Problem 13 - 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. $$ y=\frac{1}{(1+x)^{2}}, x \neq-1 $$

Step-by-Step Solution

Answer

The function is increasing for \( x < -1 \) and decreasing for \( x > -1 \). It is concave up for all \( x \neq -1 \).

Step 1: Find the First Derivative

The function is given as \( y = \frac{1}{(1+x)^2} \). To find the first derivative \( y' \), apply the chain rule. Rewrite the function as \( y = (1+x)^{-2} \). The derivative of \( (1+x)^{-2} \) is \( y' = -2(1+x)^{-3} \).

Step 2: Set the First Derivative Equal to Zero

To find critical points, set \( y' = -2(1+x)^{-3} = 0 \). Since the derivative is never zero for any \( x \), there are no critical points from setting the derivative equal to zero.

Step 3: Determine Intervals of Increase and Decrease

Examine the sign of \( y' = -2(1+x)^{-3} \) for different intervals of \( x \). For \( x > -1 \), \( (1+x)^3 > 0 \) making \( y' < 0 \). Thus, the function is decreasing for \( x > -1 \). For \( x < -1 \), \( (1+x)^3 < 0 \) making \( y' > 0 \), meaning the function is increasing for \( x < -1 \).

Step 4: Find the Second Derivative

To determine concavity, find the second derivative \( y'' \). Differentiate \( y' = -2(1+x)^{-3} \) to get \( y'' = 6(1+x)^{-4} \).

Step 5: Determine Intervals of Concavity

The sign of \( y'' = 6(1+x)^{-4} \) determines concavity. For any \( x eq -1 \), \((1+x)^4 > 0\), so \( y'' > 0 \). Therefore, the function is concave up for all \( x eq -1 \). There are no intervals where the function is concave down.

Key Concepts

Second Derivative Test

The second derivative test helps us understand the concavity of the function and identify potential points for local maxima or minima.
Once we have the first derivative of a function, finding the second derivative allows us to examine how the function curves.

• If the second derivative \( y'' \) is positive for a particular region, the function is concave up in that region.
• If \( y'' \) is negative, the function is concave down.
• If \( y'' = 0 \), the test is inconclusive, and other methods might be needed.

In this exercise, we found that the second derivative is always positive because \( y'' = 6(1+x)^{-4} > 0 \) for all values of \( x eq -1 \). This outcome indicates that the function is concave up throughout its domain, with no intervals where it is concave down.

Increasing and Decreasing Functions

A function can move up (increase) or down (decrease) along the y-axis, depending on the sign of its first derivative.

• The function is increasing in regions where the first derivative \( y' \) is positive.
• Conversely, where \( y' \) is negative, the function decreases.

In this particular exercise, we derived \( y' = -2(1+x)^{-3} \), which changes its sign around the point \( x = -1 \).
Here's what we found:

• For \( x > -1 \), the value of \( y' \) is negative, so the function is decreasing.
• For \( x < -1 \), \( y' \) becomes positive, meaning the function is increasing.

By examining these patterns, you can determine how the function behaves over different intervals.

Concave Up and Concave Down

The concept of concavity describes how a curve bends as the x-values change.
This bending gives us insight into the acceleration or deceleration of growth or decline.

• A function is concave up if it bends upwards, resembling a cup, and \( y'' > 0 \).
• It is concave down if it bends downwards, as in an upside-down cup, with \( y'' < 0 \).

In our solution, we found the second derivative \( y'' = 6(1+x)^{-4} \) to be consistently positive across the domain.
This finding means the function is always concave up, offering a nice, steady upward-bending curve.
No regions exhibit a "concave down" behaviour.

Critical Points

Critical points are essential in calculus because they help identify where the function changes direction, potentially indicating a local maximum or minimum.
These points occur where the first derivative \( y' \) is zero or undefined.

• When \( y' = 0 \), we usually find peak or valley points.
• If the first derivative does not exist, it can also suggest a critical point.

In this exercise, however, setting \( y' = -2(1+x)^{-3} = 0 \) did not yield any real solutions, implying no critical points where the function ceases to increase or decrease.
As such, the critical point analysis confirms that the function’s increasing and decreasing pattern remains uninterrupted over its domain.