Problem 70 - Calculus and its applications

Question

Let \(f(x)=\frac{x}{x+1}\) and \(g(x)=\frac{-1}{x+1}\) a) Compute \(f^{\prime}(x)\). b) Compute \(g^{\prime}(x)\). c) What can you conclude about \(f\) and \(g\) on the basis of your results from parts (a) and (b)?

Step-by-Step Solution

Answer

Both functions have the same derivative, meaning they share identical slopes at any point, differing only by a constant value.

Step 1: Find the derivative of f(x) using the quotient rule

The function is given by \(f(x) = \frac{x}{x+1}\). To find the derivative of \(f(x)\), we use the quotient rule, which states that for \(h(x) = \frac{u(x)}{v(x)}\), the derivative \(h'(x)\) is given by \(\frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\), where \(u(x) = x\) and \(v(x) = x+1\).First, compute \(u'(x) = 1\) and \(v'(x) = 1\).Using the quotient rule, we have:\[f'(x) = \frac{(1)(x+1) - (x)(1)}{(x+1)^2} = \frac{x+1-x}{(x+1)^2} = \frac{1}{(x+1)^2}\].

Step 2: Find the derivative of g(x) using the quotient rule

Now consider \(g(x) = \frac{-1}{x+1}\). Similar to part (a), apply the quotient rule where \(u(x) = -1\) and \(v(x) = x+1\).First, compute \(u'(x) = 0\) and \(v'(x) = 1\).Using the quotient rule, we have:\[g'(x) = \frac{(0)(x+1) - (-1)(1)}{(x+1)^2} = \frac{1}{(x+1)^2}\].

Step 3: Compare the derivatives of f(x) and g(x)

We've calculated that both derivatives \(f'(x)\) and \(g'(x)\) equal \(\frac{1}{(x+1)^2}\). Since \(f'(x) = g'(x)\), this means that the rate of change of \(f(x)\) and \(g(x)\) with respect to \(x\) are identical at each point in their domains.

Step 4: Conclusion: Relationship between f and g

Although \(f(x)\) and \(g(x)\) are different functions, they have the same derivative, suggesting they have similar slopes at any given point \(x\) where they are defined. This indicates the functions may differ only by a constant, representing parallel transformations in the graph.

Key Concepts

Quotient Rule

The quotient rule is a fundamental technique in calculus for finding the derivative of a function that is expressed as a quotient of two differentiable functions. Suppose we have a function given by \( h(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable. To find the derivative \( h'(x) \), we apply the quotient rule:

• Compute \( u'(x) \), the derivative of the numerator function \( u(x) \).
• Compute \( v'(x) \), the derivative of the denominator function \( v(x) \).
• Plug these into the formula:
  \[ h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

This formula tells us how the rate of change of the entire quotient behaves concerning the changes in both the numerator and the denominator.In our exercise, for \( f(x) = \frac{x}{x+1} \), \( u(x) = x \) and \( v(x) = x+1 \). By applying the quotient rule, we find \( f'(x) = \frac{1}{(x+1)^2} \). Similarly, for \( g(x) = \frac{-1}{x+1} \), despite a different numerator, the derivative is also \( \frac{1}{(x+1)^2} \). This shows how useful the quotient rule is for such functions.

Function Derivatives

The derivative of a function represents the rate at which the function's value changes as its input changes. Understanding function derivatives is crucial in mathematical analysis because it provides insight into the behavior and properties of functions.When differentiating a function, we essentially measure its slope or steepness at any given point. For polynomial functions, the process is straightforward, often involving the power rule. However, with the quotient of functions, like in our exercise, the quotient rule becomes essential.For \( f(x) = \frac{x}{x+1} \), the derivative \( f'(x) = \frac{1}{(x+1)^2} \) indicates that the slope of \( f(x) \) is dependent on the position \( x \) in the domain. Importantly, the positive derivative suggests that \( f(x) \) is an increasing function wherever \( x+1 eq 0 \). Likewise, \( g(x) = \frac{-1}{x+1} \) also yields a derivative of \( \frac{1}{(x+1)^2} \), showing similar increasing behavior at every point in its domain. This congruence leads to significant insights into their graphical representational differences and similarities.

Mathematical Analysis

Mathematical analysis is the discipline of mathematics that deals with limits and related theories, such as differentiation, integration, measure, sequences, and series. It provides rigorous methods for assessing the behavior of functions, crucial for understanding their properties and relationships.In our exercise, the mathematical analysis of the functions \( f(x) = \frac{x}{x+1} \) and \( g(x) = \frac{-1}{x+1} \) reveals interesting properties. Both functions have the same derivative, \( \frac{1}{(x+1)^2} \). This analysis shows that although \( f(x) \) and \( g(x) \) might appear different at first glance, they share similar slopes at every point in their domains. This aspect of mathematical analysis helps us conclude that they are not "dissimilar" as functions, and their core behavior is quite parallel. Such analysis is invaluable in graphing these functions as well as understanding transformations such as vertical shifts, which do not affect the derivative but noticeably alter the graph's position.This analytical approach emphasizes how calculus provides us with tools to discern deeper functional relationships and behaviors beyond mere algebraic manipulation.