Problem 75 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the relative extrema, if any, of each function. Use the second derivative test, if applicable. $$ g(s)=\frac{s}{1+s^{2}} $$

Step-by-Step Solution

Answer

The function \(g(s)=\frac{s}{1+s^{2}}\) has a relative maximum at \(s=1\) and a relative minimum at \(s=-1\).

Step 1: Find the first derivative

To find the critical points of the function, we first need to find its first derivative. Using the quotient rule, we have: $$ g'(s) = \frac{(1 + s^2)(1) - s(2s)}{(1 + s^2)^2} = \frac{1- s^2}{(1+s^2)^2} $$

Step 2: Find the critical points

To find the critical points of the function, set the first derivative equal to zero and solve for \(s\): $$ g'(s) = \frac{1- s^2}{(1+s^2)^2} = 0 $$ Either the numerator is zero, or the denominator is infinite. The denominator will be infinite when \(1+s^2=0\) which means \(s=\pm i\), these are complex numbers and can't be the critical points of the function. Now we have to solve for the numerator equal to zero: $$ 1- s^2=0 $$ $$ s^2=1 $$ Therefore, \(s=\pm 1\) are the critical points.

Step 3: Find the second derivative

Now, find the second derivative by differentiating the first derivative: $$ g''(s) = \frac{d}{ds}[\frac{1- s^2}{(1+s^2)^2}] $$ Using the quotient rule, we obtain: $$ g''(s) = \frac{-2s(1+s^2)^2 - (1+s^2)(4s^2(1+s^2) - 4s^3)(1+s^2)}{(1+s^2)^4} $$ Simplifying the second derivative, we get: $$ g''(s) = \frac{-2s}{(1+s^2)^3} $$

Step 4: Find the nature of critical points using the second-derivative test

By using the second derivative test, we can determine if the critical points are relative extrema. For a critical point \(s\): - If \(g''(s) > 0\), the function has a relative minimum at \(s\) - If \(g''(s) < 0\), the function has a relative maximum at \(s\) - If \(g''(s) = 0\), the test is inconclusive. For \(s=1\): $$ g''(1) = \frac{-2}{(1+1^2)^3} = -\frac{1}{8} < 0 $$ Therefore, the function has a relative maximum at \(s=1\). For \(s=-1\): $$ g''(-1) = \frac{2}{(1+(-1)^2)^3} = \frac{1}{8} > 0 $$ Therefore, the function has a relative minimum at \(s=-1\).

Step 5: Conclusion

The function \(g(s)=\frac{s}{1+s^{2}}\) has a relative maximum at \(s=1\) and a relative minimum at \(s=-1\).

Key Concepts

Second Derivative Test

The Second Derivative Test is a useful tool for determining the nature of critical points of a function. Once you've found the critical points, you can use this test to figure out if each point is a relative minimum, a relative maximum, or if it's inconclusive.

Here's how it works:

• Calculate the second derivative of the function.
• Plug each critical point into the second derivative.

Based on the sign of the second derivative, you can draw conclusions:

• If the second derivative is greater than zero at a critical point, the function has a relative minimum there.
• If the second derivative is less than zero, the function has a relative maximum at that point.
• If the second derivative equals zero, the test is inconclusive, and other methods may be needed to determine the nature of the point.

For the function provided, at the critical points of 1 and -1, the second derivative is negative for 1, indicating a relative maximum, and positive for -1, indicating a relative minimum. This method is straightforward and effective for understanding relative extrema.

Quotient Rule

Differentiation often involves functions where one is divided by another. In such cases, the Quotient Rule is your go-to method to find derivatives. The rule provides a systematic way to differentiate fractions.

Here's the formula:\[\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]Where:

• \( u \) and \( v \) are functions of \( x \).
• \( u' \) and \( v' \) are their respective derivatives.

In the given problem, \( g(s) = \frac{s}{1 + s^2} \), you can apply the Quotient Rule easily:

• Derive the numerator \( s \) which is \( 1 \).
• Derive the denominator \( 1 + s^2 \), which results in \( 2s \).

Following the Quotient Rule formula provides the first derivative, which is crucial in finding critical points.

Critical Points

Critical points are the values of \( s \) where the first derivative of the function is zero or undefined. These points are essential because they help identify where potential extrema (maximum or minimum values) occur.

To find critical points:

• First, take the derivative of the function.
• Set the derivative equal to zero and solve for \( s \).
• Also consider where the derivative does not exist, if applicable.

For the function \( g(s) = \frac{s}{1+s^2} \), setting the first derivative \( g'(s) \) equal to zero allows solving for the critical points. Here, the critical points found were \( s = 1 \) and \( s = -1 \). These points are particularly significant because they help determine where the relative extrema can possibly occur.

Differentiation

In calculus, differentiation is the process of finding the derivative of a function. Derivatives play a vital role in various fields such as physics, engineering, and economics, providing a way to analyze rates of change.

Your primary focus when differentiating is to apply various rules appropriately:

• The Power Rule
• The Product Rule
• The Quotient Rule
• The Chain Rule

For this exercise, the Quotient Rule was utilized to differentiate \( g(s) = \frac{s}{1+s^2} \). This differentiation step is foundational to finding critical points and applying tests like the Second Derivative Test. Differentiation not only helps in finding extrema but also in understanding the behavior of graphs, optimizing functions, and solving equations analytically.