Calculus Examples

$y = - 2 \sec (4 x)$

Step 1

Write $y = - 2 \sec (4 x)$ as a function.

$f (x) = - 2 \sec (4 x)$

Step 2

Find the first derivative of the function.

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Step 2.1

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

$- 2 \frac{d}{d x} [\sec (4 x)]$

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 $f (x) = \sec (x)$ and $g (x) = 4 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $4 x$ .

$- 2 (\frac{d}{d u} [\sec (u)] \frac{d}{d x} [4 x])$

Step 2.2.2

The derivative of $\sec (u)$ with respect to $u$ is $\sec (u) \tan (u)$ .

$- 2 (\sec (u) \tan (u) \frac{d}{d x} [4 x])$

Step 2.2.3

Replace all occurrences of $u$ with $4 x$ .

$- 2 (\sec (4 x) \tan (4 x) \frac{d}{d x} [4 x])$

Step 2.3

Differentiate.

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Step 2.3.1

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

$- 2 \sec (4 x) \tan (4 x) (4 \frac{d}{d x} [x])$

Step 2.3.2

Multiply $4$ by $- 2$ .

$- 8 \sec (4 x) \tan (4 x) \frac{d}{d x} [x]$

Step 2.3.3

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

$- 8 \sec (4 x) \tan (4 x) \cdot 1$

Step 2.3.4

Multiply $- 8$ by $1$ .

$- 8 \sec (4 x) \tan (4 x)$

Step 3

Find the second derivative of the function.

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Step 3.1

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8 \sec (4 x) \tan (4 x)$ with respect to $x$ is $- 8 \frac{d}{d x} [\sec (4 x) \tan (4 x)]$ .

$f'' (x) = - 8 \frac{d}{d x} (\sec (4 x) \tan (4 x))$

Step 3.2

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 $f (x) = \sec (4 x)$ and $g (x) = \tan (4 x)$ .

$f'' (x) = - 8 (\sec (4 x) \frac{d}{d x} (\tan (4 x)) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \tan (x)$ and $g (x) = 4 x$ .

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Step 3.3.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$f'' (x) = - 8 (\sec (4 x) (\frac{d}{d u (1)} (\tan (u_{1})) \frac{d}{d x} (4 x)) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.3.2

The derivative of $\tan (u_{1})$ with respect to $u_{1}$ is $\sec^{2} (u_{1})$ .

$f'' (x) = - 8 (\sec (4 x) (\sec^{2} (u_{1}) \frac{d}{d x} (4 x)) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.3.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$f'' (x) = - 8 (\sec (4 x) (\sec^{2} (4 x) \frac{d}{d x} (4 x)) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.4

Raise $\sec (4 x)$ to the power of $1$ .

$f'' (x) = - 8 (\sec^{2} (4 x) \sec (4 x) \frac{d}{d x} (4 x) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.5

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

$f'' (x) = - 8 ({\sec (4 x)}^{2 + 1} \frac{d}{d x} (4 x) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.6

Differentiate.

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Step 3.6.1

Add $2$ and $1$ .

$f'' (x) = - 8 (\sec^{3} (4 x) \frac{d}{d x} (4 x) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.6.2

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

$f'' (x) = - 8 (\sec^{3} (4 x) (4 \frac{d}{d x} (x)) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.6.3

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

$f'' (x) = - 8 (\sec^{3} (4 x) (4 \cdot 1) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.6.4

Simplify the expression.

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Step 3.6.4.1

Multiply $4$ by $1$ .

$f'' (x) = - 8 (\sec^{3} (4 x) \cdot 4 + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.6.4.2

Move $4$ to the left of $\sec^{3} (4 x)$ .

$f'' (x) = - 8 (4 \cdot \sec^{3} (4 x) + \tan (4 x) \frac{d}{d x} (\sec (4 x)))$

Step 3.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sec (x)$ and $g (x) = 4 x$ .

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Step 3.7.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan (4 x) (\frac{d}{d u (2)} (\sec (u_{2})) \frac{d}{d x} (4 x)))$

Step 3.7.2

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

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan (4 x) (\sec (u_{2}) \tan (u_{2}) \frac{d}{d x} (4 x)))$

Step 3.7.3

Replace all occurrences of $u_{2}$ with $4 x$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan (4 x) (\sec (4 x) \tan (4 x) \frac{d}{d x} (4 x)))$

Step 3.8

Raise $\tan (4 x)$ to the power of $1$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan (4 x) \tan (4 x) (\sec (4 x) \frac{d}{d x} (4 x)))$

Step 3.9

Raise $\tan (4 x)$ to the power of $1$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan (4 x) \tan (4 x) (\sec (4 x) \frac{d}{d x} (4 x)))$

Step 3.10

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

$f'' (x) = - 8 (4 \sec^{3} (4 x) + {\tan (4 x)}^{1 + 1} (\sec (4 x) \frac{d}{d x} (4 x)))$

Step 3.11

Add $1$ and $1$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan^{2} (4 x) (\sec (4 x) \frac{d}{d x} (4 x)))$

Step 3.12

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

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan^{2} (4 x) \sec (4 x) (4 \frac{d}{d x} (x)))$

Step 3.13

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

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan^{2} (4 x) \sec (4 x) (4 \cdot 1))$

Step 3.14

Simplify the expression.

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Step 3.14.1

Multiply $4$ by $1$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + \tan^{2} (4 x) \sec (4 x) \cdot 4)$

Step 3.14.2

Move $4$ to the left of $\tan^{2} (4 x) \sec (4 x)$ .

$f'' (x) = - 8 (4 \sec^{3} (4 x) + 4 \cdot (\tan^{2} (4 x) \sec (4 x)))$

Step 3.15

Simplify.

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Step 3.15.1

Apply the distributive property.

$f'' (x) = - 8 (4 \sec^{3} (4 x)) - 8 (4 \tan^{2} (4 x) \sec (4 x))$

Step 3.15.2

Combine terms.

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Step 3.15.2.1

Multiply $4$ by $- 8$ .

$f'' (x) = - 32 \sec^{3} (4 x) - 8 (4 \tan^{2} (4 x) \sec (4 x))$

Step 3.15.2.2

Multiply $4$ by $- 8$ .

$f'' (x) = - 32 \sec^{3} (4 x) - 32 \tan^{2} (4 x) \sec (4 x)$

Step 3.15.3

Reorder terms.

$f'' (x) = - 32 \tan^{2} (4 x) \sec (4 x) - 32 \sec^{3} (4 x)$

Step 4

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

$- 8 \sec (4 x) \tan (4 x) = 0$

Step 5

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

$\sec (4 x) = 0$

$\tan (4 x) = 0$

Step 6

Set $\sec (4 x)$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $\sec (4 x)$ equal to $0$ .

$\sec (4 x) = 0$

Step 6.2

The range of secant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

Step 7

Set $\tan (4 x)$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $\tan (4 x)$ equal to $0$ .

$\tan (4 x) = 0$

Step 7.2

Solve $\tan (4 x) = 0$ for $x$ .

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Step 7.2.1

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

$4 x = \arctan (0)$

Step 7.2.2

Simplify the right side.

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Step 7.2.2.1

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

$4 x = 0$

Step 7.2.3

Divide each term in $4 x = 0$ by $4$ and simplify.

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Step 7.2.3.1

Divide each term in $4 x = 0$ by $4$ .

$\frac{4 x}{4} = \frac{0}{4}$

Step 7.2.3.2

Simplify the left side.

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Step 7.2.3.2.1

Cancel the common factor of $4$ .

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Step 7.2.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{0}{4}$

Step 7.2.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{0}{4}$

Step 7.2.3.3

Simplify the right side.

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Step 7.2.3.3.1

Divide $0$ by $4$ .

$x = 0$

Step 7.2.4

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.

$4 x = π + 0$

Step 7.2.5

Solve for $x$ .

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Step 7.2.5.1

Add $π$ and $0$ .

$4 x = π$

Step 7.2.5.2

Divide each term in $4 x = π$ by $4$ and simplify.

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Step 7.2.5.2.1

Divide each term in $4 x = π$ by $4$ .

$\frac{4 x}{4} = \frac{π}{4}$

Step 7.2.5.2.2

Simplify the left side.

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Step 7.2.5.2.2.1

Cancel the common factor of $4$ .

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Step 7.2.5.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{π}{4}$

Step 7.2.5.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{π}{4}$

Step 7.2.6

The solution to the equation $4 x = 0$ .

$x = 0, \frac{π}{4}$

Step 8

The final solution is all the values that make $- 8 \sec (4 x) \tan (4 x) = 0$ true.

$x = 0, \frac{π}{4}$

Step 9

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.

$- 32 \tan^{2} (4 (0)) \sec (4 (0)) - 32 \sec^{3} (4 (0))$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

Multiply $4$ by $0$ .

$- 32 \tan^{2} (0) \sec (4 (0)) - 32 \sec^{3} (4 (0))$

Step 10.1.2

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

$- 32 \cdot 0^{2} \sec (4 (0)) - 32 \sec^{3} (4 (0))$

Step 10.1.3

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

$- 32 \cdot 0 \sec (4 (0)) - 32 \sec^{3} (4 (0))$

Step 10.1.4

Multiply $- 32$ by $0$ .

$0 \sec (4 (0)) - 32 \sec^{3} (4 (0))$

Step 10.1.5

Multiply $4$ by $0$ .

$0 \sec (0) - 32 \sec^{3} (4 (0))$

Step 10.1.6

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

$0 \cdot 1 - 32 \sec^{3} (4 (0))$

Step 10.1.7

Multiply $0$ by $1$ .

$0 - 32 \sec^{3} (4 (0))$

Step 10.1.8

Multiply $4$ by $0$ .

$0 - 32 \sec^{3} (0)$

Step 10.1.9

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

$0 - 32 \cdot 1^{3}$

Step 10.1.10

One to any power is one.

$0 - 32 \cdot 1$

Step 10.1.11

Multiply $- 32$ by $1$ .

$0 - 32$

Step 10.2

Subtract $32$ from $0$ .

$- 32$

Step 11

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

$x = 0$ is a local maximum

Step 12

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

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Step 12.1

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

$f (0) = - 2 \sec (4 (0))$

Step 12.2

Simplify the result.

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Step 12.2.1

Multiply $4$ by $0$ .

$f (0) = - 2 \sec (0)$

Step 12.2.2

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

$f (0) = - 2 \cdot 1$

Step 12.2.3

Multiply $- 2$ by $1$ .

$f (0) = - 2$

Step 12.2.4

The final answer is $- 2$ .

$y = - 2$

Step 13

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.

$- 32 \tan^{2} (4 (\frac{π}{4})) \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14

Evaluate the second derivative.

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Step 14.1

Simplify each term.

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Step 14.1.1

Cancel the common factor of $4$ .

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Step 14.1.1.1

Cancel the common factor.

$- 32 \tan^{2} (4 \frac{π}{4}) \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.1.2

Rewrite the expression.

$- 32 \tan^{2} (π) \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because tangent is negative in the second quadrant.

$- 32 {(- \tan (0))}^{2} \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.3

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

$- 32 {(- 0)}^{2} \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.4

Multiply $- 1$ by $0$ .

$- 32 \cdot 0^{2} \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.5

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

$- 32 \cdot 0 \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.6

Multiply $- 32$ by $0$ .

$0 \sec (4 (\frac{π}{4})) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.7

Cancel the common factor of $4$ .

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Step 14.1.7.1

Cancel the common factor.

$0 \sec (4 \frac{π}{4}) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.7.2

Rewrite the expression.

$0 \sec (π) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.8

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$0 (- \sec (0)) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.9

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

$0 (- 1 \cdot 1) - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.10

Multiply $0 (- 1 \cdot 1)$ .

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Step 14.1.10.1

Multiply $- 1$ by $1$ .

$0 \cdot - 1 - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.10.2

Multiply $0$ by $- 1$ .

$0 - 32 \sec^{3} (4 (\frac{π}{4}))$

Step 14.1.11

Cancel the common factor of $4$ .

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Step 14.1.11.1

Cancel the common factor.

$0 - 32 \sec^{3} (4 \frac{π}{4})$

Step 14.1.11.2

Rewrite the expression.

$0 - 32 \sec^{3} (π)$

Step 14.1.12

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$0 - 32 {(- \sec (0))}^{3}$

Step 14.1.13

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

$0 - 32 {(- 1 \cdot 1)}^{3}$

Step 14.1.14

Multiply $- 1$ by $1$ .

$0 - 32 {(- 1)}^{3}$

Step 14.1.15

Raise $- 1$ to the power of $3$ .

$0 - 32 \cdot - 1$

Step 14.1.16

Multiply $- 32$ by $- 1$ .

$0 + 32$

Step 14.2

Add $0$ and $32$ .

$32$

Step 15

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

$x = \frac{π}{4}$ is a local minimum

Step 16

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

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Step 16.1

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

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

Step 16.2

Simplify the result.

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Step 16.2.1

Cancel the common factor of $4$ .

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Step 16.2.1.1

Cancel the common factor.

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

Step 16.2.1.2

Rewrite the expression.

$f (\frac{π}{4}) = - 2 \sec (π)$

Step 16.2.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because secant is negative in the second quadrant.

$f (\frac{π}{4}) = - 2 (- \sec (0))$

Step 16.2.3

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

$f (\frac{π}{4}) = - 2 (- 1 \cdot 1)$

Step 16.2.4

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

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Step 16.2.4.1

Multiply $- 1$ by $1$ .

$f (\frac{π}{4}) = - 2 \cdot - 1$

Step 16.2.4.2

Multiply $- 2$ by $- 1$ .

$f (\frac{π}{4}) = 2$

Step 16.2.5

The final answer is $2$ .

$y = 2$

Step 17

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

$(0, - 2)$ is a local maxima

$(\frac{π}{4}, 2)$ is a local minima

Step 18