Problem 26 - Thomas Calculus in SI Units

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}

Step-by-Step Solution

Answer

Local extrema at \( x = \pm\frac{\pi}{6} \), and inflection point at \( x = 0 \).

Step 1: Find the derivative

To locate critical points, we first need to find the derivative of the given function, which is \( y = \frac{4}{3}x - \tan x \). The derivative \( y' \) is calculated as follows: \[ y' = \frac{d}{dx} \left( \frac{4}{3}x - \tan x \right) = \frac{4}{3} - \sec^2 x \]

Step 2: Identify critical points

Set the derivative equal to zero to find critical points where the function may attain local extrema:\[ \frac{4}{3} - \sec^2 x = 0 \]Solve for \( x \):\[ \sec^2 x = \frac{4}{3} \]Use \( \sec x = \frac{1}{\cos x} \), we find:\[ \cos^2 x = \frac{3}{4} \right) \rightarrow \cos x = \pm\frac{\sqrt{3}}{2} \]Within \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), the solution is \( x = \pm\frac{\pi}{6} \).

Step 3: Determine local extrema

Substitute the critical points back into the original function to find the function values:For \( x = \frac{\pi}{6} \):\[ y = \frac{4}{3} \cdot \frac{\pi}{6} - \tan \left( \frac{\pi}{6} \right) \approx \frac{2\pi}{9} - \frac{1}{\sqrt{3}} \equiv g_1\]For \( x = -\frac{\pi}{6} \):\[ y = \frac{4}{3} \cdot \left(-\frac{\pi}{6}\right) - \tan \left(-\frac{\pi}{6}\right) \approx -\frac{2\pi}{9} + \frac{1}{\sqrt{3}} \equiv g_2\]Compare \( g_1 \) and \( g_2 \) to determine which are minima/maxima.

Step 4: Find the second derivative

To determine concavity and inflection points, find the second derivative:\[ y'' = \frac{d}{dx} \left(-\sec^2 x\right) = -2\sec^2 x \tan x \]

Step 5: Determine points of inflection

Points of inflection occur where the second derivative changes sign. We solve:\[ -2\sec^2 x \tan x = 0 \]This implies \( \tan x = 0 \), so \( x = 0 \) within the interval.

Step 6: Graph the function

Plot the function \( y = \frac{4}{3}x - \tan x \) and highlight critical points and inflection points on the graph. Ensure to note that the local extrema are found at \( x = \pm\frac{\pi}{6} \) and inflection at \( x = 0 \).

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function changes at any given point. The derivative of a function, often noted as \( y' \) or \( f'(x) \), provides the slope of the tangent line to the curve at a certain point. For our function, \( y = \frac{4}{3}x - \tan x \), we calculated the derivative to be \( y' = \frac{4}{3} - \sec^2 x \).
This expression tells us how the function changes as \( x \) changes.
The \( \sec^2 x \) term, being positive, modifies the linear change introduced by \( \frac{4}{3}x \), adding a degree of curvature depending on the angle.

Critical Points

Critical points occur where a function's derivative equals zero or is undefined. These points are important because they can indicate where the function reaches local maximums or minimums—collectively called local extrema. To locate these for our function, \( y' = \frac{4}{3} - \sec^2 x \) was set to zero, leading to the equation \( \sec^2 x = \frac{4}{3} \).

By solving this, we find critical values at \( x = \pm\frac{\pi}{6} \) within the given interval.
Substituting these values into the original function helps determine the actual local maximum and minimum values.

• For \( x = \frac{\pi}{6} \), a calculation gives a certain value of \( y \).
• For \( x = -\frac{\pi}{6} \), another distinct \( y \) value is obtained.

Comparing these tells us whether each point is a maximum or a minimum.

Second Derivative

The second derivative, noted as \( y'' \), helps us understand the curvature or "concavity" of the function at a point. If \( y'' > 0 \), the function is convex, indicating a local minimum, and if \( y'' < 0 \), it's concave, suggesting a local maximum. For our function, the second derivative is calculated as \( y'' = -2\sec^2 x \tan x \).
This expression's complexity arises from the trigonometric functions involved, meaning the curvature depends heavily on \( x \)'s value.

• Concavity information can confirm the nature of critical points.
• An undefined or zero \( y'' \) could indicate a point of inflection.

Solving \( -2\sec^2 x \tan x = 0 \) specifically looks for where this derivative changes sign.

Inflection Points

An inflection point is where the function's curve changes from concave (curving upward) to convex (curving downward) or vice versa. This occurs where the second derivative is zero or undefined and changes sign at these points. For our function, the key inflection point is found at \( x = 0 \), determined from solving \( -2\sec^2 x \tan x = 0 \), where \( \tan x \) is zero.

Examining the interval \( -\frac{\pi}{2} < x < \frac{\pi}{2} \), zero is the point at which \( y'' \) transitions signifying a shift in concavity.

• This makes \( x = 0 \) a vital inflection point of the function.
• Understanding this helps visualize the changing slope around the origin.

By graphing, these points highlight the dynamic change in the curve's direction.