Problem 28 - Calculus A New Horizon

Question

Analyze the trigonometric function \(f\) over the specified interval, stating where \(f\) is increasing. decreasing, concave up, and concave down, and stating the \(x\) coordinates of all inflection points. Confirm that your results are consistent with the graph of \(f\) generated with a graphing utility. $$f(x)=2 x+\cot x ;(0, \pi)$$

Step-by-Step Solution

Answer

f is increasing on (0, π/4) and (3π/4, π); decreasing on (π/4, 3π/4). f is concave down on (0, π/2) and concave up on (π/2, π) with an inflection point at x = π/2.

Step 1: Find the First Derivative

The first derivative of a function helps determine where it is increasing or decreasing. The given function is \( f(x) = 2x + \cot x \). The derivative of \( 2x \) is 2, and the derivative of \( \cot x \) is \( -\csc^2 x \). Therefore, the first derivative is \( f'(x) = 2 - \csc^2 x \).

Step 2: Analyze the First Derivative

To determine where \( f(x) \) is increasing or decreasing, set \( f'(x) > 0 \) for increasing and \( f'(x) < 0 \) for decreasing. Solve \( 2 - \csc^2 x > 0 \), leading to \( 2 > \csc^2 x \) or equivalently \( \csc^2 x < 2 \). The solutions for this inequality can be found between the values where sine is around \( \pm 1/\sqrt{2} \). This condition holds true in the intervals \( (0, \pi/4) \) and \( (3\pi/4, \pi) \). Consequently, \( f \) is increasing on these intervals and decreasing on \( (\pi/4, 3\pi/4) \).

Step 3: Find the Second Derivative

The second derivative helps in finding concavity and inflection points. Calculate the second derivative of \( f(x) \), where \( f''(x) = \frac{d}{dx}(-\csc^2 x) = 2\csc^2 x \cot x \).

Step 4: Analyze the Second Derivative

To determine the intervals of concavity and inflection points, set \( f''(x) = 0 \). However, \( f''(x) = 2\csc^2 x \cot x \) cannot be zero as \( \cot x \) is undefined at the same points \( x \) where \( \csc^2 x \) leads to a zero product. Check the sign of \( f''(x) \) in each interval: \( f \) is concave up where \( f''(x) > 0 \) and concave down where \( f''(x) < 0 \). Solving this, we find that \( f \) is concave down on \( (0, \pi/2) \) and \( f \) is concave up on \( (\pi/2, \pi) \).

Step 5: Identify Inflection Points

Inflection points occur where the concavity changes. There is a change in concavity at \( x = \pi/2 \) from concave down to concave up. Therefore, there is an inflection point at \( x = \pi/2 \).

Key Concepts

First Derivative

The first derivative of a function provides crucial insights into the intervals where the function is increasing or decreasing. To find the first derivative of the given function \( f(x) = 2x + \cot x \), we start by differentiating each term individually. The derivative of \( 2x \) is simply \( 2 \). For \( \cot x \), the derivative is \( -\csc^2 x \). Bringing them together, we obtain the first derivative: \( f'(x) = 2 - \csc^2 x \). Knowing \( f'(x) \), we can explore where the function increases or decreases by examining the sign of \( f'(x) \). To determine this:

• Set \( f'(x) > 0 \) to find intervals where the function is increasing.
• Set \( f'(x) < 0 \) to find intervals where the function is decreasing.

Solving \( 2 > \csc^2 x \) tells us where the function increases. Specifically, the function increases in the intervals \( (0, \pi/4) \) and \( (3\pi/4, \pi) \). On the other hand, the function decreases in the interval \( (\pi/4, 3\pi/4) \). Understanding these intervals helps to sketch the function's general shape.

Second Derivative

The second derivative of a function provides insight into the function's concavity and inflection points. For the given function, we need to differentiate the first derivative \( f'(x) = 2 - \csc^2 x \) to find \( f''(x) \). Differentiating \( f'(x) \), we get \( f''(x) = \frac{d}{dx}(-\csc^2 x) = 2\csc^2 x \cot x \). This expression will allow us to analyze concavity. The second derivative is instrumental in determining:

• Concave up intervals, where \( f''(x) > 0 \).
• Concave down intervals, where \( f''(x) < 0 \).

For this particular function, note that the product \( 2\csc^2 x \cot x \) relates to varying sine and cotangent behaviors, impacting concavity. Analyzing the sign changes reveals that the function is concave down on \( (0, \pi/2) \), Conversely, it is concave up on \( (\pi/2, \pi) \). Recognizing these intervals is key in predicting the function's curve.

Concavity and Inflection Points

Concavity determines the nature of the curve's bend and where it changes direction. Inflection points occur where the concavity changes. As we determined previously, \( f''(x) \) helps find these points, particularly at sign changes from positive to negative or vice versa. For the function \( f(x) = 2x + \cot x \), the second derivative analysis showed shifts:

• From concave down \((0, \pi/2)\) to concave up \((\pi/2, \pi)\).

This change occurs at \( x = \pi/2 \), marking an inflection point. Inflection points are crucial because they represent a switch in the concave nature of the graph, providing insight into the overall shape and behavior of the curve over its domain. Identifying them allows us to understand how the function's slope increases or decreases over different intervals.