Problem 25 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. Determine whether or not \(f\) attains an absolute maximum and absolute minimum value. If it does, determine the absolute maximum and/or minimum value. \(f(x)=\frac{x^{3}}{x^{2}+1}\)

Step-by-Step Solution

Answer

The function \(f(x) = \frac{x^3}{x^2 + 1}\) has one critical point \(x = 0\). However, it can't be determined whether this point is a local minimum, maximum or a saddle point from the second derivative test. Other methods such as the first derivative test has to be used. The function has no absolute maximum or minimum, as it is not bounded.

Step 1: Find the Critical Points

The derivative of the function \(f(x) = \frac{x^3}{x^2 + 1}\) is found by the quotient rule as: \(f'(x) = \frac{(x^2 + 1)3x^2 - x^3(2x)}{(x^2 + 1)^2} = \frac{x^2}{(x^2 + 1)^2}\). Setting this equal to zero gives \(x = 0\).

Step 2: Classify the Critical Points

The second derivative of \(f\) is \(f''(x) = \frac{-4x^3}{(x^2+1)^3}\). Substitute \(x = 0\) into \(f''(x)\), we get \(f''(0) = 0\). Therefore, it can't be determined whether \(x = 0\) is a local minimum or maximum or saddle point from the second derivative test. Thus, other methods such as the first derivative test has to be used.

Step 3: Determine Absolute Maximum and Minimum Values

This function is defined for all real numbers, therefore, there is no boundary where it would possibly attain the maximum or minimum value. As it is not bounded, the function has no absolute minimum or maximum, The function tends to -∞ as x tends to -∞ and to +∞ as x tends to +∞.

Key Concepts

Quotient Rule

When dealing with functions that are defined as quotients, like our function \( f(x) = \frac{x^3}{x^2 + 1} \), we use the quotient rule to find the derivative. The quotient rule provides a systematic way to find the derivative of a function that is a ratio of two other functions. To apply the quotient rule, if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then its derivative \( f'(x) \) is given by the formula:

• \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \)

In our case, \( u(x) = x^3 \) and \( v(x) = x^2 + 1 \). Calculating their derivatives gives \( u'(x) = 3x^2 \) and \( v'(x) = 2x \). We plug these into the formula to get the derivative of \( f(x) \). Understanding the quotient rule is critical for solving problems involving rational functions, as it lets us determine the rates of change and critical points efficiently.

Second Derivative Test

The second derivative test is a tool used to classify critical points found using the first derivative. Once you have a critical point, you can plug it into the second derivative to determine the nature of the point. Essentially, the second derivative test tells us about the concavity of the function at a given point.
If the second derivative \( f''(x) \) at a critical point is positive, then the function is concave up at that point, indicating a local minimum.
If \( f''(x) \) is negative, the function is concave down, indicating a local maximum.
If \( f''(x) = 0 \), the test is inconclusive, which happened in our case when evaluating the critical point at \( x=0 \). This means we need to try another method, like the first derivative test, to determine if it's a minimum, maximum, or other.

First Derivative Test

When the second derivative test is inconclusive, such as when \( f''(x) = 0 \), the first derivative test becomes useful for classifying critical points. This test involves looking at the sign of \( f'(x) \) around the critical point.

• If \( f'(x) \) changes from negative to positive at a critical point \( x = c \), then \( c \) is a local minimum.
• If \( f'(x) \) changes from positive to negative, \( c \) is a local maximum.
• If \( f'(x) \) does not change signs, there is no local extremum at \( c \).

The first derivative test gives us much-needed insight when the second derivative doesn't provide enough information, fitting perfectly to present a complete analysis of those tricky critical points.

Absolute Maximum and Minimum

Finding the absolute maximum and minimum values of a function involves identifying the highest and lowest values on the function's entire domain. This is slightly different from finding local extrema, which looks at relative peaks and troughs in specific regions.
For unbounded continuous functions, like \( f(x) = \frac{x^3}{x^2+1} \), an absolute maximum or minimum may not exist. This is because the function's behavior as \( x \) approaches infinity or negative infinity could make it head towards infinity itself.
In the given problem, since the function is not restricted to a closed interval, the nature of polynomial functions and rational expressions means that \( f(x) \) can grow indefinitely. Therefore, \( f(x) \) does not maintain absolute boundaries, showing once again that understanding how a function behaves at extremes is crucial for establishing the existence of absolute extrema.