Calculus Examples

$f (x) = 3 \cos^{2} (x) - 6 \sin (x)$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $3 \cos^{2} (x) - 6 \sin (x)$ with respect to $x$ is $\frac{d}{d x} [3 \cos^{2} (x)] + \frac{d}{d x} [- 6 \sin (x)]$ .

$\frac{d}{d x} [3 \cos^{2} (x)] + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2

Evaluate $\frac{d}{d x} [3 \cos^{2} (x)]$ .

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Step 1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 \cos^{2} (x)$ with respect to $x$ is $3 \frac{d}{d x} [\cos^{2} (x)]$ .

$3 \frac{d}{d x} [\cos^{2} (x)] + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2.2

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

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Step 1.2.2.1

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

$3 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [\cos (x)]) + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2.2.2

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

$3 (2 u \frac{d}{d x} [\cos (x)]) + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2.2.3

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

$3 (2 \cos (x) \frac{d}{d x} [\cos (x)]) + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2.3

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

$3 (2 \cos (x) (- \sin (x))) + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2.4

Multiply $- 1$ by $2$ .

$3 (- 2 \cos (x) \sin (x)) + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.2.5

Multiply $- 2$ by $3$ .

$- 6 \cos (x) \sin (x) + \frac{d}{d x} [- 6 \sin (x)]$

Step 1.3

Evaluate $\frac{d}{d x} [- 6 \sin (x)]$ .

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Step 1.3.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 \sin (x)$ with respect to $x$ is $- 6 \frac{d}{d x} [\sin (x)]$ .

$- 6 \cos (x) \sin (x) - 6 \frac{d}{d x} [\sin (x)]$

Step 1.3.2

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

$- 6 \cos (x) \sin (x) - 6 \cos (x)$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $- 6 \cos (x) \sin (x) - 6 \cos (x)$ with respect to $x$ is $\frac{d}{d x} [- 6 \cos (x) \sin (x)] + \frac{d}{d x} [- 6 \cos (x)]$ .

$f'' (x) = \frac{d}{d x} (- 6 \cos (x) \sin (x)) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2

Evaluate $\frac{d}{d x} [- 6 \cos (x) \sin (x)]$ .

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Step 2.2.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 \cos (x) \sin (x)$ with respect to $x$ is $- 6 \frac{d}{d x} [\cos (x) \sin (x)]$ .

$f'' (x) = - 6 \frac{d}{d x} (\cos (x) \sin (x)) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.2

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

$f'' (x) = - 6 (\cos (x) \frac{d}{d x} (\sin (x)) + \sin (x) \frac{d}{d x} (\cos (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.3

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

$f'' (x) = - 6 (\cos (x) \cos (x) + \sin (x) \frac{d}{d x} (\cos (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.4

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

$f'' (x) = - 6 (\cos (x) \cos (x) + \sin (x) (- \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.5

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

$f'' (x) = - 6 (\cos (x) \cos (x) + \sin (x) (- \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.6

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

$f'' (x) = - 6 (\cos (x) \cos (x) + \sin (x) (- \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.7

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

$f'' (x) = - 6 ({\cos (x)}^{1 + 1} + \sin (x) (- \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.8

Add $1$ and $1$ .

$f'' (x) = - 6 (\cos^{2} (x) + \sin (x) (- \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.9

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

$f'' (x) = - 6 (\cos^{2} (x) - (\sin (x) \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.10

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

$f'' (x) = - 6 (\cos^{2} (x) - (\sin (x) \sin (x))) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.11

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

$f'' (x) = - 6 (\cos^{2} (x) - {\sin (x)}^{1 + 1}) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.2.12

Add $1$ and $1$ .

$f'' (x) = - 6 (\cos^{2} (x) - \sin^{2} (x)) + \frac{d}{d x} (- 6 \cos (x))$

Step 2.3

Evaluate $\frac{d}{d x} [- 6 \cos (x)]$ .

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Step 2.3.1

Since $- 6$ is constant with respect to $x$ , the derivative of $- 6 \cos (x)$ with respect to $x$ is $- 6 \frac{d}{d x} [\cos (x)]$ .

$f'' (x) = - 6 (\cos^{2} (x) - \sin^{2} (x)) - 6 \frac{d}{d x} \cos (x)$

Step 2.3.2

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

$f'' (x) = - 6 (\cos^{2} (x) - \sin^{2} (x)) - 6 (- \sin (x))$

Step 2.3.3

Multiply $- 1$ by $- 6$ .

$f'' (x) = - 6 (\cos^{2} (x) - \sin^{2} (x)) + 6 \sin (x)$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$f'' (x) = - 6 \cos^{2} (x) - 6 (- \sin^{2} (x)) + 6 \sin (x)$

Step 2.4.2

Multiply $- 1$ by $- 6$ .

$f'' (x) = - 6 \cos^{2} (x) + 6 \sin^{2} (x) + 6 \sin (x)$

Step 3

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

$- 6 \cos (x) \sin (x) - 6 \cos (x) = 0$

Step 4

Factor $- 6 \cos (x) \sin (x) - 6 \cos (x)$ .

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Step 4.1

Factor $6 \cos (x)$ out of $- 6 \cos (x) \sin (x) - 6 \cos (x)$ .

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Step 4.1.1

Factor $6 \cos (x)$ out of $- 6 \cos (x) \sin (x)$ .

$6 \cos (x) (- 1 \sin (x)) - 6 \cos (x) = 0$

Step 4.1.2

Factor $6 \cos (x)$ out of $- 6 \cos (x)$ .

$6 \cos (x) (- 1 \sin (x)) + 6 \cos (x) \cdot - 1 = 0$

Step 4.1.3

Factor $6 \cos (x)$ out of $6 \cos (x) (- 1 \sin (x)) + 6 \cos (x) \cdot - 1$ .

$6 \cos (x) (- 1 \sin (x) - 1) = 0$

Step 4.2

Rewrite $- 1 \sin (x)$ as $- \sin (x)$ .

$6 \cos (x) (- \sin (x) - 1) = 0$

Step 5

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

$\cos (x) = 0$

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

Step 6

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

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Step 6.1

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

$\cos (x) = 0$

Step 6.2

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

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Step 6.2.1

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

$x = \arccos (0)$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 6.2.3

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.

$x = 2 π - \frac{π}{2}$

Step 6.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 6.2.4.1

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

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

Step 6.2.4.2

Combine fractions.

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Step 6.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 6.2.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 2 - π}{2}$

Step 6.2.4.3

Simplify the numerator.

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Step 6.2.4.3.1

Multiply $2$ by $2$ .

$x = \frac{4 π - π}{2}$

Step 6.2.4.3.2

Subtract $π$ from $4 π$ .

$x = \frac{3 π}{2}$

Step 6.2.5

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

$x = \frac{π}{2}, \frac{3 π}{2}$

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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Step 7.2.1

Add $1$ to both sides of the equation.

$- \sin (x) = 1$

Step 7.2.2

Divide each term in $- \sin (x) = 1$ by $- 1$ and simplify.

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Step 7.2.2.1

Divide each term in $- \sin (x) = 1$ by $- 1$ .

$\frac{- \sin (x)}{- 1} = \frac{1}{- 1}$

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{\sin (x)}{1} = \frac{1}{- 1}$

Step 7.2.2.2.2

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

$\sin (x) = \frac{1}{- 1}$

Step 7.2.2.3

Simplify the right side.

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Step 7.2.2.3.1

Divide $1$ by $- 1$ .

$\sin (x) = - 1$

Step 7.2.3

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

$x = \arcsin (- 1)$

Step 7.2.4

Simplify the right side.

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Step 7.2.4.1

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

$x = - \frac{π}{2}$

Step 7.2.5

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 7.2.6

Simplify the expression to find the second solution.

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Step 7.2.6.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 7.2.6.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 7.2.7

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

$x = - \frac{π}{2}, \frac{3 π}{2}$

Step 8

The final solution is all the values that make $6 \cos (x) (- \sin (x) - 1) = 0$ true.

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

Step 9

Evaluate the second derivative at $x = \frac{π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 6 \cos^{2} (\frac{π}{2}) + 6 \sin^{2} (\frac{π}{2}) + 6 \sin (\frac{π}{2})$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$- 6 \cdot 0^{2} + 6 \sin^{2} (\frac{π}{2}) + 6 \sin (\frac{π}{2})$

Step 10.1.2

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

$- 6 \cdot 0 + 6 \sin^{2} (\frac{π}{2}) + 6 \sin (\frac{π}{2})$

Step 10.1.3

Multiply $- 6$ by $0$ .

$0 + 6 \sin^{2} (\frac{π}{2}) + 6 \sin (\frac{π}{2})$

Step 10.1.4

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

$0 + 6 \cdot 1^{2} + 6 \sin (\frac{π}{2})$

Step 10.1.5

One to any power is one.

$0 + 6 \cdot 1 + 6 \sin (\frac{π}{2})$

Step 10.1.6

Multiply $6$ by $1$ .

$0 + 6 + 6 \sin (\frac{π}{2})$

Step 10.1.7

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

$0 + 6 + 6 \cdot 1$

Step 10.1.8

Multiply $6$ by $1$ .

$0 + 6 + 6$

Step 10.2

Simplify by adding numbers.

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Step 10.2.1

Add $0$ and $6$ .

$6 + 6$

Step 10.2.2

Add $6$ and $6$ .

$12$

Step 11

$x = \frac{π}{2}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{π}{2}$ is a local minimum

Step 12

Find the y-value when $x = \frac{π}{2}$ .

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Step 12.1

Replace the variable $x$ with $\frac{π}{2}$ in the expression.

$f (\frac{π}{2}) = 3 \cos^{2} (\frac{π}{2}) - 6 \sin (\frac{π}{2})$

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$f (\frac{π}{2}) = 3 \cdot 0^{2} - 6 \sin (\frac{π}{2})$

Step 12.2.1.2

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

$f (\frac{π}{2}) = 3 \cdot 0 - 6 \sin (\frac{π}{2})$

Step 12.2.1.3

Multiply $3$ by $0$ .

$f (\frac{π}{2}) = 0 - 6 \sin (\frac{π}{2})$

Step 12.2.1.4

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

$f (\frac{π}{2}) = 0 - 6 \cdot 1$

Step 12.2.1.5

Multiply $- 6$ by $1$ .

$f (\frac{π}{2}) = 0 - 6$

Step 12.2.2

Subtract $6$ from $0$ .

$f (\frac{π}{2}) = - 6$

Step 12.2.3

The final answer is $- 6$ .

$y = - 6$

Step 13

Evaluate the second derivative at $x = \frac{3 π}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 6 \cos^{2} (\frac{3 π}{2}) + 6 \sin^{2} (\frac{3 π}{2}) + 6 \sin (\frac{3 π}{2})$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

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

$- 6 \cos^{2} (\frac{π}{2}) + 6 \sin^{2} (\frac{3 π}{2}) + 6 \sin (\frac{3 π}{2})$

Step 14.1.2

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$- 6 \cdot 0^{2} + 6 \sin^{2} (\frac{3 π}{2}) + 6 \sin (\frac{3 π}{2})$

Step 14.1.3

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

$- 6 \cdot 0 + 6 \sin^{2} (\frac{3 π}{2}) + 6 \sin (\frac{3 π}{2})$

Step 14.1.4

Multiply $- 6$ by $0$ .

$0 + 6 \sin^{2} (\frac{3 π}{2}) + 6 \sin (\frac{3 π}{2})$

Step 14.1.5

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.

$0 + 6 {(- \sin (\frac{π}{2}))}^{2} + 6 \sin (\frac{3 π}{2})$

Step 14.1.6

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

$0 + 6 {(- 1 \cdot 1)}^{2} + 6 \sin (\frac{3 π}{2})$

Step 14.1.7

Multiply $- 1$ by $1$ .

$0 + 6 {(- 1)}^{2} + 6 \sin (\frac{3 π}{2})$

Step 14.1.8

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

$0 + 6 \cdot 1 + 6 \sin (\frac{3 π}{2})$

Step 14.1.9

Multiply $6$ by $1$ .

$0 + 6 + 6 \sin (\frac{3 π}{2})$

Step 14.1.10

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.

$0 + 6 + 6 (- \sin (\frac{π}{2}))$

Step 14.1.11

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

$0 + 6 + 6 (- 1 \cdot 1)$

Step 14.1.12

Multiply $6 (- 1 \cdot 1)$ .

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Step 14.1.12.1

Multiply $- 1$ by $1$ .

$0 + 6 + 6 \cdot - 1$

Step 14.1.12.2

Multiply $6$ by $- 1$ .

$0 + 6 - 6$

Step 14.2

Simplify by adding and subtracting.

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Step 14.2.1

Add $0$ and $6$ .

$6 - 6$

Step 14.2.2

Subtract $6$ from $6$ .

$0$

Step 15

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

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Step 15.1

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

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

Step 15.2

Substitute any number, such as $- 4$ , from the interval $(- \infty, - \frac{π}{2})$ in the first derivative $- 6 \cos (x) \sin (x) - 6 \cos (x)$ to check if the result is negative or positive.

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Step 15.2.1

Replace the variable $x$ with $- 4$ in the expression.

$f' (- 4) = - 6 \cos (- 4) \sin (- 4) - 6 \cos (- 4)$

Step 15.2.2

Simplify the result.

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Step 15.2.2.1

Simplify each term.

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Step 15.2.2.1.1

Evaluate $\cos (- 4)$ .

$f' (- 4) = - 6 \cdot (- 0.65364362 \sin (- 4)) - 6 \cos (- 4)$

Step 15.2.2.1.2

Multiply $- 6$ by $- 0.65364362$ .

$f' (- 4) = 3.92186172 \sin (- 4) - 6 \cos (- 4)$

Step 15.2.2.1.3

Evaluate $\sin (- 4)$ .

$f' (- 4) = 3.92186172 \cdot 0.75680249 - 6 \cos (- 4)$

Step 15.2.2.1.4

Multiply $3.92186172$ by $0.75680249$ .

$f' (- 4) = 2.96807473 - 6 \cos (- 4)$

Step 15.2.2.1.5

Evaluate $\cos (- 4)$ .

$f' (- 4) = 2.96807473 - 6 \cdot - 0.65364362$

Step 15.2.2.1.6

Multiply $- 6$ by $- 0.65364362$ .

$f' (- 4) = 2.96807473 + 3.92186172$

Step 15.2.2.2

Add $2.96807473$ and $3.92186172$ .

$f' (- 4) = 6.88993646$

Step 15.2.2.3

The final answer is $6.88993646$ .

$6.88993646$

Step 15.3

Substitute any number, such as $0$ , from the interval $(- \frac{π}{2}, \frac{π}{2})$ in the first derivative $- 6 \cos (x) \sin (x) - 6 \cos (x)$ to check if the result is negative or positive.

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Step 15.3.1

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

$f' (0) = - 6 \cos (0) \sin (0) - 6 \cos (0)$

Step 15.3.2

Simplify the result.

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Step 15.3.2.1

Simplify each term.

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Step 15.3.2.1.1

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

$f' (0) = - 6 \cdot (1 \sin (0)) - 6 \cos (0)$

Step 15.3.2.1.2

Multiply $- 6$ by $1$ .

$f' (0) = - 6 \sin (0) - 6 \cos (0)$

Step 15.3.2.1.3

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

$f' (0) = - 6 \cdot 0 - 6 \cos (0)$

Step 15.3.2.1.4

Multiply $- 6$ by $0$ .

$f' (0) = 0 - 6 \cos (0)$

Step 15.3.2.1.5

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

$f' (0) = 0 - 6 \cdot 1$

Step 15.3.2.1.6

Multiply $- 6$ by $1$ .

$f' (0) = 0 - 6$

Step 15.3.2.2

Subtract $6$ from $0$ .

$f' (0) = - 6$

Step 15.3.2.3

The final answer is $- 6$ .

$- 6$

Step 15.4

Substitute any number, such as $3$ , from the interval $(\frac{π}{2}, \frac{3 π}{2})$ in the first derivative $- 6 \cos (x) \sin (x) - 6 \cos (x)$ to check if the result is negative or positive.

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Step 15.4.1

Replace the variable $x$ with $3$ in the expression.

$f' (3) = - 6 \cos (3) \sin (3) - 6 \cos (3)$

Step 15.4.2

Simplify the result.

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Step 15.4.2.1

Simplify each term.

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Step 15.4.2.1.1

Evaluate $\cos (3)$ .

$f' (3) = - 6 \cdot (- 0.98999249 \sin (3)) - 6 \cos (3)$

Step 15.4.2.1.2

Multiply $- 6$ by $- 0.98999249$ .

$f' (3) = 5.93995497 \sin (3) - 6 \cos (3)$

Step 15.4.2.1.3

Evaluate $\sin (3)$ .

$f' (3) = 5.93995497 \cdot 0.14112 - 6 \cos (3)$

Step 15.4.2.1.4

Multiply $5.93995497$ by $0.14112$ .

$f' (3) = 0.83824649 - 6 \cos (3)$

Step 15.4.2.1.5

Evaluate $\cos (3)$ .

$f' (3) = 0.83824649 - 6 \cdot - 0.98999249$

Step 15.4.2.1.6

Multiply $- 6$ by $- 0.98999249$ .

$f' (3) = 0.83824649 + 5.93995497$

Step 15.4.2.2

Add $0.83824649$ and $5.93995497$ .

$f' (3) = 6.77820147$

Step 15.4.2.3

The final answer is $6.77820147$ .

$6.77820147$

Step 15.5

Substitute any number, such as $7$ , from the interval $(\frac{3 π}{2}, \infty)$ in the first derivative $- 6 \cos (x) \sin (x) - 6 \cos (x)$ to check if the result is negative or positive.

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Step 15.5.1

Replace the variable $x$ with $7$ in the expression.

$f' (7) = - 6 \cos (7) \sin (7) - 6 \cos (7)$

Step 15.5.2

Simplify the result.

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Step 15.5.2.1

Simplify each term.

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Step 15.5.2.1.1

Evaluate $\cos (7)$ .

$f' (7) = - 6 \cdot (0.75390225 \sin (7)) - 6 \cos (7)$

Step 15.5.2.1.2

Multiply $- 6$ by $0.75390225$ .

$f' (7) = - 4.52341352 \sin (7) - 6 \cos (7)$

Step 15.5.2.1.3

Evaluate $\sin (7)$ .

$f' (7) = - 4.52341352 \cdot 0.65698659 - 6 \cos (7)$

Step 15.5.2.1.4

Multiply $- 4.52341352$ by $0.65698659$ .

$f' (7) = - 2.97182206 - 6 \cos (7)$

Step 15.5.2.1.5

Evaluate $\cos (7)$ .

$f' (7) = - 2.97182206 - 6 \cdot 0.75390225$

Step 15.5.2.1.6

Multiply $- 6$ by $0.75390225$ .

$f' (7) = - 2.97182206 - 4.52341352$

Step 15.5.2.2

Subtract $4.52341352$ from $- 2.97182206$ .

$f' (7) = - 7.49523559$

Step 15.5.2.3

The final answer is $- 7.49523559$ .

$- 7.49523559$

Step 15.6

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 15.7

Since the first derivative changed signs from negative to positive around $x = \frac{π}{2}$ , then $x = \frac{π}{2}$ is a local minimum.

$x = \frac{π}{2}$ is a local minimum

Step 15.8

Since the first derivative changed signs from positive to negative around $x = \frac{3 π}{2}$ , then $x = \frac{3 π}{2}$ is a local maximum.

$x = \frac{3 π}{2}$ is a local maximum

Step 15.9

These are the local extrema for $f (x) = 3 \cos^{2} (x) - 6 \sin (x)$ .

$x = - \frac{π}{2}$ is a local maximum

$x = \frac{π}{2}$ is a local minimum

$x = \frac{3 π}{2}$ is a local maximum

$x = - \frac{π}{2}$ is a local maximum

$x = \frac{π}{2}$ is a local minimum

$x = \frac{3 π}{2}$ is a local maximum

Step 16