify the left side.

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Step 1.2.4.2.1

Cancel the common factor of

2

$2$ .

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Step 1.2.4.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{0}{2}$

Step 1.2.4.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{0}{2}$

Step 1.2.4.3

Simplify the right side.

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Step 1.2.4.3.1

Divide

$0$ by

$2$ .

$x = 0$

Step 1.2.5

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.

$2 x = π - 0$

Step 1.2.6

Solve for

$x$ .

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Step 1.2.6.1

Simplify.

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Step 1.2.6.1.1

Multiply

$- 1$ by

$0$ .

$2 x = π + 0$

Step 1.2.6.1.2

Add

$π$ and

$0$ .

$2 x = π$

Step 1.2.6.2

Divide each term in

$2 x = π$ by

$2$ and simplify.

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Step 1.2.6.2.1

Divide each term in

$2 x = π$ by

$2$ .

$\frac{2 x}{2} = \frac{π}{2}$

Step 1.2.6.2.2

Simplify the left side.

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Step 1.2.6.2.2.1

Cancel the common factor of

$2$ .

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Step 1.2.6.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{π}{2}$

Step 1.2.6.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{π}{2}$

Step 1.2.7

Find the period of

$\sin (2 x)$ .

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Step 1.2.7.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 1.2.7.2

Replace

$b$ with

$2$ in the formula for period.

$\frac{2 π}{| 2 |}$

Step 1.2.7.3

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

$0$ and

$2$ is

$2$ .

$\frac{2 π}{2}$

Step 1.2.7.4

Cancel the common factor of

$2$ .

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Step 1.2.7.4.1

Cancel the common factor.

$\frac{2 π}{2}$

Step 1.2.7.4.2

Divide

$π$ by

$1$ .

$π$

Step 1.2.8

The period of the

$\sin (2 x)$ function is

$π$ so values will repeat every

$π$ radians in both directions.

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

$n$

Step 1.2.9

Consolidate the answers.

$x = \frac{π n}{2}$ , for any integer

$n$

$x = \frac{π n}{2}$ , 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

$\sin^{2} (x)$ at each

$x$ value where the derivative is

$0$ or undefined.

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Step 1.4.1

Evaluate at

$x = 0$ .

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Step 1.4.1.1

Substitute

$0$ for

$x$ .

$\sin^{2} (0)$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

The exact value of

$\sin (0)$ is

$0$ .

$0^{2}$

Step 1.4.1.2.2

Raising

$0$ to any positive power yields

$0$ .

$0$

Step 1.4.2

Evaluate at

$x = \frac{π}{2}$ .

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Step 1.4.2.1

Substitute

$\frac{π}{2}$ for

$x$ .

$\sin^{2} (\frac{π}{2})$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

The exact value of

$\sin (\frac{π}{2})$ is

$1$ .

$1^{2}$

Step 1.4.2.2.2

One to any power is one.

$1$

Step 1.4.3

List all of the points.

$(0 + π n, 0), (\frac{π}{2} + π n, 1)$ , for any integer

$n$

$(0 + π n, 0), (\frac{π}{2} + π n, 1)$ , for any integer

$n$

$(0 + π n, 0), (\frac{π}{2} + π n, 1)$ , for any integer

$n$

Step 2

Exclude the points that are not on the interval.

$(\frac{π}{2}, 1)$

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

$\sin (2 x)$ 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.

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

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Multiply

$2$ by

$- 2$ .

$f' (- 2) = \sin (- 4)$

Step 3.2.2.2

Evaluate

$\sin (- 4)$ .

$f' (- 2) = 0.75680249$

Step 3.2.2.3

The final answer is

$0.75680249$ .

$0.75680249$

Step 3.3

Substitute any number, such as

$1$ , from the interval

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

$\sin (2 x)$ 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.

$f' (1) = \sin (2 (1))$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Multiply

$2$ by

$1$ .

$f' (1) = \sin (2)$

Step 3.3.2.2

Evaluate

$\sin (2)$ .

$f' (1) = 0.90929742$

Step 3.3.2.3

The final answer is

$0.90929742$ .

$0.90929742$

Step 3.4

Substitute any number, such as

$2$ , from the interval

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

$\sin (2 x)$ 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.

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

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Multiply

$2$ by

$2$ .

$f' (2) = \sin (4)$

Step 3.4.2.2

Evaluate

$\sin (4)$ .

$f' (2) = - 0.75680249$

Step 3.4.2.3

The final answer is

$- 0.75680249$ .

$- 0.75680249$

Step 3.5

Substitute any number, such as

$6$ , from the interval

$(π, \infty)$ in the first derivative

$\sin (2 x)$ 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.

$f' (6) = \sin (2 (6))$

Step 3.5.2

Simplify the result.

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Step 3.5.2.1

Multiply

$2$ by

$6$ .

$f' (6) = \sin (12)$

Step 3.5.2.2

Evaluate

$\sin (12)$ .

$f' (6) = - 0.53657291$

Step 3.5.2.3

The final answer is

$- 0.53657291$ .

$- 0.53657291$

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 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 3.8

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

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

$x = \frac{π}{2}$ is a local maximum

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:

$(\frac{π}{2}, 1)$

No absolute minimum

Step 5