Problem 44 - Essential Calculus

Question

Find differentiable functions \(f\) and \(g\) such that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow \infty} g(x)=\infty \quad\) and \(\lim _{x \rightarrow \infty}[f(x)-g(x)]=25\). Explain how you obtained your answers. (Note: There are many correct answers.)

Step-by-Step Solution

Answer

The functions are \(f(x) = x + 25\) and \(g(x) = x\).

Step 1: Identify appropriate functions

To solve this problem, let's consider simple linear functions. Based on the given conditions, one could take \(f(x)\) as \(x + c\) and \(g(x)\) as \(x\), where \(c\) is a constant.

Step 2: Find the constant

Given that \(\lim _{x \rightarrow \infty}[f(x)-g(x)]=25\), substituting our chosen functions we get: \(\lim _{x \rightarrow \infty}[(x + c) - x] = 25\). From this, we can see that \(c=25\).

Step 3: Check the conditions

Now that we've found \(c\), let's plug it back into our functions: \(f(x) = x+25\), \(g(x) = x\). It is clear that both functions are differentiable everywhere. Also, as \(x\) becomes large, both \(f(x)\) and \(g(x)\) tend to infinity. Therefore, they meet all the set conditions.

Key Concepts

Differentiable Functions

Differentiable functions are an essential concept in calculus. They are functions that have a derivative at every point within their domain. This means that at any point on the function's curve, it is possible to draw a tangent line that touches the curve at that exact point. Differentiability is critical because it gives us the ability to understand the rate of change of the function's values with respect to changes in the input variable.
For a function to be differentiable everywhere, its derivative must exist for all input values. In the provided exercise, both functions \(f(x) = x+25\) and \(g(x) = x\) are differentiable because they are linear functions. Linear functions, as outlined next, have a constant rate of change, meaning their derivatives are constants. For any linear function \( f(x) = mx + b \), its derivative, \( f'(x) = m \), tells us that its slope is always \(m\), and in this case, it is 1. Thus, differentiability is easily confirmed.

Linear Functions

Linear functions are among the simplest forms of functions encountered in algebra and calculus. They are functions that can be expressed in the form \( f(x) = mx + b \), where \(m\) and \(b\) are constants. The graph of a linear function is a straight line, which makes it easy to handle, especially when solving calculus problems such as the one given in the exercise.

• In the function form \(f(x) = x + 25\), \(m = 1\), which means the slope of the line is 1.
• The constant \(b = 25\) indicates the y-intercept, or where the line crosses the y-axis.
• Similarly, \(g(x) = x\) is another linear function with a slope \(m = 1\) and y-intercept \(b = 0\).

The straightforward nature of linear functions simplifies the calculation of limits, as shown in the step-by-step solution. Because they grow infinitely along both axes, they naturally tend to infinity as \(x\) grows, meeting one of the crucial conditions set in the problem.

Calculus Problem Solving

Solving calculus problems often involves breaking down the problem into manageable steps and applying fundamental calculus principles. In this exercise, it involved choosing suitable functions, checking their limits, and ensuring they met all required conditions.
The first step in solving the exercise was to choose functions that are known to extend to infinity, making linear functions a perfect candidate due to their simplicity and infinite growth characteristics. The next step was computing the limit of the difference \(f(x) - g(x)\). When doing so, the expertise lies in algebraic manipulation like evaluating limits, which in many calculus problems should be straightforward if the functions are well-chosen. For instance, in this case:

• The limit \( \lim_{x \rightarrow \infty} [f(x) - g(x)] = 25 \) was achieved by setting \(f(x) = x + 25\) and \(g(x) = x\).
• The choice of \(c = 25\) ensured that the condition \(\lim_{x \rightarrow \infty} (x+c-x) = 25\) was satisfied neatly.

By carefully selecting and verifying that the conditions met the requirement of the problem, the solution demonstrated the essence of calculus problem-solving techniques.