Calculus Examples

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

Step 1

Find the first derivative of the function.

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Step 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 1.2

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

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Step 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 1.2.1.1

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 1.2.1.2

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 1.2.1.3

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

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

Step 1.2.2

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

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

Step 1.3

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

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

Step 1.4

Simplify.

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Step 1.4.1

Reorder terms.

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

Step 1.4.2

Simplify each term.

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Step 1.4.2.1

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

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

Step 1.4.2.2

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

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

Step 1.4.2.3

Apply the sine double-angle identity.

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

Step 2

Find the second derivative of the function.

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Step 2.1

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

$f'' (x) = \frac{d}{d x} (\sin (2 x)) + \frac{d}{d x} (- \sin (x))$

Step 2.2

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

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Step 2.2.1

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

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Step 2.2.1.1

To apply the Chain Rule, set $u$ as $2 x$ .

$f'' (x) = \frac{d}{d u} (\sin (u)) \frac{d}{d x} (2 x) + \frac{d}{d x} (- \sin (x))$

Step 2.2.1.2

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

$f'' (x) = \cos (u) \frac{d}{d x} (2 x) + \frac{d}{d x} (- \sin (x))$

Step 2.2.1.3

Replace all occurrences of $u$ with $2 x$ .

$f'' (x) = \cos (2 x) \frac{d}{d x} (2 x) + \frac{d}{d x} (- \sin (x))$

Step 2.2.2

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

$f'' (x) = \cos (2 x) (2 \frac{d}{d x} (x)) + \frac{d}{d x} (- \sin (x))$

Step 2.2.3

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

$f'' (x) = \cos (2 x) (2 \cdot 1) + \frac{d}{d x} (- \sin (x))$

Step 2.2.4

Multiply $2$ by $1$ .

$f'' (x) = \cos (2 x) \cdot 2 + \frac{d}{d x} (- \sin (x))$

Step 2.2.5

Move $2$ to the left of $\cos (2 x)$ .

$f'' (x) = 2 \cos (2 x) + \frac{d}{d x} (- \sin (x))$

Step 2.3

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

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Step 2.3.1

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

$f'' (x) = 2 \cos (2 x) - \frac{d}{d x} \sin (x)$

Step 2.3.2

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

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

Step 3

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

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

Step 4

Apply the sine double-angle identity.

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

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 6

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 7

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

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Step 7.1

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

$\sin (x) = 0$

Step 7.2

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

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Step 7.2.1

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

$x = \arcsin (0)$

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

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

$x = 0$

Step 7.2.3

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 7.2.4

Subtract $0$ from $π$ .

$x = π$

Step 7.2.5

The solution to the equation $x = 0$ .

$x = 0, π$

Step 8

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

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Step 8.1

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

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

Step 8.2

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

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Step 8.2.1

Add $1$ to both sides of the equation.

$2 \cos (x) = 1$

Step 8.2.2

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

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Step 8.2.2.1

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

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

Step 8.2.2.2

Simplify the left side.

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Step 8.2.2.2.1

Cancel the common factor of $2$ .

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Step 8.2.2.2.1.1

Cancel the common factor.

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

Step 8.2.2.2.1.2

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

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

Step 8.2.3

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

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

Step 8.2.4

Simplify the right side.

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Step 8.2.4.1

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

$x = \frac{π}{3}$

Step 8.2.5

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 8.2.6

Simplify $2 π - \frac{π}{3}$ .

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Step 8.2.6.1

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 8.2.6.2

Combine fractions.

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Step 8.2.6.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 8.2.6.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 8.2.6.3

Simplify the numerator.

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Step 8.2.6.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 8.2.6.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 8.2.7

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

$x = \frac{π}{3}, \frac{5 π}{3}$

Step 9

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

$x = 0, π, \frac{π}{3}, \frac{5 π}{3}$

Step 10

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \cos (2 (0)) - \cos (0)$

Step 11

Evaluate the second derivative.

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Step 11.1

Simplify each term.

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Step 11.1.1

Multiply $2$ by $0$ .

$2 \cos (0) - \cos (0)$

Step 11.1.2

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

$2 \cdot 1 - \cos (0)$

Step 11.1.3

Multiply $2$ by $1$ .

$2 - \cos (0)$

Step 11.1.4

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

$2 - 1 \cdot 1$

Step 11.1.5

Multiply $- 1$ by $1$ .

$2 - 1$

Step 11.2

Subtract $1$ from $2$ .

$1$

Step 12

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 13

Find the y-value when $x = 0$ .

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Step 13.1

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

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

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

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

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

Step 13.2.1.2

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

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

Step 13.2.1.3

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

$f (0) = 0 + 1$

Step 13.2.2

Add $0$ and $1$ .

$f (0) = 1$

Step 13.2.3

The final answer is $1$ .

$y = 1$

Step 14

Evaluate the second derivative at $x = π$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \cos (2 (π)) - \cos (π)$

Step 15

Evaluate the second derivative.

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Step 15.1

Simplify each term.

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Step 15.1.1

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$2 \cos (0) - \cos (π)$

Step 15.1.2

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

$2 \cdot 1 - \cos (π)$

Step 15.1.3

Multiply $2$ by $1$ .

$2 - \cos (π)$

Step 15.1.4

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.

$2 - - \cos (0)$

Step 15.1.5

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

$2 - (- 1 \cdot 1)$

Step 15.1.6

Multiply $- (- 1 \cdot 1)$ .

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Step 15.1.6.1

Multiply $- 1$ by $1$ .

$2 - - 1$

Step 15.1.6.2

Multiply $- 1$ by $- 1$ .

$2 + 1$

Step 15.2

Add $2$ and $1$ .

$3$

Step 16

$x = π$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = π$ is a local minimum

Step 17

Find the y-value when $x = π$ .

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Step 17.1

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

$f (π) = \sin^{2} (π) + \cos (π)$

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.

$f (π) = \sin^{2} (0) + \cos (π)$

Step 17.2.1.2

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

$f (π) = 0^{2} + \cos (π)$

Step 17.2.1.3

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

$f (π) = 0 + \cos (π)$

Step 17.2.1.4

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.

$f (π) = 0 - \cos (0)$

Step 17.2.1.5

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

$f (π) = 0 - 1 \cdot 1$

Step 17.2.1.6

Multiply $- 1$ by $1$ .

$f (π) = 0 - 1$

Step 17.2.2

Subtract $1$ from $0$ .

$f (π) = - 1$

Step 17.2.3

The final answer is $- 1$ .

$y = - 1$

Step 18

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

$2 \cos (2 (\frac{π}{3})) - \cos (\frac{π}{3})$

Step 19

Evaluate the second derivative.

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Step 19.1

Simplify each term.

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Step 19.1.1

Combine $2$ and $\frac{π}{3}$ .

$2 \cos (\frac{2 π}{3}) - \cos (\frac{π}{3})$

Step 19.1.2

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.

$2 (- \cos (\frac{π}{3})) - \cos (\frac{π}{3})$

Step 19.1.3

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

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

Step 19.1.4

Cancel the common factor of $2$ .

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Step 19.1.4.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Step 19.1.4.2

Cancel the common factor.

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

Step 19.1.4.3

Rewrite the expression.

$- 1 - \cos (\frac{π}{3})$

Step 19.1.5

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

$- 1 - \frac{1}{2}$

Step 19.2

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

$- 1 \cdot \frac{2}{2} - \frac{1}{2}$

Step 19.3

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

$\frac{- 1 \cdot 2}{2} - \frac{1}{2}$

Step 19.4

Combine the numerators over the common denominator.

$\frac{- 1 \cdot 2 - 1}{2}$

Step 19.5

Simplify the numerator.

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Step 19.5.1

Multiply $- 1$ by $2$ .

$\frac{- 2 - 1}{2}$

Step 19.5.2

Subtract $1$ from $- 2$ .

$\frac{- 3}{2}$

Step 19.6

Move the negative in front of the fraction.

$- \frac{3}{2}$

Step 20

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

$x = \frac{π}{3}$ is a local maximum

Step 21

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

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Step 21.1

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

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

Step 21.2

Simplify the result.

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Step 21.2.1

Simplify each term.

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Step 21.2.1.1

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

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

Step 21.2.1.2

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

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

Step 21.2.1.3

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

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Step 21.2.1.3.1

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

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

Step 21.2.1.3.2

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

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

Step 21.2.1.3.3

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

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

Step 21.2.1.3.4

Cancel the common factor of $2$ .

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Step 21.2.1.3.4.1

Cancel the common factor.

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

Step 21.2.1.3.4.2

Rewrite the expression.

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

Step 21.2.1.3.5

Evaluate the exponent.

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

Step 21.2.1.4

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

$f (\frac{π}{3}) = \frac{3}{4} + \cos (\frac{π}{3})$

Step 21.2.1.5

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

$f (\frac{π}{3}) = \frac{3}{4} + \frac{1}{2}$

Step 21.2.2

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

$f (\frac{π}{3}) = \frac{3}{4} + \frac{1}{2} \cdot \frac{2}{2}$

Step 21.2.3

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

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Step 21.2.3.1

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

$f (\frac{π}{3}) = \frac{3}{4} + \frac{2}{2 \cdot 2}$

Step 21.2.3.2

Multiply $2$ by $2$ .

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

Step 21.2.4

Combine the numerators over the common denominator.

$f (\frac{π}{3}) = \frac{3 + 2}{4}$

Step 21.2.5

Add $3$ and $2$ .

$f (\frac{π}{3}) = \frac{5}{4}$

Step 21.2.6

The final answer is $\frac{5}{4}$ .

$y = \frac{5}{4}$

Step 22

Evaluate the second derivative at $x = \frac{5 π}{3}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$2 \cos (2 (\frac{5 π}{3})) - \cos (\frac{5 π}{3})$

Step 23

Evaluate the second derivative.

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Step 23.1

Simplify each term.

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Step 23.1.1

Multiply $2 (\frac{5 π}{3})$ .

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Step 23.1.1.1

Combine $2$ and $\frac{5 π}{3}$ .

$2 \cos (\frac{2 (5 π)}{3}) - \cos (\frac{5 π}{3})$

Step 23.1.1.2

Multiply $5$ by $2$ .

$2 \cos (\frac{10 π}{3}) - \cos (\frac{5 π}{3})$

Step 23.1.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$2 \cos (\frac{4 π}{3}) - \cos (\frac{5 π}{3})$

Step 23.1.3

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 third quadrant.

$2 (- \cos (\frac{π}{3})) - \cos (\frac{5 π}{3})$

Step 23.1.4

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

$2 (- \frac{1}{2}) - \cos (\frac{5 π}{3})$

Step 23.1.5

Cancel the common factor of $2$ .

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Step 23.1.5.1

Move the leading negative in $- \frac{1}{2}$ into the numerator.

$2 (\frac{- 1}{2}) - \cos (\frac{5 π}{3})$

Step 23.1.5.2

Cancel the common factor.

$2 (\frac{- 1}{2}) - \cos (\frac{5 π}{3})$

Step 23.1.5.3

Rewrite the expression.

$- 1 - \cos (\frac{5 π}{3})$

Step 23.1.6

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

$- 1 - \cos (\frac{π}{3})$

Step 23.1.7

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

$- 1 - \frac{1}{2}$

Step 23.2

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

$- 1 \cdot \frac{2}{2} - \frac{1}{2}$

Step 23.3

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

$\frac{- 1 \cdot 2}{2} - \frac{1}{2}$

Step 23.4

Combine the numerators over the common denominator.

$\frac{- 1 \cdot 2 - 1}{2}$

Step 23.5

Simplify the numerator.

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Step 23.5.1

Multiply $- 1$ by $2$ .

$\frac{- 2 - 1}{2}$

Step 23.5.2

Subtract $1$ from $- 2$ .

$\frac{- 3}{2}$

Step 23.6

Move the negative in front of the fraction.

$- \frac{3}{2}$

Step 24

$x = \frac{5 π}{3}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{5 π}{3}$ is a local maximum

Step 25

Find the y-value when $x = \frac{5 π}{3}$ .

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Step 25.1

Replace the variable $x$ with $\frac{5 π}{3}$ in the expression.

$f (\frac{5 π}{3}) = \sin^{2} (\frac{5 π}{3}) + \cos (\frac{5 π}{3})$

Step 25.2

Simplify the result.

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Step 25.2.1

Simplify each term.

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Step 25.2.1.1

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.

$f (\frac{5 π}{3}) = {(- \sin (\frac{π}{3}))}^{2} + \cos (\frac{5 π}{3})$

Step 25.2.1.2

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

$f (\frac{5 π}{3}) = {(- \frac{\sqrt{3}}{2})}^{2} + \cos (\frac{5 π}{3})$

Step 25.2.1.3

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 25.2.1.3.1

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

$f (\frac{5 π}{3}) = {(- 1)}^{2} {(\frac{\sqrt{3}}{2})}^{2} + \cos (\frac{5 π}{3})$

Step 25.2.1.3.2

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

$f (\frac{5 π}{3}) = {(- 1)}^{2} (\frac{{\sqrt{3}}^{2}}{2^{2}}) + \cos (\frac{5 π}{3})$

Step 25.2.1.4

Raise $- 1$ to the power of $2$ .

$f (\frac{5 π}{3}) = 1 (\frac{{\sqrt{3}}^{2}}{2^{2}}) + \cos (\frac{5 π}{3})$

Step 25.2.1.5

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

$f (\frac{5 π}{3}) = \frac{{\sqrt{3}}^{2}}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.6

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

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Step 25.2.1.6.1

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

$f (\frac{5 π}{3}) = \frac{{(3^{\frac{1}{2}})}^{2}}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.6.2

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

$f (\frac{5 π}{3}) = \frac{3^{\frac{1}{2} \cdot 2}}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.6.3

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

$f (\frac{5 π}{3}) = \frac{3^{\frac{2}{2}}}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.6.4

Cancel the common factor of $2$ .

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Step 25.2.1.6.4.1

Cancel the common factor.

$f (\frac{5 π}{3}) = \frac{3^{\frac{2}{2}}}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.6.4.2

Rewrite the expression.

$f (\frac{5 π}{3}) = \frac{3}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.6.5

Evaluate the exponent.

$f (\frac{5 π}{3}) = \frac{3}{2^{2}} + \cos (\frac{5 π}{3})$

Step 25.2.1.7

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

$f (\frac{5 π}{3}) = \frac{3}{4} + \cos (\frac{5 π}{3})$

Step 25.2.1.8

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

$f (\frac{5 π}{3}) = \frac{3}{4} + \cos (\frac{π}{3})$

Step 25.2.1.9

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

$f (\frac{5 π}{3}) = \frac{3}{4} + \frac{1}{2}$

Step 25.2.2

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

$f (\frac{5 π}{3}) = \frac{3}{4} + \frac{1}{2} \cdot \frac{2}{2}$

Step 25.2.3

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

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Step 25.2.3.1

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

$f (\frac{5 π}{3}) = \frac{3}{4} + \frac{2}{2 \cdot 2}$

Step 25.2.3.2

Multiply $2$ by $2$ .

$f (\frac{5 π}{3}) = \frac{3}{4} + \frac{2}{4}$

Step 25.2.4

Combine the numerators over the common denominator.

$f (\frac{5 π}{3}) = \frac{3 + 2}{4}$

Step 25.2.5

Add $3$ and $2$ .

$f (\frac{5 π}{3}) = \frac{5}{4}$

Step 25.2.6

The final answer is $\frac{5}{4}$ .

$y = \frac{5}{4}$

Step 26

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

$(0, 1)$ is a local minima

$(π, - 1)$ is a local minima

$(\frac{π}{3}, \frac{5}{4})$ is a local maxima

$(\frac{5 π}{3}, \frac{5}{4})$ is a local maxima

Step 27