Problem 8 - Calculus for Biology and Medicine

Question

Use the first derivative test and the second derivative test to determine where each function is increasing, decreasing, concave up, and concave down. You do not need to use a graphing calculator for these exercises. $$ y=(3 x-1)^{1 / 3}, x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function is increasing on \( (-\infty, \frac{1}{3}) \cup (\frac{1}{3}, \infty) \) and concave down everywhere except possibly at \( x = \frac{1}{3} \).

Step 1: Find the First Derivative

Differentiate the function \( y = (3x - 1)^{1/3} \) with respect to \( x \). We use the chain rule: \( y' = \frac{1}{3}(3x - 1)^{-2/3} \times 3 = (3x - 1)^{-2/3} \).

Step 2: Find Critical Points

Critical points occur where the first derivative is zero or undefined. The derivative \( y' = (3x - 1)^{-2/3} \) never equals zero, but is undefined when \( 3x - 1 = 0 \). Solving gives \( x = \frac{1}{3} \). Thus, \( x = \frac{1}{3} \) is a critical point.

Step 3: Apply the First Derivative Test

Choose test points in the intervals \((-, \frac{1}{3})\) and \((\frac{1}{3}, )\). For \( x = 0 \) in the first interval, \( y' > 0 \), indicating increasing; for \( x = 1 \) in the second interval, \( y' > 0 \), indicating increasing. Hence, \( y \) is increasing everywhere except at \( x = \frac{1}{3} \).

Step 4: Find the Second Derivative

Differentiate \( y' = (3x - 1)^{-2/3} \): \( y'' = \frac{d}{dx}[(3x-1)^{-2/3}] = -\frac{2}{3} (3x - 1)^{-5/3} \times 3 = -2(3x - 1)^{-5/3} \).

Step 5: Find Points of Inflection

The second derivative \( y'' = -2(3x - 1)^{-5/3} \) is undefined at \( x = \frac{1}{3} \), but no zeros exist. However, the sign of \( y'' \) helps determine concavity.

Step 6: Apply the Second Derivative Test

Select test points to determine concavity. At \( x = 0 \), \( y'' < 0 \), indicating concave down; at \( x = 1 \), \( y'' < 0 \), indicating concave down. Function \( y \) is concave down except possibly at \( x = \frac{1}{3} \).

Key Concepts

Second Derivative Test

The Second Derivative Test is a valuable tool used to assess the concavity of a function and locate points of inflection. By analyzing the second derivative of a function, we can gain insight into the shape of the graph and predict its behavior.
To use the Second Derivative Test, we first compute the second derivative. For the function given, the second derivative is obtained by differentiating the first derivative:- The first derivative, given as \((3x - 1)^{-2/3}\)- The second derivative, \(y'' = -2(3x - 1)^{-5/3}\)
The Second Derivative Test follows these steps:

• Evaluate the second derivative at points of interest, typically the critical points where the first derivative is zero or undefined. In this exercise, the second derivative is undefined at x = \( \frac{1}{3} \).
• If the second derivative is positive at a point, the graph is concave up (like a cup); if negative, concave down (like a frown).

In our example, testing for values around \( x = \frac{1}{3} \), we find that for both \( x = 0 \) and \( x = 1 \), the second derivative is negative, indicating the function is concave down for these intervals.

Increasing and Decreasing Functions

Understanding where a function is increasing or decreasing is crucial in sketching its graph and grasping its behavior.
To determine these intervals, we apply the First Derivative Test. It involves the following steps:

• Find the first derivative of the function to determine the rate of change. Here, \( y' = (3x - 1)^{-2/3} \).
• Identify critical points where the derivative is zero or undefined. For this function, it is undefined at x = \( \frac{1}{3} \).
• Analyze the intervals formed by critical points using the first derivative to determine the sign and hence the behavior of the function.

In the given exercise, by selecting test points from intervals \((- \infty, \frac{1}{3})\) and \((\frac{1}{3}, \infty)\):

• At \( x = 0 \), first derivative is positive, indicating an increasing function.
• At \( x = 1 \), first derivative is positive, indicating an increasing function.

Thus, the function is increasing everywhere except possibly at the critical point. Being aware of the increasing nature between these intervals helps in visualizing the general direction of the graph.

Concavity and Points of Inflection

Concavity tells us how a function bends, which is important for understanding a graph's shape. When we speak of concavity, it means how the function curves around its points.
To define where a function is concave up or down, the second derivative is used:

• When the second derivative is positive, the function is concave up, resembling a U-shape.
• When the second derivative is negative, the function is concave down, resembling an inverted U-shape.

In our function, \(y'' = -2(3x - 1)^{-5/3}\), we tested the intervals and observed:

• At points \( x = 0 \) and \( x = 1 \), the second derivative is negative, meaning the function is concave down.

Points of Inflection are locations where the graph changes concavity. Although the exercise identifies \(x = \frac{1}{3} \) as a special point, it’s mainly a point of undefined behavior rather than a traditional inflection due to second derivative being undefined there. This understanding shapes the way we interpret and sketch function graphs.