Calculus Examples

$2 \cos (θ) + \cos^{2} (θ)$

Step 1

Write $2 \cos (θ) + \cos^{2} (θ)$ as a function.

$f (θ) = 2 \cos (θ) + \cos^{2} (θ)$

Step 2

Find the first derivative of the function.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $- 1$ by $2$ .

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

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

To apply the Chain Rule, set $u$ as $\cos (θ)$ .

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

Step 2.3.1.2

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

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

Step 2.3.1.3

Replace all occurrences of $u$ with $\cos (θ)$ .

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

Step 2.3.2

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

$- 2 \sin (θ) + 2 \cos (θ) (- \sin (θ))$

Step 2.3.3

Multiply $- 1$ by $2$ .

$- 2 \sin (θ) - 2 \cos (θ) \sin (θ)$

Step 2.4

Reorder terms.

$- 2 \cos (θ) \sin (θ) - 2 \sin (θ)$

Step 3

Find the second derivative of the function.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

Differentiate using the Product Rule which states that $\frac{d}{d θ} [f (θ) g (θ)]$ is $f (θ) \frac{d}{d θ} [g (θ)] + g (θ) \frac{d}{d θ} [f (θ)]$ where $f (θ) = \cos (θ)$ and $g (θ) = \sin (θ)$ .

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

Raise $\cos (θ)$ to the power of $1$ .

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

Step 3.2.6

Raise $\cos (θ)$ to the power of $1$ .

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

Step 3.2.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.2.8

Add $1$ and $1$ .

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

Step 3.2.9

Raise $\sin (θ)$ to the power of $1$ .

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

Step 3.2.10

Raise $\sin (θ)$ to the power of $1$ .

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

Step 3.2.11

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 3.2.12

Add $1$ and $1$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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

$f'' (θ) = - 2 (\cos^{2} (θ) - \sin^{2} (θ)) - 2 \cos (θ)$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$f'' (θ) = - 2 \cos^{2} (θ) - 2 (- \sin^{2} (θ)) - 2 \cos (θ)$

Step 3.4.2

Multiply $- 1$ by $- 2$ .

$f'' (θ) = - 2 \cos^{2} (θ) + 2 \sin^{2} (θ) - 2 \cos (θ)$

Step 4

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

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

Step 5

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

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Step 5.1

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

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

Step 5.2

Factor $2 \sin (θ)$ out of $- 2 \sin (θ)$ .

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

Step 5.3

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

$2 \sin (θ) (- \cos (θ) - 1) = 0$

Step 6

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

$\sin (θ) = 0$

$- \cos (θ) - 1 = 0$

Step 7

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

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Step 7.1

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Step 7.2

Solve $\sin (θ) = 0$ for $θ$ .

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Step 7.2.1

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

$θ = \arcsin (0)$

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

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

$θ = 0$

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.

$θ = π - 0$

Step 7.2.4

Subtract $0$ from $π$ .

$θ = π$

Step 7.2.5

The solution to the equation $θ = 0$ .

$θ = 0, π$

Step 8

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

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Step 8.1

Set $- \cos (θ) - 1$ equal to $0$ .

$- \cos (θ) - 1 = 0$

Step 8.2

Solve $- \cos (θ) - 1 = 0$ for $θ$ .

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Step 8.2.1

Add $1$ to both sides of the equation.

$- \cos (θ) = 1$

Step 8.2.2

Divide each term in $- \cos (θ) = 1$ by $- 1$ and simplify.

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Step 8.2.2.1

Divide each term in $- \cos (θ) = 1$ by $- 1$ .

$\frac{- \cos (θ)}{- 1} = \frac{1}{- 1}$

Step 8.2.2.2

Simplify the left side.

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Step 8.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{\cos (θ)}{1} = \frac{1}{- 1}$

Step 8.2.2.2.2

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{- 1}$

Step 8.2.2.3

Simplify the right side.

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Step 8.2.2.3.1

Divide $1$ by $- 1$ .

$\cos (θ) = - 1$

Step 8.2.3

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

$θ = \arccos (- 1)$

Step 8.2.4

Simplify the right side.

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Step 8.2.4.1

The exact value of $\arccos (- 1)$ is $π$ .

$θ = π$

Step 8.2.5

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$θ = 2 π - π$

Step 8.2.6

Subtract $π$ from $2 π$ .

$θ = π$

Step 8.2.7

The solution to the equation $θ = π$ .

$θ = π, π$

Step 9

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

$θ = 0, π$

Step 10

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

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

Step 11

Evaluate the second derivative.

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Step 11.1

Simplify each term.

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Step 11.1.1

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

$- 2 \cdot 1^{2} + 2 \sin^{2} (0) - 2 \cos (0)$

Step 11.1.2

One to any power is one.

$- 2 \cdot 1 + 2 \sin^{2} (0) - 2 \cos (0)$

Step 11.1.3

Multiply $- 2$ by $1$ .

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

Step 11.1.4

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

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

Step 11.1.5

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

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

Step 11.1.6

Multiply $2$ by $0$ .

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

Step 11.1.7

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

$- 2 + 0 - 2 \cdot 1$

Step 11.1.8

Multiply $- 2$ by $1$ .

$- 2 + 0 - 2$

Step 11.2

Simplify by adding and subtracting.

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Step 11.2.1

Add $- 2$ and $0$ .

$- 2 - 2$

Step 11.2.2

Subtract $2$ from $- 2$ .

$- 4$

Step 12

$θ = 0$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$θ = 0$ is a local maximum

Step 13

Find the y-value when $θ = 0$ .

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Step 13.1

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

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

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 $\cos (0)$ is $1$ .

$f (0) = 2 \cdot 1 + \cos^{2} (0)$

Step 13.2.1.2

Multiply $2$ by $1$ .

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

Step 13.2.1.3

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

$f (0) = 2 + 1^{2}$

Step 13.2.1.4

One to any power is one.

$f (0) = 2 + 1$

Step 13.2.2

Add $2$ and $1$ .

$f (0) = 3$

Step 13.2.3

The final answer is $3$ .

$y = 3$

Step 14

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

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

Step 15

Evaluate the second derivative.

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Step 15.1

Simplify each term.

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Step 15.1.1

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.

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

Step 15.1.2

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

$- 2 {(- 1 \cdot 1)}^{2} + 2 \sin^{2} (π) - 2 \cos (π)$

Step 15.1.3

Multiply $- 1$ by $1$ .

$- 2 {(- 1)}^{2} + 2 \sin^{2} (π) - 2 \cos (π)$

Step 15.1.4

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

$- 2 \cdot 1 + 2 \sin^{2} (π) - 2 \cos (π)$

Step 15.1.5

Multiply $- 2$ by $1$ .

$- 2 + 2 \sin^{2} (π) - 2 \cos (π)$

Step 15.1.6

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

$- 2 + 2 \sin^{2} (0) - 2 \cos (π)$

Step 15.1.7

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

$- 2 + 2 \cdot 0^{2} - 2 \cos (π)$

Step 15.1.8

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

$- 2 + 2 \cdot 0 - 2 \cos (π)$

Step 15.1.9

Multiply $2$ by $0$ .

$- 2 + 0 - 2 \cos (π)$

Step 15.1.10

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.

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

Step 15.1.11

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

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

Step 15.1.12

Multiply $- 2 (- 1 \cdot 1)$ .

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Step 15.1.12.1

Multiply $- 1$ by $1$ .

$- 2 + 0 - 2 \cdot - 1$

Step 15.1.12.2

Multiply $- 2$ by $- 1$ .

$- 2 + 0 + 2$

Step 15.2

Simplify by adding numbers.

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Step 15.2.1

Add $- 2$ and $0$ .

$- 2 + 2$

Step 15.2.2

Add $- 2$ and $2$ .

$0$

Step 16

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

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Step 16.1

Split $(- \infty, \infty)$ into separate intervals around the $θ$ values that make the first derivative $0$ or undefined.

$(- \infty, 0) \cup (0, π) \cup (π, \infty)$

Step 16.2

Substitute any number, such as $- 2$ , from the interval $(- \infty, 0)$ in the first derivative $- 2 \cos (θ) \sin (θ) - 2 \sin (θ)$ to check if the result is negative or positive.

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Step 16.2.1

Replace the variable $θ$ with $- 2$ in the expression.

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

Step 16.2.2

Simplify the result.

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Step 16.2.2.1

Simplify each term.

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Step 16.2.2.1.1

Evaluate $\cos (- 2)$ .

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

Step 16.2.2.1.2

Multiply $- 2$ by $- 0.41614683$ .

$f' (- 2) = 0.83229367 \sin (- 2) - 2 \sin (- 2)$

Step 16.2.2.1.3

Evaluate $\sin (- 2)$ .

$f' (- 2) = 0.83229367 \cdot - 0.90929742 - 2 \sin (- 2)$

Step 16.2.2.1.4

Multiply $0.83229367$ by $- 0.90929742$ .

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

Step 16.2.2.1.5

Evaluate $\sin (- 2)$ .

$f' (- 2) = - 0.75680249 - 2 \cdot - 0.90929742$

Step 16.2.2.1.6

Multiply $- 2$ by $- 0.90929742$ .

$f' (- 2) = - 0.75680249 + 1.81859485$

Step 16.2.2.2

Add $- 0.75680249$ and $1.81859485$ .

$f' (- 2) = 1.06179235$

Step 16.2.2.3

The final answer is $1.06179235$ .

$1.06179235$

Step 16.3

Substitute any number, such as $2$ , from the interval $(0, π)$ in the first derivative $- 2 \cos (θ) \sin (θ) - 2 \sin (θ)$ to check if the result is negative or positive.

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Step 16.3.1

Replace the variable $θ$ with $2$ in the expression.

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

Step 16.3.2

Simplify the result.

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Step 16.3.2.1

Simplify each term.

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Step 16.3.2.1.1

Evaluate $\cos (2)$ .

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

Step 16.3.2.1.2

Multiply $- 2$ by $- 0.41614683$ .

$f' (2) = 0.83229367 \sin (2) - 2 \sin (2)$

Step 16.3.2.1.3

Evaluate $\sin (2)$ .

$f' (2) = 0.83229367 \cdot 0.90929742 - 2 \sin (2)$

Step 16.3.2.1.4

Multiply $0.83229367$ by $0.90929742$ .

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

Step 16.3.2.1.5

Evaluate $\sin (2)$ .

$f' (2) = 0.75680249 - 2 \cdot 0.90929742$

Step 16.3.2.1.6

Multiply $- 2$ by $0.90929742$ .

$f' (2) = 0.75680249 - 1.81859485$

Step 16.3.2.2

Subtract $1.81859485$ from $0.75680249$ .

$f' (2) = - 1.06179235$

Step 16.3.2.3

The final answer is $- 1.06179235$ .

$- 1.06179235$

Step 16.4

Substitute any number, such as $6$ , from the interval $(π, \infty)$ in the first derivative $- 2 \cos (θ) \sin (θ) - 2 \sin (θ)$ to check if the result is negative or positive.

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Step 16.4.1

Replace the variable $θ$ with $6$ in the expression.

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

Step 16.4.2

Simplify the result.

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Step 16.4.2.1

Simplify each term.

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Step 16.4.2.1.1

Evaluate $\cos (6)$ .

$f' (6) = - 2 \cdot (0.96017028 \sin (6)) - 2 \sin (6)$

Step 16.4.2.1.2

Multiply $- 2$ by $0.96017028$ .

$f' (6) = - 1.92034057 \sin (6) - 2 \sin (6)$

Step 16.4.2.1.3

Evaluate $\sin (6)$ .

$f' (6) = - 1.92034057 \cdot - 0.27941549 - 2 \sin (6)$

Step 16.4.2.1.4

Multiply $- 1.92034057$ by $- 0.27941549$ .

$f' (6) = 0.53657291 - 2 \sin (6)$

Step 16.4.2.1.5

Evaluate $\sin (6)$ .

$f' (6) = 0.53657291 - 2 \cdot - 0.27941549$

Step 16.4.2.1.6

Multiply $- 2$ by $- 0.27941549$ .

$f' (6) = 0.53657291 + 0.55883099$

Step 16.4.2.2

Add $0.53657291$ and $0.55883099$ .

$f' (6) = 1.09540391$

Step 16.4.2.3

The final answer is $1.09540391$ .

$1.09540391$

Step 16.5

Since the first derivative changed signs from positive to negative around $θ = 0$ , then $θ = 0$ is a local maximum.

$θ = 0$ is a local maximum

Step 16.6

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

$θ = π$ is a local minimum

Step 16.7

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

$θ = 0$ is a local maximum

$θ = π$ is a local minimum

$θ = 0$ is a local maximum

$θ = π$ is a local minimum

Step 17