Calculus Examples

$g (x) = - 5 \sec (x)$ , $- \frac{π}{2} < x < \frac{3 π}{2}$

Step 1

Find the critical points.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Find the first derivative.

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Step 1.1.1.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 \sec (x)$ with respect to $x$ is $- 5 \frac{d}{d x} [\sec (x)]$ .

$- 5 \frac{d}{d x} [\sec (x)]$

Step 1.1.1.2

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

$f' (x) = - 5 \sec (x) \tan (x)$

Step 1.1.2

The first derivative of $g (x)$ with respect to $x$ is $- 5 \sec (x) \tan (x)$ .

$- 5 \sec (x) \tan (x)$

Step 1.2

Set the first derivative equal to $0$ then solve the equation $- 5 \sec (x) \tan (x) = 0$ .

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Step 1.2.1

Set the first derivative equal to $0$ .

$- 5 \sec (x) \tan (x) = 0$

Step 1.2.2

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

$\sec (x) = 0$

$\tan (x) = 0$

Step 1.2.3

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

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Step 1.2.3.1

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

$\sec (x) = 0$

Step 1.2.3.2

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 1.2.4

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

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Step 1.2.4.1

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

$\tan (x) = 0$

Step 1.2.4.2

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

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Step 1.2.4.2.1

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

$x = \arctan (0)$

Step 1.2.4.2.2

Simplify the right side.

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Step 1.2.4.2.2.1

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

$x = 0$

Step 1.2.4.2.3

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.

$x = π + 0$

Step 1.2.4.2.4

Add $π$ and $0$ .

$x = π$

Step 1.2.4.2.5

Find the period of $\tan (x)$ .

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Step 1.2.4.2.5.1

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Step 1.2.4.2.5.2

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

Step 1.2.4.2.5.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Step 1.2.4.2.5.4

Divide $π$ by $1$ .

$π$

Step 1.2.4.2.6

The period of the $\tan (x)$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = π n, π + π n$ , for any integer $n$

Step 1.2.5

The final solution is all the values that make $- 5 \sec (x) \tan (x) = 0$ true.

$x = π n, π + π n$ , for any integer $n$

Step 1.2.6

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 1.3

Find the values where the derivative is undefined.

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Step 1.3.1

Set the argument in $\sec (x)$ equal to $\frac{π}{2} + π n$ to find where the expression is undefined.

$x = \frac{π}{2} + π n$ , for any integer $n$

Step 1.3.2

The equation is undefined where the denominator equals $0$ , the argument of a square root is less than $0$ , or the argument of a logarithm is less than or equal to $0$ .

${x | x = \frac{π}{2} + π n}$ $n$ , for any integer $n$

Step 1.4

Evaluate $- 5 \sec (x)$ at each $x$ value where the derivative is $0$ or undefined.

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Step 1.4.1

Evaluate at $x = 0$ .

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Step 1.4.1.1

Substitute $0$ for $x$ .

$- 5 \sec (0)$

Step 1.4.1.2

Simplify.

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Step 1.4.1.2.1

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

$- 5 \cdot 1$

Step 1.4.1.2.2

Multiply $- 5$ by $1$ .

$- 5$

Step 1.4.2

Evaluate at $x = π$ .

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Step 1.4.2.1

Substitute $π$ for $x$ .

$- 5 \sec (π)$

Step 1.4.2.2

Simplify.

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Step 1.4.2.2.1

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.

$- 5 (- \sec (0))$

Step 1.4.2.2.2

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

$- 5 (- 1 \cdot 1)$

Step 1.4.2.2.3

Multiply $- 5 (- 1 \cdot 1)$ .

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Step 1.4.2.2.3.1

Multiply $- 1$ by $1$ .

$- 5 \cdot - 1$

Step 1.4.2.2.3.2

Multiply $- 5$ by $- 1$ .

$5$

Step 1.4.3

List all of the points.

$(0 + 2 π n, - 5), (π + 2 π n, 5)$ , for any integer $n$

Step 2

Exclude the points that are not on the interval.

$(0, - 5), (π, 5)$

Step 3

Use the second derivative test to determine which points can be maxima or minima.

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Step 3.1

Find the second derivative.

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Step 3.1.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 \sec (x) \tan (x)$ with respect to $x$ is $- 5 \frac{d}{d x} [\sec (x) \tan (x)]$ .

$- 5 \frac{d}{d x} [\sec (x) \tan (x)]$

Step 3.1.2

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

$- 5 (\sec (x) \frac{d}{d x} [\tan (x)] + \tan (x) \frac{d}{d x} [\sec (x)])$

Step 3.1.3

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

$- 5 (\sec (x) \sec^{2} (x) + \tan (x) \frac{d}{d x} [\sec (x)])$

Step 3.1.4

Multiply $\sec (x)$ by $\sec^{2} (x)$ by adding the exponents.

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Step 3.1.4.1

Multiply $\sec (x)$ by $\sec^{2} (x)$ .

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Step 3.1.4.1.1

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

$- 5 (\sec^{1} (x) \sec^{2} (x) + \tan (x) \frac{d}{d x} [\sec (x)])$

Step 3.1.4.1.2

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

$- 5 ({\sec (x)}^{1 + 2} + \tan (x) \frac{d}{d x} [\sec (x)])$

Step 3.1.4.2

Add $1$ and $2$ .

$- 5 (\sec^{3} (x) + \tan (x) \frac{d}{d x} [\sec (x)])$

Step 3.1.5

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

$- 5 (\sec^{3} (x) + \tan (x) (\sec (x) \tan (x)))$

Step 3.1.6

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

$- 5 (\sec^{3} (x) + \tan^{1} (x) \tan (x) \sec (x))$

Step 3.1.7

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

$- 5 (\sec^{3} (x) + \tan^{1} (x) \tan^{1} (x) \sec (x))$

Step 3.1.8

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

$- 5 (\sec^{3} (x) + {\tan (x)}^{1 + 1} \sec (x))$

Step 3.1.9

Add $1$ and $1$ .

$- 5 (\sec^{3} (x) + \tan^{2} (x) \sec (x))$

Step 3.1.10

Simplify.

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Step 3.1.10.1

Apply the distributive property.

$- 5 \sec^{3} (x) - 5 (\tan^{2} (x) \sec (x))$

Step 3.1.10.2

Reorder terms.

$- 5 \tan^{2} (x) \sec (x) - 5 \sec^{3} (x)$

Step 3.2

Substitute $0$ for $x$ and simplify.

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Step 3.2.1

Substitute $0$ for $x$ .

$- 5 \tan^{2} (0) \sec (0) - 5 \sec^{3} (0)$

Step 3.2.2

Evaluate $\tan (0)$ .

$- 5 \cdot 0^{2} \sec (0) - 5 \sec^{3} (0)$

Step 3.2.3

Raise $0$ to the power of $2$ .

$- 5 \cdot 0 \sec (0) - 5 \sec^{3} (0)$

Step 3.2.4

Multiply $- 5$ by $0$ .

$0 \sec (0) - 5 \sec^{3} (0)$

Step 3.2.5

Evaluate $\sec (0)$ .

$0 \cdot 1 - 5 \sec^{3} (0)$

Step 3.2.6

Multiply $0$ by $1$ .

$0 - 5 \sec^{3} (0)$

Step 3.2.7

Evaluate $\sec (0)$ .

$0 - 5 \cdot 1^{3}$

Step 3.2.8

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

$0 - 5 \cdot 1$

Step 3.2.9

Multiply $- 5$ by $1$ .

$0 - 5$

Step 3.2.10

Subtract $5$ from $0$ .

$- 5$

Step 3.3

Because the second derivative is negative at $0$ , it is a maximum.

$0$ is a local maximum

Step 3.4

Substitute $π$ for $x$ and simplify.

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Step 3.4.1

Substitute $π$ for $x$ .

$- 5 \tan^{2} (π) \sec (π) - 5 \sec^{3} (π)$

Step 3.4.2

Evaluate $\tan (π)$ .

$- 5 {(0)}^{2} \sec (π) - 5 \sec^{3} (π)$

Step 3.4.3

Raise $0$ to the power of $2$ .

$- 5 \cdot 0 \sec (π) - 5 \sec^{3} (π)$

Step 3.4.4

Multiply $- 5$ by $0$ .

$0 \sec (π) - 5 \sec^{3} (π)$

Step 3.4.5

Evaluate $\sec (π)$ .

$0 \cdot - 1 - 5 \sec^{3} (π)$

Step 3.4.6

Multiply $0$ by $- 1$ .

$0 - 5 \sec^{3} (π)$

Step 3.4.7

Evaluate $\sec (π)$ .

$0 - 5 {(- 1)}^{3}$

Step 3.4.8

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

$0 - 5 \cdot - 1$

Step 3.4.9

Multiply $- 5$ by $- 1$ .

$0 + 5$

Step 3.4.10

Add $0$ and $5$ .

$5$

Step 3.5

Because the second derivative is positive at $π$ , it is a minimum.

$π$ is a local minimum

Step 3.6

List the local extrema

$0$ is a local maximum

$π$ is a local minimum

$0$ is a local maximum

$π$ is a local minimum

Step 4

Compare the $g (x)$ values found for each value of $x$ in order to determine the absolute maximum and minimum over the given interval. The maximum will occur at the highest $g (x)$ value and the minimum will occur at the lowest $g (x)$ value.

Absolute Maximum: $(0, - 5)$

Absolute Minimum: $(π, 5)$

Step 5