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

Step-by-Step Solution

Answer

The function is decreasing on \((-\infty, \frac{1}{2})\), increasing on \((\frac{1}{2}, \infty)\), and concave up everywhere.

Step 1: Find the first derivative

To determine where the function is increasing or decreasing, we first need the first derivative of the function. The given function is \( y = x^2 - x - 4 \). The derivative with respect to \( x \) is found using basic differentiation rules: \( y' = 2x - 1 \).

Step 2: Solve for critical points

Critical points occur where the first derivative is zero or undefined. Set the first derivative equal to zero and solve for \( x \): \( 2x - 1 = 0 \). Solving this gives \( x = \frac{1}{2} \). Since the first derivative is a linear polynomial, it is defined for all \( x \). Thus, the only critical point is at \( x = \frac{1}{2} \).

Step 3: Apply the First Derivative Test

Use the first derivative test to determine intervals of increasing and decreasing functions. Test points around the critical point \( x = \frac{1}{2} \): Choose \( x = 0 \) and \( x = 1 \).- For \( x = 0 \), \( y'(0) = 2(0) - 1 = -1 \), the function is decreasing.- For \( x = 1 \), \( y'(1) = 2(1) - 1 = 1 \), the function is increasing.Thus, \( y(x) \) is decreasing on \( (-\infty, \frac{1}{2}) \) and increasing on \( (\frac{1}{2}, \infty) \).

Step 4: Find the second derivative

The second derivative will help determine concavity. Differentiate the first derivative \( y' = 2x - 1 \) to find the second derivative: \( y'' = 2 \).

Step 5: Determine concavity using the second derivative

For all \( x \), the second derivative \( y'' = 2 \) is positive, indicating that the function is concave up across its entire domain \(( -\infty, \infty )\) and there are no points where it is concave down.

Key Concepts

First Derivative Test

The first derivative test is a handy method in calculus to determine where a function is increasing or decreasing. To apply this test, we look at the sign of the first derivative, which is the slope of the tangent line at any given point on the function. The basic idea is to:

• Find the first derivative of the function.
• Locate the critical points by setting the first derivative equal to zero or where it is undefined.
• Choose test points on either side of each critical point to determine the sign of the derivative in those intervals.

By analyzing these signs, if the first derivative changes from negative to positive at a critical point, the function changes from decreasing to increasing, indicating a local minimum. Conversely, if it changes from positive to negative, we identify a local maximum. The function \(y = x^2 - x - 4\) was found to be decreasing on \((-\infty, \frac{1}{2})\) and increasing on \((\frac{1}{2}, \infty)\) from this test.

Second Derivative Test

The second derivative test is used to determine the concavity of a function and verify if a critical point is a local minimum or maximum. Here’s how you can execute this test effectively:

• Compute the second derivative of the function. The second derivative gives us the rate of change of the slope.
• Check the sign of the second derivative at the critical points.
  • If \(y'' > 0\), the function is concave up, resembling a cup, which suggests a local minimum at the critical point.
  • If \(y'' < 0\), the function is concave down, resembling a cap, indicating a local maximum.

In our function example, the second derivative is always positive (\(y'' = 2\)), which tells us that the function is concave up across its entire domain.

Critical Points

Critical points are the points on a graph where the first derivative is zero or undefined. These points are crucial as they have the potential to be locations of local maxima, minima, or points of inflection. To effectively find and use critical points:

• Derive the first derivative of the function.
• Set this derivative to zero and solve for \(x\), identifying locations where the slope of the tangent is flat (horizontal).
• Consider points where the derivative might be undefined, as these can also be critical points.

For the function \(y = x^2 - x - 4\), the only critical point is found at \(x = \frac{1}{2}\). Using this point with the first and second derivative tests, we can predict the behavior of the function around it.

Concavity

Concavity describes how a curve bends; whether it bends upwards like a cup or downwards like a cap. In calculus, concavity is determined through the second derivative:

• Second derivative, \(y''\), further confirms whether a segment of the function is concave up or down.
• Concave up (\(y'' > 0\) ) indicates that the function curves away from the x-axis upward.
• Concave down (\(y'' < 0\) ) indicates that it curves towards the x-axis downward.

For \(y = x^2 - x - 4\), the second derivative is constantly positive (\(y'' = 2\)), so this function is consistently concave up for all real numbers \(x\). This creates a parabolic shape that opens upwards.