Problem 32 - Calculus for Biology and Medicine

Question

Determine whether the functions have absolute maxima and minima, and, if so, find their coordinates. Find inflection points. Find the intervals on which the function is increasing, on which it is decreasing, on which it is concave up, and on which it is concave down. Sketch the graph of each function. $$ y=\tan x-x, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) $$

Step-by-Step Solution

Answer

The function is increasing on \((-\frac{\pi}{2}, \frac{\pi}{2})\) with no absolute extrema. It is concave up on \((0, \frac{\pi}{2})\) and concave down on \((-\frac{\pi}{2}, 0)\), with a local minimum at \( x=0 \).

Step 1: Find Critical Points

To find the critical points, we need to find where the first derivative of the function is zero or undefined. Given \( y = \tan x - x \), first find \( y' = \sec^2 x - 1 \). We set \( \sec^2 x - 1 = 0 \) and solve for \( x \). This simplifies to \( \tan^2 x = 0 \), so \( \tan x = 0 \), which happens at \( x = 0 \) within the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step 2: Determine Increasing/Decreasing Intervals

Analyze the sign of the first derivative \( y' = \sec^2 x - 1 \). Since \( \sec^2 x \geq 1 \) for all \( x \), \( y' = \sec^2 x - 1 \geq 0 \). Therefore, the function is increasing on the entire interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).

Step 3: Check for Absolute Maxima and Minima

Since the function is increasing throughout the interval without any endpoints, there are no absolute maxima or minima. The critical point at \( x = 0 \) is a local minimum, given the increasing behavior on either side within the interval.

Step 4: Find Inflection Points

To find inflection points, we need the second derivative \( y'' = 2\sec^2 x \tan x \). An inflection point occurs where \( y'' = 0 \). In this function, as \( \tan x = 0 \) at \( x = 0 \), the second derivative does not change sign, therefore, \( x = 0 \) is not an inflection point. There are no inflection points in this interval.

Step 5: Determine Concavity Intervals

The second derivative \( y'' = 2\sec^2 x \tan x \) indicates concavity when \( x > 0 \), \( y'' > 0 \) (concave up) and when \( x < 0 \), \( y'' < 0 \) (concave down). Hence, it's concave up on \((0, \frac{\pi}{2})\) and concave down on \((-\frac{\pi}{2}, 0)\).

Step 6: Sketch the Graph

To sketch the graph: note the behavior. The function is increasing through the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), with a local minimum at \( x = 0 \). The nature of the curve is concave down approaching zero from the left and concave up moving away from zero to the right. Consider the vertical asymptotes as \( x \rightarrow \pm \frac{\pi}{2} \).

Key Concepts

Critical Points

In calculus, critical points of a function are values of \( x \) where the first derivative \( y' \) is either zero or undefined. These points are essential because they help identify maxima, minima, or points where the function's behavior changes. For the function \( y = \tan x - x \) on the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\), the first derivative is \( y' = \sec^2 x - 1 \).

To find critical points, set \( y' = 0 \). This simplifies to \( \sec^2 x - 1 = 0 \), leading to \( \tan^2 x = 0 \), and hence \( \tan x = 0 \). The only value of \( x \) that satisfies this condition within the interval is \( x = 0 \). This point is crucial for further analysis, like determining the nature of maxima and minima or the function's trajectory in a given interval.

First Derivative

The first derivative of a function provides information about its monotonicity, indicating where the function is increasing or decreasing. It gives the slope of the tangent line to the curve at any point. For the function \( y = \tan x - x \), the derivative is \( y' = \sec^2 x - 1 \).

Knowing that \( \sec^2 x \geq 1 \) for all \( x \), it follows that \( y' = \sec^2 x - 1 \geq 0 \). This insight tells us that the function is increasing over the entire interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Thus, the absence of any segments where the derivative is negative indicates that there are no sections where the function decreases.

Second Derivative

The second derivative of a function, denoted as \( y'' \), reveals information about the concavity of the function and helps in identifying potential inflection points where the concavity changes. Calculating \( y'' \) for \( y = \tan x - x \), we get \( y'' = 2\sec^2 x \tan x \).

This derivative being dependent on \( \tan x \) means its sign is tied to whether \( x \) is positive or negative. Since \( \tan x = 0 \) at \( x = 0 \), we look near this point to assess concavity changes, but as \( \tan x \) remains zero, \( x = 0 \) is not an inflection point. Therefore, no inflection points exist where the concavity transitions.

Inflection Points

Inflection points occur where the curve changes concavity. This means the second derivative \( y'' \) must change sign. For the function \( y = \tan x - x \), with \( y'' = 2\sec^2 x \tan x \), looking for sign changes in \( y'' \) helps find these points.

In this specific function, \( \tan x = 0 \) at \( x = 0 \), so \( y'' = 0 \). However, because there is no sign change in the second derivative at \( x = 0 \), it confirms that \( x = 0 \) is not an inflection point. Therefore, within the entire interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), there are no inflection points, maintaining consistent concavity segments.

Concavity

Concavity describes how a curve bends. It informs whether a function looks like an upward or downward opening cup. The second derivative tells us about concavity. If \( y'' > 0 \), the function is concave up, and if \( y'' < 0 \), it is concave down.

For the function \( y = \tan x - x \), the second derivative \( y'' = 2\sec^2 x \tan x \) offers insight. Within the interval \((0, \frac{\pi}{2})\), \( \tan x \) is positive, making \( y'' > 0 \); therefore, the function is concave up. Meanwhile, for \( x < 0 \), \( \tan x \) is negative, leading to \( y'' < 0 \) and indicating a region of concave down. Understanding these intervals is key when sketching the graph and analyzing changes in the bending directions of the curve.