Problem 27 - Essential Calculus

Question

Writing In Exercises 27 and \(28,\) explain why the Mean Value Theorem does not apply to the function \(f\) on the interval [0,6]. $$ f(x)=\frac{1}{x-3} $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply to the function \(f(x) = \frac{1}{x-3}\) on the interval [0,6] because the function is not continuous at x=3, which is within the interval. This violates one of the conditions for the MVT, which says the function must be continuous on the whole interval.

Step 1: Understand the Mean Value Theorem

The Mean Value Theorem (MVT) says that if a function is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), there's at least one 'c' in (a,b) for which the function's derivative at 'c' equals the average rate of change on the interval [a,b].

Step 2: Identify the function and interval

Here, the function is \(f(x) = \frac{1}{x-3}\) and the interval is [0,6].

Step 3: Check if the function satisfies the conditions of the MVT

This function \(f(x)\) is not continuous at x=3 (within the interval [0,6]), as it has a vertical asymptote there. Because the function is not continuous on the whole interval [0,6], it doesn't meet the MVT conditions, so the MVT doesn't apply.

Key Concepts

Functional Continuity

Understanding the concept of continuity is essential in calculus and helps determine where the Mean Value Theorem (MVT) can be applied. Continuity of a function refers to it behaving in a predictable manner, where there are no jumps, breaks, or holes within a specified interval. For a function to be continuous on a closed interval - Every point within the interval should be defined- The function should not have any abrupt changes
- The left-hand and right-hand limits at every point in the interval should be equal to the value of the function at that pointIn our particular problem, we have a function defined as \[f(x) = \frac{1}{x-3}\]This function has an issue at the point \(x = 3\) because the denominator becomes zero, creating an undefined point. This results in the function having a vertical asymptote at \(x = 3\), breaking continuity. Therefore, the function is not continuous on the interval [0,6], making it unsuitable for the MVT to apply. Identifying such breaks in continuity is crucial in determining if functions can be analyzed using tools like the Mean Value Theorem.

Calculus Problem Solving

When solving calculus problems, understanding the requirements of theorems like the Mean Value Theorem (MVT) is vital. The MVT can be seen as a powerful tool that requires the function to be continuous on a closed interval and differentiable on an open interval within. This means the function must not only behave predictably but also be smooth enough to have a derivative at every point within the interval without exceptions.In calculus problem-solving, we often:- Validate that all conditions for using a theorem are met- Check for any asymptotes or discontinuities within the interval
- Calculate derivatives and check differentiabilityIn our problem, we see that the function \(f(x) = \frac{1}{x-3}\) fails the continuity condition due to the presence of a vertical asymptote at \(x = 3\). This means the function cannot satisfy the conditions of the MVT, leading to its inapplicability. Reviewing the breakdown helps ensure a problem is approached analytically and methodically, paving the way for solving more complex calculus problems.

Asymptotes in Functions

Asymptotes play a significant role in understanding the behavior of functions, especially when discussing their limits and continuity. An asymptote is a line that a graph approaches but never touches, creating points where the function is undefined. Vertical asymptotes often arise where the function's denominator equals zero.For the function in our problem, \[f(x) = \frac{1}{x-3}\]The vertical asymptote appears at \(x=3\). Understanding this behavior is crucial as it highlights points of discontinuity. Vertical asymptotes indicate that the function's value grows infinitely as it approaches a specific value of \(x\). This particular asymptotic behavior results in the Mean Value Theorem's conditions not being satisfied, as it requires the function to be continuous across the entire interval.In general, when investigating functions:- Identify any vertical, horizontal, or oblique asymptotes
- Determine the impact these asymptotes have on the function's domain and continuityComprehending the influence of asymptotes enables thorough analysis of functions, central to forming a foundation for advanced calculus topics.