Precalculus Examples

$f (x) = 4 \sin (2 x) - 3$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $4 \sin (2 x) - 3$ with respect to $x$ is $\frac{d}{d x} [4 \sin (2 x)] + \frac{d}{d x} [- 3]$ .

$\frac{d}{d x} [4 \sin (2 x)] + \frac{d}{d x} [- 3]$

Step 1.2

Evaluate $\frac{d}{d x} [4 \sin (2 x)]$ .

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Step 1.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 \sin (2 x)$ with respect to $x$ is $4 \frac{d}{d x} [\sin (2 x)]$ .

$4 \frac{d}{d x} [\sin (2 x)] + \frac{d}{d x} [- 3]$

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

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Step 1.2.2.1

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

$4 (\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 3]$

Step 1.2.2.2

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

$4 (\cos (u) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 3]$

Step 1.2.2.3

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

$4 (\cos (2 x) \frac{d}{d x} [2 x]) + \frac{d}{d x} [- 3]$

Step 1.2.3

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

$4 (\cos (2 x) (2 \frac{d}{d x} [x])) + \frac{d}{d x} [- 3]$

Step 1.2.4

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

$4 (\cos (2 x) (2 \cdot 1)) + \frac{d}{d x} [- 3]$

Step 1.2.5

Multiply $2$ by $1$ .

$4 (\cos (2 x) \cdot 2) + \frac{d}{d x} [- 3]$

Step 1.2.6

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

$4 (2 \cdot \cos (2 x)) + \frac{d}{d x} [- 3]$

Step 1.2.7

Multiply $2$ by $4$ .

$8 \cos (2 x) + \frac{d}{d x} [- 3]$

Step 1.3

Differentiate using the Constant Rule.

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Step 1.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$8 \cos (2 x) + 0$

Step 1.3.2

Add $8 \cos (2 x)$ and $0$ .

$8 \cos (2 x)$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $8$ is constant with respect to $x$ , the derivative of $8 \cos (2 x)$ with respect to $x$ is $8 \frac{d}{d x} [\cos (2 x)]$ .

$f'' (x) = 8 \frac{d}{d x} (\cos (2 x))$

Step 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 2.2.1

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

$f'' (x) = 8 (\frac{d}{d u} (\cos (u)) \frac{d}{d x} (2 x))$

Step 2.2.2

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

$f'' (x) = 8 (- \sin (u) \frac{d}{d x} (2 x))$

Step 2.2.3

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

$f'' (x) = 8 (- \sin (2 x) \frac{d}{d x} (2 x))$

Step 2.3

Differentiate.

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Step 2.3.1

Multiply $- 1$ by $8$ .

$f'' (x) = - 8 (\sin (2 x) \frac{d}{d x} (2 x))$

Step 2.3.2

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

$f'' (x) = - 8 \sin (2 x) (2 \frac{d}{d x} (x))$

Step 2.3.3

Multiply $2$ by $- 8$ .

$f'' (x) = - 16 \sin (2 x) \frac{d}{d x} x$

Step 2.3.4

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

$f'' (x) = - 16 \sin (2 x) \cdot 1$

Step 2.3.5

Multiply $- 16$ by $1$ .

$f'' (x) = - 16 \sin (2 x)$

Step 3

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

$8 \cos (2 x) = 0$

Step 4

Divide each term in $8 \cos (2 x) = 0$ by $8$ and simplify.

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Step 4.1

Divide each term in $8 \cos (2 x) = 0$ by $8$ .

$\frac{8 \cos (2 x)}{8} = \frac{0}{8}$

Step 4.2

Simplify the left side.

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Step 4.2.1

Cancel the common factor of $8$ .

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Step 4.2.1.1

Cancel the common factor.

$\frac{8 \cos (2 x)}{8} = \frac{0}{8}$

Step 4.2.1.2

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

$\cos (2 x) = \frac{0}{8}$

Step 4.3

Simplify the right side.

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Step 4.3.1

Divide $0$ by $8$ .

$\cos (2 x) = 0$

Step 5

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

$2 x = \arccos (0)$

Step 6

Simplify the right side.

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Step 6.1

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

$2 x = \frac{π}{2}$

Step 7

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

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Step 7.1

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

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

Step 7.2

Simplify the left side.

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Step 7.2.1

Cancel the common factor of $2$ .

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Step 7.2.1.1

Cancel the common factor.

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

Step 7.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{2}}{2}$

Step 7.3

Simplify the right side.

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Step 7.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{2} \cdot \frac{1}{2}$

Step 7.3.2

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

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Step 7.3.2.1

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

$x = \frac{π}{2 \cdot 2}$

Step 7.3.2.2

Multiply $2$ by $2$ .

$x = \frac{π}{4}$

Step 8

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.

$2 x = 2 π - \frac{π}{2}$

Step 9

Solve for $x$ .

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Step 9.1

Simplify.

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Step 9.1.1

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

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

Step 9.1.2

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 9.1.3

Combine the numerators over the common denominator.

$2 x = \frac{2 π \cdot 2 - π}{2}$

Step 9.1.4

Multiply $2$ by $2$ .

$2 x = \frac{4 π - π}{2}$

Step 9.1.5

Subtract $π$ from $4 π$ .

$2 x = \frac{3 π}{2}$

Step 9.2

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ and simplify.

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Step 9.2.1

Divide each term in $2 x = \frac{3 π}{2}$ by $2$ .

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

Step 9.2.2

Simplify the left side.

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Step 9.2.2.1

Cancel the common factor of $2$ .

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Step 9.2.2.1.1

Cancel the common factor.

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

Step 9.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{3 π}{2}}{2}$

Step 9.2.3

Simplify the right side.

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Step 9.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{3 π}{2} \cdot \frac{1}{2}$

Step 9.2.3.2

Multiply $\frac{3 π}{2} \cdot \frac{1}{2}$ .

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Step 9.2.3.2.1

Multiply $\frac{3 π}{2}$ by $\frac{1}{2}$ .

$x = \frac{3 π}{2 \cdot 2}$

Step 9.2.3.2.2

Multiply $2$ by $2$ .

$x = \frac{3 π}{4}$

Step 10

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

$x = \frac{π}{4}, \frac{3 π}{4}$

Step 11

Evaluate the second derivative at $x = \frac{π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 16 \sin (2 (\frac{π}{4}))$

Step 12

Evaluate the second derivative.

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Step 12.1

Cancel the common factor of $2$ .

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Step 12.1.1

Factor $2$ out of $4$ .

$- 16 \sin (2 \frac{π}{2 (2)})$

Step 12.1.2

Cancel the common factor.

$- 16 \sin (2 \frac{π}{2 \cdot 2})$

Step 12.1.3

Rewrite the expression.

$- 16 \sin (\frac{π}{2})$

Step 12.2

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

$- 16 \cdot 1$

Step 12.3

Multiply $- 16$ by $1$ .

$- 16$

Step 13

$x = \frac{π}{4}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = \frac{π}{4}$ is a local maximum

Step 14

Find the y-value when $x = \frac{π}{4}$ .

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Step 14.1

Replace the variable $x$ with $\frac{π}{4}$ in the expression.

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

Step 14.2

Simplify the result.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

Cancel the common factor of $2$ .

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Step 14.2.1.1.1

Factor $2$ out of $4$ .

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

Step 14.2.1.1.2

Cancel the common factor.

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

Step 14.2.1.1.3

Rewrite the expression.

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

Step 14.2.1.2

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

$f (\frac{π}{4}) = 4 \cdot 1 - 3$

Step 14.2.1.3

Multiply $4$ by $1$ .

$f (\frac{π}{4}) = 4 - 3$

Step 14.2.2

Subtract $3$ from $4$ .

$f (\frac{π}{4}) = 1$

Step 14.2.3

The final answer is $1$ .

$y = 1$

Step 15

Evaluate the second derivative at $x = \frac{3 π}{4}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 16 \sin (2 (\frac{3 π}{4}))$

Step 16

Evaluate the second derivative.

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Step 16.1

Cancel the common factor of $2$ .

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Step 16.1.1

Factor $2$ out of $4$ .

$- 16 \sin (2 \frac{3 π}{2 (2)})$

Step 16.1.2

Cancel the common factor.

$- 16 \sin (2 \frac{3 π}{2 \cdot 2})$

Step 16.1.3

Rewrite the expression.

$- 16 \sin (\frac{3 π}{2})$

Step 16.2

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 fourth quadrant.

$- 16 (- \sin (\frac{π}{2}))$

Step 16.3

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

$- 16 (- 1 \cdot 1)$

Step 16.4

Multiply $- 16 (- 1 \cdot 1)$ .

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Step 16.4.1

Multiply $- 1$ by $1$ .

$- 16 \cdot - 1$

Step 16.4.2

Multiply $- 16$ by $- 1$ .

$16$

Step 17

$x = \frac{3 π}{4}$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = \frac{3 π}{4}$ is a local minimum

Step 18

Find the y-value when $x = \frac{3 π}{4}$ .

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Step 18.1

Replace the variable $x$ with $\frac{3 π}{4}$ in the expression.

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

Step 18.2

Simplify the result.

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Step 18.2.1

Simplify each term.

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Step 18.2.1.1

Cancel the common factor of $2$ .

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Step 18.2.1.1.1

Factor $2$ out of $4$ .

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

Step 18.2.1.1.2

Cancel the common factor.

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

Step 18.2.1.1.3

Rewrite the expression.

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

Step 18.2.1.2

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 fourth quadrant.

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

Step 18.2.1.3

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

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

Step 18.2.1.4

Multiply $4 (- 1 \cdot 1)$ .

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Step 18.2.1.4.1

Multiply $- 1$ by $1$ .

$f (\frac{3 π}{4}) = 4 \cdot - 1 - 3$

Step 18.2.1.4.2

Multiply $4$ by $- 1$ .

$f (\frac{3 π}{4}) = - 4 - 3$

Step 18.2.2

Subtract $3$ from $- 4$ .

$f (\frac{3 π}{4}) = - 7$

Step 18.2.3

The final answer is $- 7$ .

$y = - 7$

Step 19

These are the local extrema for $f (x) = 4 \sin (2 x) - 3$ .

$(\frac{π}{4}, 1)$ is a local maxima

$(\frac{3 π}{4}, - 7)$ is a local minima

Step 20