Calculus Examples

$h (x) = \sin^{2} (x) + \cos (x)$ , $0 < x < 2 π$

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

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sin (x)$ .

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Step 1.1.1.2.1.1

To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

Step 1.1.1.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 2$ .

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

Step 1.1.1.2.1.3

Replace all occurrences of $u$ with $\sin (x)$ .

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

Step 1.1.1.2.2

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

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

Step 1.1.1.3

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

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

Step 1.1.1.4

Simplify.

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Step 1.1.1.4.1

Reorder terms.

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

Step 1.1.1.4.2

Simplify each term.

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Step 1.1.1.4.2.1

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

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

Step 1.1.1.4.2.2

Reorder $\sin (x)$ and $2$ .

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

Step 1.1.1.4.2.3

Apply the sine double-angle identity.

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

Step 1.1.2

The first derivative of $h (x)$ with respect to $x$ is $\sin (2 x) - \sin (x)$ .

$\sin (2 x) - \sin (x)$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $\sin (2 x) - \sin (x) = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$\sin (2 x) - \sin (x) = 0$

Step 1.2.2

Apply the sine double-angle identity.

$2 \sin (x) \cos (x) - \sin (x) = 0$

Step 1.2.3

Factor $\sin (x)$ out of $2 \sin (x) \cos (x) - \sin (x)$ .

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Step 1.2.3.1

Factor $\sin (x)$ out of $2 \sin (x) \cos (x)$ .

$\sin (x) (2 \cos (x)) - \sin (x) = 0$

Step 1.2.3.2

Factor $\sin (x)$ out of $- \sin (x)$ .

$\sin (x) (2 \cos (x)) + \sin (x) \cdot - 1 = 0$

Step 1.2.3.3

Factor $\sin (x)$ out of $\sin (x) (2 \cos (x)) + \sin (x) \cdot - 1$ .

$\sin (x) (2 \cos (x) - 1) = 0$

Step 1.2.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\sin (x) = 0$

$2 \cos (x) - 1 = 0$

Step 1.2.5

Set $\sin (x)$ equal to $0$ and solve for $x$ .

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Step 1.2.5.1

Set $\sin (x)$ equal to $0$ .

$\sin (x) = 0$

Step 1.2.5.2

Solve $\sin (x) = 0$ for $x$ .

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Step 1.2.5.2.1

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

$x = \arcsin (0)$

Step 1.2.5.2.2

Simplify the right side.

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Step 1.2.5.2.2.1

The exact value of $\arcsin (0)$ is $0$ .

$x = 0$

Step 1.2.5.2.3

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.

$x = π - 0$

Step 1.2.5.2.4

Subtract $0$ from $π$ .

$x = π$

Step 1.2.5.2.5

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

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Step 1.2.5.2.5.1

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

$\frac{2 π}{| b |}$

Step 1.2.5.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 1.2.5.2.5.3

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

$\frac{2 π}{1}$

Step 1.2.5.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.5.2.6

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

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

Step 1.2.6

Set $2 \cos (x) - 1$ equal to $0$ and solve for $x$ .

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Step 1.2.6.1

Set $2 \cos (x) - 1$ equal to $0$ .

$2 \cos (x) - 1 = 0$

Step 1.2.6.2

Solve $2 \cos (x) - 1 = 0$ for $x$ .

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Step 1.2.6.2.1

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Step 1.2.6.2.2

Divide each term in $2 \cos (x) = 1$ by $2$ and simplify.

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Step 1.2.6.2.2.1

Divide each term in $2 \cos (x) = 1$ by $2$ .

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

Step 1.2.6.2.2.2

Simplify the left side.

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Step 1.2.6.2.2.2.1

Cancel the common factor of $2$ .

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Step 1.2.6.2.2.2.1.1

Cancel the common factor.

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

Step 1.2.6.2.2.2.1.2

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

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

Step 1.2.6.2.3

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

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

Step 1.2.6.2.4

Simplify the right side.

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Step 1.2.6.2.4.1

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

$x = \frac{π}{3}$

Step 1.2.6.2.5

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.

$x = 2 π - \frac{π}{3}$

Step 1.2.6.2.6

Simplify $2 π - \frac{π}{3}$ .

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Step 1.2.6.2.6.1

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

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

Step 1.2.6.2.6.2

Combine fractions.

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Step 1.2.6.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 1.2.6.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 1.2.6.2.6.3

Simplify the numerator.

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Step 1.2.6.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 1.2.6.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 1.2.6.2.7

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

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Step 1.2.6.2.7.1

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

$\frac{2 π}{| b |}$

Step 1.2.6.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 1.2.6.2.7.3

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

$\frac{2 π}{1}$

Step 1.2.6.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 1.2.6.2.8

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

$x = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 1.2.7

The final solution is all the values that make $\sin (x) (2 \cos (x) - 1) = 0$ true.

$x = 2 π n, π + 2 π n, \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

Step 1.2.8

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

$x = π n, \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , 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 $\sin^{2} (x) + \cos (x)$ 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 $x$ .

$\sin^{2} (0) + \cos (0)$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

Simplify each term.

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Step 1.4.1.2.1.1

The exact value of $\sin (0)$ is $0$ .

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

Step 1.4.1.2.1.2

Raising $0$ to any positive power yields $0$ .

$0 + \cos (0)$

Step 1.4.1.2.1.3

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

$0 + 1$

Step 1.4.1.2.2

Add $0$ and $1$ .

$1$

Step 1.4.2

Evaluate at $x = π$ .

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Step 1.4.2.1

Substitute $π$ for $x$ .

$\sin^{2} (π) + \cos (π)$

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.

$\sin^{2} (0) + \cos (π)$

Step 1.4.2.2.1.2

The exact value of $\sin (0)$ is $0$ .

$0^{2} + \cos (π)$

Step 1.4.2.2.1.3

Raising $0$ to any positive power yields $0$ .

$0 + \cos (π)$

Step 1.4.2.2.1.4

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

$0 - \cos (0)$

Step 1.4.2.2.1.5

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

$0 - 1 \cdot 1$

Step 1.4.2.2.1.6

Multiply $- 1$ by $1$ .

$0 - 1$

Step 1.4.2.2.2

Subtract $1$ from $0$ .

$- 1$

Step 1.4.3

Evaluate at $x = \frac{π}{3}$ .

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Step 1.4.3.1

Substitute $\frac{π}{3}$ for $x$ .

$\sin^{2} (\frac{π}{3}) + \cos (\frac{π}{3})$

Step 1.4.3.2

Simplify.

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Step 1.4.3.2.1

Simplify each term.

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Step 1.4.3.2.1.1

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

${(\frac{\sqrt{3}}{2})}^{2} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.2

Apply the product rule to $\frac{\sqrt{3}}{2}$ .

$\frac{{\sqrt{3}}^{2}}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.3

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 1.4.3.2.1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\frac{{(3^{\frac{1}{2}})}^{2}}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.3.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3^{\frac{1}{2} \cdot 2}}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.3.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{3^{\frac{2}{2}}}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.3.4

Cancel the common factor of $2$ .

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Step 1.4.3.2.1.3.4.1

Cancel the common factor.

$\frac{3^{\frac{2}{2}}}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.3.4.2

Rewrite the expression.

$\frac{3^{1}}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.3.5

Evaluate the exponent.

$\frac{3}{2^{2}} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.4

Raise $2$ to the power of $2$ .

$\frac{3}{4} + \cos (\frac{π}{3})$

Step 1.4.3.2.1.5

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

$\frac{3}{4} + \frac{1}{2}$

Step 1.4.3.2.2

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{3}{4} + \frac{1}{2} \cdot \frac{2}{2}$

Step 1.4.3.2.3

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

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Step 1.4.3.2.3.1

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

$\frac{3}{4} + \frac{2}{2 \cdot 2}$

Step 1.4.3.2.3.2

Multiply $2$ by $2$ .

$\frac{3}{4} + \frac{2}{4}$

Step 1.4.3.2.4

Combine the numerators over the common denominator.

$\frac{3 + 2}{4}$

Step 1.4.3.2.5

Add $3$ and $2$ .

$\frac{5}{4}$

Step 1.4.4

List all of the points.

$(0 + 2 π n, 1), (π + 2 π n, - 1), (\frac{π}{3} + 2 π n, \frac{5}{4})$ , for any integer $n$

Step 2

Exclude the points that are not on the interval.

$(π, - 1), (\frac{π}{3}, \frac{5}{4})$

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.

$(- \infty, 0) \cup (0, \frac{π}{3}) \cup (\frac{π}{3}, π) \cup (π, 2 π) \cup (2 π, \infty)$

Step 3.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $\sin (2 x) - \sin (x)$ 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.

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

Step 3.2.2

Simplify the result.

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Step 3.2.2.1

Simplify each term.

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Step 3.2.2.1.1

Multiply $2$ by $- 2$ .

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

Step 3.2.2.1.2

Evaluate $\sin (- 4)$ .

$h' (- 2) = 0.75680249 - \sin (- 2)$

Step 3.2.2.1.3

Evaluate $\sin (- 2)$ .

$h' (- 2) = 0.75680249 + 0.90929742$

Step 3.2.2.1.4

Multiply $- 1$ by $- 0.90929742$ .

$h' (- 2) = 0.75680249 + 0.90929742$

Step 3.2.2.2

Add $0.75680249$ and $0.90929742$ .

$h' (- 2) = 1.66609992$

Step 3.2.2.3

The final answer is $1.66609992$ .

$1.66609992$

Step 3.3

Substitute any number, such as $1$ , from the interval $(0, \frac{π}{3})$ in the first derivative $\sin (2 x) - \sin (x)$ 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.

$h' (1) = \sin (2 (1)) - \sin (1)$

Step 3.3.2

Simplify the result.

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Step 3.3.2.1

Simplify each term.

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Step 3.3.2.1.1

Multiply $2$ by $1$ .

$h' (1) = \sin (2) - \sin (1)$

Step 3.3.2.1.2

Evaluate $\sin (2)$ .

$h' (1) = 0.90929742 - \sin (1)$

Step 3.3.2.1.3

Evaluate $\sin (1)$ .

$h' (1) = 0.90929742 - 1 \cdot 0.84147098$

Step 3.3.2.1.4

Multiply $- 1$ by $0.84147098$ .

$h' (1) = 0.90929742 - 0.84147098$

Step 3.3.2.2

Subtract $0.84147098$ from $0.90929742$ .

$h' (1) = 0.06782644$

Step 3.3.2.3

The final answer is $0.06782644$ .

$0.06782644$

Step 3.4

Substitute any number, such as $2$ , from the interval $(\frac{π}{3}, π)$ in the first derivative $\sin (2 x) - \sin (x)$ 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.

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

Step 3.4.2

Simplify the result.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Multiply $2$ by $2$ .

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

Step 3.4.2.1.2

Evaluate $\sin (4)$ .

$h' (2) = - 0.75680249 - \sin (2)$

Step 3.4.2.1.3

Evaluate $\sin (2)$ .

$h' (2) = - 0.75680249 - 1 \cdot 0.90929742$

Step 3.4.2.1.4

Multiply $- 1$ by $0.90929742$ .

$h' (2) = - 0.75680249 - 0.90929742$

Step 3.4.2.2

Subtract $0.90929742$ from $- 0.75680249$ .

$h' (2) = - 1.66609992$

Step 3.4.2.3

The final answer is $- 1.66609992$ .

$- 1.66609992$

Step 3.5

Substitute any number, such as $5$ , from the interval $(π, 2 π)$ in the first derivative $\sin (2 x) - \sin (x)$ to check if the result is negative or positive.

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Step 3.5.1

Replace the variable $x$ with $5$ in the expression.

$h' (5) = \sin (2 (5)) - \sin (5)$

Step 3.5.2

Simplify the result.

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Step 3.5.2.1

Simplify each term.

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Step 3.5.2.1.1

Multiply $2$ by $5$ .

$h' (5) = \sin (10) - \sin (5)$

Step 3.5.2.1.2

Evaluate $\sin (10)$ .

$h' (5) = - 0.54402111 - \sin (5)$

Step 3.5.2.1.3

Evaluate $\sin (5)$ .

$h' (5) = - 0.54402111 + 0.95892427$

Step 3.5.2.1.4

Multiply $- 1$ by $- 0.95892427$ .

$h' (5) = - 0.54402111 + 0.95892427$

Step 3.5.2.2

Add $- 0.54402111$ and $0.95892427$ .

$h' (5) = 0.41490316$

Step 3.5.2.3

The final answer is $0.41490316$ .

$0.41490316$

Step 3.6

Substitute any number, such as $9$ , from the interval $(2 π, \infty)$ in the first derivative $\sin (2 x) - \sin (x)$ to check if the result is negative or positive.

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Step 3.6.1

Replace the variable $x$ with $9$ in the expression.

$h' (9) = \sin (2 (9)) - \sin (9)$

Step 3.6.2

Simplify the result.

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Step 3.6.2.1

Simplify each term.

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Step 3.6.2.1.1

Multiply $2$ by $9$ .

$h' (9) = \sin (18) - \sin (9)$

Step 3.6.2.1.2

Evaluate $\sin (18)$ .

$h' (9) = - 0.75098724 - \sin (9)$

Step 3.6.2.1.3

Evaluate $\sin (9)$ .

$h' (9) = - 0.75098724 - 1 \cdot 0.41211848$

Step 3.6.2.1.4

Multiply $- 1$ by $0.41211848$ .

$h' (9) = - 0.75098724 - 0.41211848$

Step 3.6.2.2

Subtract $0.41211848$ from $- 0.75098724$ .

$h' (9) = - 1.16310573$

Step 3.6.2.3

The final answer is $- 1.16310573$ .

$- 1.16310573$

Step 3.7

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 3.8

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

$x = \frac{π}{3}$ is a local maximum

Step 3.9

Since the first derivative changed signs from negative to positive around $x = π$ , then $x = π$ is a local minimum.

$x = π$ is a local minimum

Step 3.10

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

$x = 2 π$ is a local maximum

Step 3.11

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

$x = \frac{π}{3}$ is a local maximum

$x = π$ is a local minimum

$x = 2 π$ is a local maximum

$x = \frac{π}{3}$ is a local maximum

$x = π$ is a local minimum

$x = 2 π$ is a local maximum

Step 4

Compare the $h (x)$ 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 $h (x)$ value and the minimum will occur at the lowest $h (x)$ value.

Absolute Maximum: $(\frac{π}{3}, \frac{5}{4})$

Absolute Minimum: $(π, - 1)$

Step 5