Problem 27 - 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. $$f(x)=\tan x ;(-\pi / 2, \pi / 2)$$

Step-by-Step Solution

Answer

The function \(f(x) = \tan x\) is increasing on \((-\pi/2, \pi/2)\), concave up on \((0, \pi/2)\), concave down on \((-\pi/2, 0)\), and has an inflection point at \(x = 0\).

Step 1: Understanding the Function

The function given is a trigonometric function, specifically the tangent function, \(f(x) = \tan x\). This function has a period of \(\pi\) and has asymptotes at \(x = \pm \frac{\pi}{2}\). We will analyze its behavior on the interval \((-\pi/2, \pi/2)\).

Step 2: Finding the First Derivative

To determine where \(f(x) = \tan x\) is increasing or decreasing, we find the first derivative. The derivative of \(\tan x\) is \(f'(x) = \sec^2 x\). Since \(\sec^2 x > 0\) for all \(x\) in the domain, the function is increasing everywhere on the interval \((-\pi/2, \pi/2)\).

Step 3: Analyzing Concavity with the Second Derivative

To find concavity, we calculate the second derivative of \(f(x)\). The second derivative is \(f''(x) = 2 \sec^2 x \tan x\). To determine concavity, we need to analyze the sign of \(f''(x)\).

Step 4: Determining Concavity Intervals

Since \(\tan x\) is greater than zero for \(x \in (0, \pi/2)\) and less than zero for \(x \in (-\pi/2, 0)\), \(f''(x) = 2 \sec^2 x \tan x\) is positive in \( (0, \pi/2) \) (concave up) and negative in \((-\pi/2, 0)\) (concave down).

Step 5: Finding Inflection Points

Inflection points occur where \(f''(x)\) changes sign. \(f''(x)\) changes sign at \(x = 0\), so there is an inflection point at \(x = 0\).

Step 6: Confirming with Graph

To confirm our analysis, graph the function \(f(x) = \tan x\) over the interval \((-\pi/2, \pi/2)\). The graph should show the function increasing throughout, concave down from \(-\pi/2\) to 0 and concave up from 0 to \(\pi/2\), with an inflection point at \(x = 0\).

Key Concepts

Tangent function

The tangent function, represented as \( f(x) = \tan x \), is one of the basic trigonometric functions. It is defined as the ratio of the sine and cosine of angle \( x \). This function is periodic with a period of \( \pi \), which means it repeats its pattern every \( \pi \) units. The unique aspect of the tangent function is its vertical asymptotes, which occur at \( x = \pm \frac{\pi}{2}, \pm \frac{3\pi}{2}, \dots \) where the function is undefined. For the interval \( (-\pi/2, \pi/2) \), which is a standard period of the tangent, the function increases rapidly from \(-\infty\) to \(+\infty\). It does not plateau or decrease, as the slope is always positive. The infinite nature of the tangent function around its asymptotes offers insights into its interesting properties such as abrupt changes between positive and negative values.

First derivative

Derivatives are central in understanding the behavior of functions. The first derivative, \( f'(x) \), provides a measure of a function's rate of change or slope. For the tangent function, the first derivative \( f'(x) = \sec^2 x \) reveals essential details. The function \( \sec x \) or secant is related to cosine, where \( \sec x = \frac{1}{\cos x} \). Thus, \( \sec^2 x \) is always positive wherever it is defined, particularly in the interval \( (-\pi/2, \pi/2) \), because the cosine function does not reach zero in this interval. This positivity means that \( \tan x \) is always increasing in this domain, evidenced by a persistently upward slope. As such, there are no critical points, ensuring that \( f(x) \) steadily increases without any decreases.

Concavity analysis

Concavity refers to the curvature of a graph, telling us whether a function bends upwards or downwards. To analyze the concavity of \( f(x) = \tan x \), we use the second derivative \( f''(x) = 2 \sec^2 x \tan x \). The sign of this second derivative helps determine concave up (positive \( f''(x) \)) or concave down (negative \( f''(x) \)) segments. By examining \( f''(x) \), we find:

• For \( x \in (-\pi/2, 0) \), \( \tan x \) is negative, making \( f''(x) \) negative, therefore, \( f(x) \) is concave down.
• For \( x \in (0, \pi/2) \), \( \tan x \) is positive, rendering \( f''(x) \) positive, thus \( f(x) \) is concave up.

This change in concavity around \( x = 0 \) suggests a potential inflection point, where the direction of curvature shifts.

Inflection points

Inflection points for a function occur where the concavity changes, which is detected through the sign change in the second derivative \( f''(x) \). In the case of \( f(x) = \tan x \), the second derivative \( f''(x) = 2 \sec^2 x \tan x \) shifts from negative to positive as \( x \) crosses 0. This transition marks \( x = 0 \) as an inflection point, where the function changes from concaving down to concaving up. Notably, this does not affect the overall increasing nature of \( \tan x \) underlined by its first derivative but provides a pivotal point on the graph marking a distinct change in its shape. Graphically, this point manifests as \( \tan x \)'s smooth transition through zero, accommodating both negative and positive curvature sections.