Problem 4 - 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. $$ y=2 x^{2}-x+3, x \in \mathbf{R} $$

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, \frac{1}{4})\), increasing on \((\frac{1}{4}, \infty)\), and concave up everywhere.

Step 1: Find the First Derivative

Given the function \( y = 2x^2 - x + 3 \), we need to find the first derivative: \( y' = \frac{d}{dx}(2x^2 - x + 3) = 4x - 1 \).

Step 2: Critical Points and the First Derivative Test

Set the first derivative equal to zero to find critical points: \( 4x - 1 = 0 \). Solving, we have \( x = \frac{1}{4} \). This is the only critical point. We test intervals around this point to determine where the function increases or decreases. Test \( x = 0 \) (left of \( \frac{1}{4} \)): \( 4(0) - 1 = -1 \), which is negative, so the function is decreasing on \( (-\infty, \frac{1}{4}) \). Test \( x = 1 \) (right of \( \frac{1}{4} \)): \( 4(1) - 1 = 3 \), which is positive, so the function is increasing on \( (\frac{1}{4}, \infty) \).

Step 3: Find the Second Derivative

Find the second derivative to determine concavity: \( y'' = \frac{d^2}{dx^2}(2x^2 - x + 3) = 4 \).

Step 4: Analyzing Concavity with the Second Derivative Test

Since the second derivative \( y'' = 4 \) is positive for all \( x \), the function is concave up everywhere in its domain \( (-\infty, \infty) \).

Step 5: Summary

The function \( y = 2x^2 - x + 3 \) is decreasing on \( (-\infty, \frac{1}{4}) \) and increasing on \( (\frac{1}{4}, \infty) \). It is concave up across its entire domain.

Key Concepts

First Derivative Test

The first derivative test is a powerful tool to determine where a function is increasing or decreasing. By calculating the first derivative of a function, we can find its rate of change. For the function given, \( y = 2x^2 - x + 3 \), the first derivative is \( y' = 4x - 1 \).
This first derivative represents the slope of the function at any point \( x \). To find critical points where the function could change direction, we set the derivative equal to zero: \( 4x - 1 = 0 \). This gives us the critical point \( x = \frac{1}{4} \).
By determining the sign of the first derivative before and after this critical point, we can tell if the function is increasing or decreasing:

• If \( y' > 0 \), the function is increasing.
• If \( y' < 0 \), the function is decreasing.

Checking right and left of \( x = \frac{1}{4} \), we find:

• For \( x < \frac{1}{4} \), \( y' = 4(0) - 1 = -1 \), the function is decreasing.
• For \( x > \frac{1}{4} \), \( y' = 4(1) - 1 = 3 \), the function is increasing.

This tells us that the function decreases on \( (-\infty, \frac{1}{4}) \) and increases on \( (\frac{1}{4}, \infty) \).

Second Derivative Test

The second derivative test helps in understanding the concavity of a function. Concavity illustrates how the slope of the function changes. If a function is concave up, it looks like a "U," and if it is concave down, it appears like an "n."
For the second derivative test, we differentiate the first derivative. Given \( y' = 4x - 1 \), its second derivative is \( y'' = 4 \). This derivative is a constant.

• If \( y'' > 0 \), the function is concave up.
• If \( y'' < 0 \), the function is concave down.

Since \( y'' = 4 \) is positive for all values of \( x \), it means the function is concave up everywhere in its domain: \( (-\infty, \infty) \).
The whole graph of the function \( y = 2x^2 - x + 3 \) displays this concave upward pattern, making it a smooth and upward-opening curve.

Critical Points

Critical points are specific points on a graph where the function's derivative is zero or undefined. These points are crucial because they can indicate a local maximum, minimum, or a saddle point—where a function may change direction from increasing to decreasing. In our function \( y = 2x^2 - x + 3 \), the first derivative \( y' = 4x - 1 \) becomes zero at \( x = \frac{1}{4} \).
This tells us a lot! Here, the slope is zero, meaning the rate of change is neither increasing nor decreasing precisely at this point.
Once we find the critical points, we can use the first derivative test to determine if these points are local maxima, minima, or neither. However, in functions like quadratics which are continuous and follow a regular pattern, this critical point at \( x = \frac{1}{4} \) generally serves as the point of inflection in directional behavior from decreasing to increasing.