Calculus Examples

$f (x) = \sin (2 x) + \cos (2 x)$

Step 1

Find the first derivative of the function.

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Step 1.1

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

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

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 $f (x) = \sin (x)$ and $g (x) = 2 x$ .

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Step 1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$\frac{d}{d u_{1}} [\sin (u_{1})] \frac{d}{d x} [2 x] + \frac{d}{d x} [\cos (2 x)]$

Step 1.2.1.2

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

$\cos (u_{1}) \frac{d}{d x} [2 x] + \frac{d}{d x} [\cos (2 x)]$

Step 1.2.1.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 1.2.2

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

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

Step 1.2.3

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

$\cos (2 x) (2 \cdot 1) + \frac{d}{d x} [\cos (2 x)]$

Step 1.2.4

Multiply $2$ by $1$ .

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

Step 1.2.5

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

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

Step 1.3

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

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Step 1.3.1

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

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Step 1.3.1.1

To apply the Chain Rule, set $u_{2}$ as $2 x$ .

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

Step 1.3.1.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 1.3.1.3

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

$2 \cos (2 x) - \sin (2 x) (2 \cdot 1)$

Step 1.3.4

Multiply $2$ by $1$ .

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

Step 1.3.5

Multiply $2$ by $- 1$ .

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

Step 2

Find the second derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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Step 2.2.2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 2.2.2.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

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

Step 2.2.2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $2$ by $1$ .

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

Step 2.2.6

Multiply $2$ by $- 1$ .

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

Step 2.2.7

Multiply $- 2$ by $2$ .

$f'' (x) = - 4 \sin (2 x) + \frac{d}{d x} (- 2 \sin (2 x))$

Step 2.3

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

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Step 2.3.1

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

$f'' (x) = - 4 \sin (2 x) - 2 \frac{d}{d x} \sin (2 x)$

Step 2.3.2

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 2.3.2.1

To apply the Chain Rule, set $u_{2}$ as $2 x$ .

$f'' (x) = - 4 \sin (2 x) - 2 (\frac{d}{d u (2)} (\sin (u_{2})) \frac{d}{d x} (2 x))$

Step 2.3.2.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$f'' (x) = - 4 \sin (2 x) - 2 (\cos (u_{2}) \frac{d}{d x} (2 x))$

Step 2.3.2.3

Replace all occurrences of $u_{2}$ with $2 x$ .

$f'' (x) = - 4 \sin (2 x) - 2 (\cos (2 x) \frac{d}{d x} (2 x))$

Step 2.3.3

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

$f'' (x) = - 4 \sin (2 x) - 2 (\cos (2 x) (2 \frac{d}{d x} (x)))$

Step 2.3.4

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

$f'' (x) = - 4 \sin (2 x) - 2 (\cos (2 x) (2 \cdot 1))$

Step 2.3.5

Multiply $2$ by $1$ .

$f'' (x) = - 4 \sin (2 x) - 2 (\cos (2 x) \cdot 2)$

Step 2.3.6

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

$f'' (x) = - 4 \sin (2 x) - 2 (2 \cdot \cos (2 x))$

Step 2.3.7

Multiply $2$ by $- 2$ .

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

Step 3

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

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

Step 4

Divide each term in the equation by $\cos (2 x)$ .

$\frac{2 \cos (2 x)}{\cos (2 x)} + \frac{- 2 \sin (2 x)}{\cos (2 x)} = \frac{0}{\cos (2 x)}$

Step 5

Cancel the common factor of $\cos (2 x)$ .

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Step 5.1

Cancel the common factor.

$\frac{2 \cos (2 x)}{\cos (2 x)} + \frac{- 2 \sin (2 x)}{\cos (2 x)} = \frac{0}{\cos (2 x)}$

Step 5.2

Divide $2$ by $1$ .

$2 + \frac{- 2 \sin (2 x)}{\cos (2 x)} = \frac{0}{\cos (2 x)}$

Step 6

Separate fractions.

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

Step 7

Convert from $\frac{\sin (2 x)}{\cos (2 x)}$ to $\tan (2 x)$ .

$2 + \frac{- 2}{1} \cdot \tan (2 x) = \frac{0}{\cos (2 x)}$

Step 8

Divide $- 2$ by $1$ .

$2 - 2 \tan (2 x) = \frac{0}{\cos (2 x)}$

Step 9

Separate fractions.

$2 - 2 \tan (2 x) = \frac{0}{1} \cdot \frac{1}{\cos (2 x)}$

Step 10

Convert from $\frac{1}{\cos (2 x)}$ to $\sec (2 x)$ .

$2 - 2 \tan (2 x) = \frac{0}{1} \cdot \sec (2 x)$

Step 11

Divide $0$ by $1$ .

$2 - 2 \tan (2 x) = 0 \sec (2 x)$

Step 12

Multiply $0$ by $\sec (2 x)$ .

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

Step 13

Subtract $2$ from both sides of the equation.

$- 2 \tan (2 x) = - 2$

Step 14

Divide each term in $- 2 \tan (2 x) = - 2$ by $- 2$ and simplify.

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Step 14.1

Divide each term in $- 2 \tan (2 x) = - 2$ by $- 2$ .

$\frac{- 2 \tan (2 x)}{- 2} = \frac{- 2}{- 2}$

Step 14.2

Simplify the left side.

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Step 14.2.1

Cancel the common factor of $- 2$ .

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Step 14.2.1.1

Cancel the common factor.

$\frac{- 2 \tan (2 x)}{- 2} = \frac{- 2}{- 2}$

Step 14.2.1.2

Divide $\tan (2 x)$ by $1$ .

$\tan (2 x) = \frac{- 2}{- 2}$

Step 14.3

Simplify the right side.

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Step 14.3.1

Divide $- 2$ by $- 2$ .

$\tan (2 x) = 1$

Step 15

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

$2 x = \arctan (1)$

Step 16

Simplify the right side.

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Step 16.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$2 x = \frac{π}{4}$

Step 17

Divide each term in $2 x = \frac{π}{4}$ by $2$ and simplify.

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Step 17.1

Divide each term in $2 x = \frac{π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{π}{4}}{2}$

Step 17.2

Simplify the left side.

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Step 17.2.1

Cancel the common factor of $2$ .

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Step 17.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{π}{4}}{2}$

Step 17.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{4}}{2}$

Step 17.3

Simplify the right side.

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Step 17.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{4} \cdot \frac{1}{2}$

Step 17.3.2

Multiply $\frac{π}{4} \cdot \frac{1}{2}$ .

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Step 17.3.2.1

Multiply $\frac{π}{4}$ by $\frac{1}{2}$ .

$x = \frac{π}{4 \cdot 2}$

Step 17.3.2.2

Multiply $4$ by $2$ .

$x = \frac{π}{8}$

Step 18

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$2 x = π + \frac{π}{4}$

Step 19

Solve for $x$ .

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Step 19.1

Simplify.

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Step 19.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 19.1.2

Combine $π$ and $\frac{4}{4}$ .

$2 x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 19.1.3

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 4 + π}{4}$

Step 19.1.4

Add $π \cdot 4$ and $π$ .

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Step 19.1.4.1

Reorder $π$ and $4$ .

$2 x = \frac{4 \cdot π + π}{4}$

Step 19.1.4.2

Add $4 \cdot π$ and $π$ .

$2 x = \frac{5 \cdot π}{4}$

Step 19.2

Divide each term in $2 x = \frac{5 \cdot π}{4}$ by $2$ and simplify.

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Step 19.2.1

Divide each term in $2 x = \frac{5 \cdot π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{4}}{2}$

Step 19.2.2

Simplify the left side.

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Step 19.2.2.1

Cancel the common factor of $2$ .

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Step 19.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{4}}{2}$

Step 19.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 \cdot π}{4}}{2}$

Step 19.2.3

Simplify the right side.

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Step 19.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 \cdot π}{4} \cdot \frac{1}{2}$

Step 19.2.3.2

Multiply $\frac{5 π}{4} \cdot \frac{1}{2}$ .

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Step 19.2.3.2.1

Multiply $\frac{5 π}{4}$ by $\frac{1}{2}$ .

$x = \frac{5 π}{4 \cdot 2}$

Step 19.2.3.2.2

Multiply $4$ by $2$ .

$x = \frac{5 π}{8}$

Step 20

The solution to the equation $2 x = \frac{π}{4}$ .

$x = \frac{π}{8}, \frac{5 π}{8}$

Step 21

Evaluate the second derivative at $x = \frac{π}{8}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \sin (2 (\frac{π}{8})) - 4 \cos (2 (\frac{π}{8}))$

Step 22

Evaluate the second derivative.

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Step 22.1

Simplify each term.

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Step 22.1.1

Cancel the common factor of $2$ .

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Step 22.1.1.1

Factor $2$ out of $8$ .

$- 4 \sin (2 \frac{π}{2 (4)}) - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.1.2

Cancel the common factor.

$- 4 \sin (2 \frac{π}{2 \cdot 4}) - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.1.3

Rewrite the expression.

$- 4 \sin (\frac{π}{4}) - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.2

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- 4 \frac{\sqrt{2}}{2} - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.3

Cancel the common factor of $2$ .

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Step 22.1.3.1

Factor $2$ out of $- 4$ .

$2 (- 2) \frac{\sqrt{2}}{2} - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.3.2

Cancel the common factor.

$2 \cdot - 2 \frac{\sqrt{2}}{2} - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.3.3

Rewrite the expression.

$- 2 \sqrt{2} - 4 \cos (2 (\frac{π}{8}))$

Step 22.1.4

Cancel the common factor of $2$ .

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Step 22.1.4.1

Factor $2$ out of $8$ .

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

Step 22.1.4.2

Cancel the common factor.

$- 2 \sqrt{2} - 4 \cos (2 \frac{π}{2 \cdot 4})$

Step 22.1.4.3

Rewrite the expression.

$- 2 \sqrt{2} - 4 \cos (\frac{π}{4})$

Step 22.1.5

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- 2 \sqrt{2} - 4 \frac{\sqrt{2}}{2}$

Step 22.1.6

Cancel the common factor of $2$ .

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Step 22.1.6.1

Factor $2$ out of $- 4$ .

$- 2 \sqrt{2} + 2 (- 2) \frac{\sqrt{2}}{2}$

Step 22.1.6.2

Cancel the common factor.

$- 2 \sqrt{2} + 2 \cdot - 2 \frac{\sqrt{2}}{2}$

Step 22.1.6.3

Rewrite the expression.

$- 2 \sqrt{2} - 2 \sqrt{2}$

Step 22.2

Subtract $2 \sqrt{2}$ from $- 2 \sqrt{2}$ .

$- 4 \sqrt{2}$

Step 23

$x = \frac{π}{8}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{8}$ is a local maximum

Step 24

Find the y-value when $x = \frac{π}{8}$ .

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Step 24.1

Replace the variable $x$ with $\frac{π}{8}$ in the expression.

$f (\frac{π}{8}) = \sin (2 (\frac{π}{8})) + \cos (2 (\frac{π}{8}))$

Step 24.2

Simplify the result.

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Step 24.2.1

Simplify each term.

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Step 24.2.1.1

Cancel the common factor of $2$ .

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Step 24.2.1.1.1

Factor $2$ out of $8$ .

$f (\frac{π}{8}) = \sin (2 (\frac{π}{2 (4)})) + \cos (2 (\frac{π}{8}))$

Step 24.2.1.1.2

Cancel the common factor.

$f (\frac{π}{8}) = \sin (2 (\frac{π}{2 \cdot 4})) + \cos (2 (\frac{π}{8}))$

Step 24.2.1.1.3

Rewrite the expression.

$f (\frac{π}{8}) = \sin (\frac{π}{4}) + \cos (2 (\frac{π}{8}))$

Step 24.2.1.2

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{8}) = \frac{\sqrt{2}}{2} + \cos (2 (\frac{π}{8}))$

Step 24.2.1.3

Cancel the common factor of $2$ .

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Step 24.2.1.3.1

Factor $2$ out of $8$ .

$f (\frac{π}{8}) = \frac{\sqrt{2}}{2} + \cos (2 (\frac{π}{2 (4)}))$

Step 24.2.1.3.2

Cancel the common factor.

$f (\frac{π}{8}) = \frac{\sqrt{2}}{2} + \cos (2 (\frac{π}{2 \cdot 4}))$

Step 24.2.1.3.3

Rewrite the expression.

$f (\frac{π}{8}) = \frac{\sqrt{2}}{2} + \cos (\frac{π}{4})$

Step 24.2.1.4

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{π}{8}) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}$

Step 24.2.2

Simplify terms.

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Step 24.2.2.1

Combine the numerators over the common denominator.

$f (\frac{π}{8}) = \frac{\sqrt{2} + \sqrt{2}}{2}$

Step 24.2.2.2

Add $\sqrt{2}$ and $\sqrt{2}$ .

$f (\frac{π}{8}) = \frac{2 \sqrt{2}}{2}$

Step 24.2.2.3

Cancel the common factor of $2$ .

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Step 24.2.2.3.1

Cancel the common factor.

$f (\frac{π}{8}) = \frac{2 \sqrt{2}}{2}$

Step 24.2.2.3.2

Divide $\sqrt{2}$ by $1$ .

$f (\frac{π}{8}) = \sqrt{2}$

Step 24.2.3

The final answer is $\sqrt{2}$ .

$y = \sqrt{2}$

Step 25

Evaluate the second derivative at $x = \frac{5 π}{8}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4 \sin (2 (\frac{5 π}{8})) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26

Evaluate the second derivative.

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Step 26.1

Simplify each term.

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Step 26.1.1

Cancel the common factor of $2$ .

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Step 26.1.1.1

Factor $2$ out of $8$ .

$- 4 \sin (2 \frac{5 π}{2 (4)}) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.1.2

Cancel the common factor.

$- 4 \sin (2 \frac{5 π}{2 \cdot 4}) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.1.3

Rewrite the expression.

$- 4 \sin (\frac{5 π}{4}) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.2

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 third quadrant.

$- 4 (- \sin (\frac{π}{4})) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.3

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$- 4 (- \frac{\sqrt{2}}{2}) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.4

Cancel the common factor of $2$ .

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Step 26.1.4.1

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

$- 4 \frac{- \sqrt{2}}{2} - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.4.2

Factor $2$ out of $- 4$ .

$2 (- 2) \frac{- \sqrt{2}}{2} - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.4.3

Cancel the common factor.

$2 \cdot - 2 \frac{- \sqrt{2}}{2} - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.4.4

Rewrite the expression.

$- 2 (- \sqrt{2}) - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.5

Multiply $- 1$ by $- 2$ .

$2 \sqrt{2} - 4 \cos (2 (\frac{5 π}{8}))$

Step 26.1.6

Cancel the common factor of $2$ .

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Step 26.1.6.1

Factor $2$ out of $8$ .

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

Step 26.1.6.2

Cancel the common factor.

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

Step 26.1.6.3

Rewrite the expression.

$2 \sqrt{2} - 4 \cos (\frac{5 π}{4})$

Step 26.1.7

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.

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

Step 26.1.8

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$2 \sqrt{2} - 4 (- \frac{\sqrt{2}}{2})$

Step 26.1.9

Cancel the common factor of $2$ .

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Step 26.1.9.1

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

$2 \sqrt{2} - 4 \frac{- \sqrt{2}}{2}$

Step 26.1.9.2

Factor $2$ out of $- 4$ .

$2 \sqrt{2} + 2 (- 2) \frac{- \sqrt{2}}{2}$

Step 26.1.9.3

Cancel the common factor.

$2 \sqrt{2} + 2 \cdot - 2 \frac{- \sqrt{2}}{2}$

Step 26.1.9.4

Rewrite the expression.

$2 \sqrt{2} - 2 (- \sqrt{2})$

Step 26.1.10

Multiply $- 1$ by $- 2$ .

$2 \sqrt{2} + 2 \sqrt{2}$

Step 26.2

Add $2 \sqrt{2}$ and $2 \sqrt{2}$ .

$4 \sqrt{2}$

Step 27

$x = \frac{5 π}{8}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{5 π}{8}$ is a local minimum

Step 28

Find the y-value when $x = \frac{5 π}{8}$ .

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Step 28.1

Replace the variable $x$ with $\frac{5 π}{8}$ in the expression.

$f (\frac{5 π}{8}) = \sin (2 (\frac{5 π}{8})) + \cos (2 (\frac{5 π}{8}))$

Step 28.2

Simplify the result.

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Step 28.2.1

Simplify each term.

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Step 28.2.1.1

Cancel the common factor of $2$ .

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Step 28.2.1.1.1

Factor $2$ out of $8$ .

$f (\frac{5 π}{8}) = \sin (2 (\frac{5 π}{2 (4)})) + \cos (2 (\frac{5 π}{8}))$

Step 28.2.1.1.2

Cancel the common factor.

$f (\frac{5 π}{8}) = \sin (2 (\frac{5 π}{2 \cdot 4})) + \cos (2 (\frac{5 π}{8}))$

Step 28.2.1.1.3

Rewrite the expression.

$f (\frac{5 π}{8}) = \sin (\frac{5 π}{4}) + \cos (2 (\frac{5 π}{8}))$

Step 28.2.1.2

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 third quadrant.

$f (\frac{5 π}{8}) = - \sin (\frac{π}{4}) + \cos (2 (\frac{5 π}{8}))$

Step 28.2.1.3

The exact value of $\sin (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{8}) = - \frac{\sqrt{2}}{2} + \cos (2 (\frac{5 π}{8}))$

Step 28.2.1.4

Cancel the common factor of $2$ .

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Step 28.2.1.4.1

Factor $2$ out of $8$ .

$f (\frac{5 π}{8}) = - \frac{\sqrt{2}}{2} + \cos (2 (\frac{5 π}{2 (4)}))$

Step 28.2.1.4.2

Cancel the common factor.

$f (\frac{5 π}{8}) = - \frac{\sqrt{2}}{2} + \cos (2 (\frac{5 π}{2 \cdot 4}))$

Step 28.2.1.4.3

Rewrite the expression.

$f (\frac{5 π}{8}) = - \frac{\sqrt{2}}{2} + \cos (\frac{5 π}{4})$

Step 28.2.1.5

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.

$f (\frac{5 π}{8}) = - \frac{\sqrt{2}}{2} - \cos (\frac{π}{4})$

Step 28.2.1.6

The exact value of $\cos (\frac{π}{4})$ is $\frac{\sqrt{2}}{2}$ .

$f (\frac{5 π}{8}) = - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}$

Step 28.2.2

Simplify terms.

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Step 28.2.2.1

Combine the numerators over the common denominator.

$f (\frac{5 π}{8}) = \frac{- \sqrt{2} - \sqrt{2}}{2}$

Step 28.2.2.2

Subtract $\sqrt{2}$ from $- \sqrt{2}$ .

$f (\frac{5 π}{8}) = \frac{- 2 \sqrt{2}}{2}$

Step 28.2.2.3

Cancel the common factor of $- 2$ and $2$ .

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Step 28.2.2.3.1

Factor $2$ out of $- 2 \sqrt{2}$ .

$f (\frac{5 π}{8}) = \frac{2 (- \sqrt{2})}{2}$

Step 28.2.2.3.2

Cancel the common factors.

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Step 28.2.2.3.2.1

Factor $2$ out of $2$ .

$f (\frac{5 π}{8}) = \frac{2 (- \sqrt{2})}{2 (1)}$

Step 28.2.2.3.2.2

Cancel the common factor.

$f (\frac{5 π}{8}) = \frac{2 (- \sqrt{2})}{2 \cdot 1}$

Step 28.2.2.3.2.3

Rewrite the expression.

$f (\frac{5 π}{8}) = \frac{- \sqrt{2}}{1}$

Step 28.2.2.3.2.4

Divide $- \sqrt{2}$ by $1$ .

$f (\frac{5 π}{8}) = - \sqrt{2}$

Step 28.2.3

The final answer is $- \sqrt{2}$ .

$y = - \sqrt{2}$

Step 29

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

$(\frac{π}{8}, \sqrt{2})$ is a local maxima

$(\frac{5 π}{8}, - \sqrt{2})$ is a local minima

Step 30