Problem 26 - University Calculus: Early Transcendentals

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\frac{4}{3} x-\tan x, \quad \frac{-\pi}{2}< x<\frac{\pi}{2}$$

Step-by-Step Solution

Answer

Local extremes at \( x = \pm\frac{\pi}{6} \); inflection point at \( x = 0 \).

Step 1: First Derivative

To find critical points, we must find the first derivative of the function. The given function is \( y = \frac{4}{3}x - \tan x \). Using the derivative rules: \( \frac{d}{dx}[\frac{4}{3}x] = \frac{4}{3} \) and \( \frac{d}{dx}[\tan x] = \sec^2 x \), we get:\[ y' = \frac{4}{3} - \sec^2 x. \]

Step 2: Find Critical Points

Set the first derivative to zero to find critical points:\[ \frac{4}{3} - \sec^2 x = 0 \] This implies \( \sec^2 x = \frac{4}{3} \). Use the identity \( \sec x = \frac{1}{\cos x} \) to find \( \cos^2 x = \frac{3}{4} \). Thus, \( \cos x = \pm \frac{\sqrt{3}}{2} \). The corresponding values of \( x \) within \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \) are \( x = \pm\frac{\pi}{6} \).

Step 3: Second Derivative for Concavity

Find the second derivative of \( y = \frac{4}{3}x - \tan x \) to determine concavity and inflection points:\( y' = \frac{4}{3} - \sec^2 x \) implies \( y'' = \frac{d}{dx}[-\sec^2 x] \). Use the derivative \( \frac{d}{dx}[\sec^2 x] = 2\sec^2 x \tan x \). Thus,\[ y'' = -2\sec^2 x \tan x. \]

Step 4: Identify Inflection Points

Set the second derivative equal to zero:\[-2\sec^2 x \tan x = 0.\]This implies \( \tan x = 0 \). Solve \( x = 0 \) within \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), indicating a possible inflection point at \( x = 0 \). Check intervals around 0 to confirm change in concavity.

Step 5: Analyze Extreme Points

Evaluate the second derivative at critical points to determine if they are local minima or maxima:For \( x = \pm \frac{\pi}{6} \): \( \sec^2(\pm\frac{\pi}{6}) = \frac{4}{3} \) shows that \( y'' eq 0 \). For \( y''(x) \), note the behavior of \( -2 \sec^2 x \tan x \) changes sign indicating local extreme points.

Step 6: Graph and Confirm Results

Graph the function over the interval \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) using critical points and inflection points. Verify local maxima/minima at \( x = \pm\frac{\pi}{6} \) and an inflection point at \( x = 0 \). The function displays concave up and down regions corresponding to intervals separated by \( x = 0 \).

Key Concepts

First Derivative

The first derivative is crucial to understanding the behavior of a function because it tells us about the slope or rate of change. By taking the derivative of the function \[ y = \frac{4}{3}x - \tan x \], we can identify where the slope is zero, which indicates potential local minimums or maximums, also known as critical points.

• The derivative of \( \frac{4}{3}x \) is \( \frac{4}{3} \).
• The derivative of \( \tan x \) is \( \sec^2 x \).

Putting it all together, the first derivative of the function is \[ y' = \frac{4}{3} - \sec^2 x \].
To find the critical points, we set this first derivative equal to zero. Solving \( \frac{4}{3} - \sec^2 x = 0 \), we find \( \sec^2 x = \frac{4}{3} \). By rearranging using the identity \( \sec x = \frac{1}{\cos x} \), it is determined that \( \cos^2 x = \frac{3}{4} \), leading to the angles \( x = \pm \frac{\pi}{6} \). These are the points where the slope of the tangent to the curve is zero. In other words, these are the potential spots for local maxima or minima.

Second Derivative

The second derivative provides insights into the concavity of the function, helping to determine whether a function is curving upwards or downwards at any given point. For the function \[ y = \frac{4}{3}x - \tan x \], the first derivative was found to be \[ y' = \frac{4}{3} - \sec^2 x \].
To find the second derivative, take the derivative of \(-\sec^2 x \).

• The derivative of \( \sec^2 x \) is \( 2\sec^2 x \tan x \).

Thus, \[ y'' = -2\sec^2 x \tan x \].
This expression will help us determine concavity and identify inflection points.

Concavity

Concavity describes the direction the function is curving. Understanding whether a portion of the graph is concave up or down helps us predict the function's behavior.

• Concave up: The graph is opening upwards like a cup, and the second derivative will be positive in this region.
• Concave down: The graph is opening downwards like a frown, and the second derivative will be negative in this region.

To assess concavity, examine the sign of the second derivative \[ y'' = -2\sec^2 x \tan x \]. In areas where \( \tan x = 0 \), the second derivative can change sign, indicating a switch in concavity.
In this function, the concavity changes nature around \( x = 0 \). To the left and right of \( x = 0 \), test values in these intervals to confirm the sign change. This will show where the graph switches from concave up to concave down, or vice versa.

Inflection Points

Inflection points occur where the concavity changes, from concave up to concave down or from concave down to concave up. These are important as they mark a change in the direction of curvature.
By finding where the second derivative \[ y'' = -2\sec^2 x \tan x \] equals 0, solve \( \tan x = 0 \). This gives us an inflection point at \( x = 0 \).To be sure it's an inflection point, confirm a change in sign of the second derivative as you move through \( x = 0 \). An easy way to test this is by selecting points slightly to the left and right of zero within the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\). The change in sign will verify that \( x = 0 \) is indeed an inflection point. This is where the function changes from curving one way to curving the opposite way.