Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

, a function is defined and a closed interval is given. Decide whether the Mean Value Theorem applies to the given function on the given interval. If it does, find all possible values of \(c ;\) if not, state the reason. In each problem, sketch the graph of the given function on the given interval. $$ f(x)=\frac{x-4}{x-3} ;[0,4] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because the function is not continuous on \([0,4]\).

Step 1: Check Continuity

The Mean Value Theorem (MVT) requires the function to be continuous on a closed interval \([a,b]\). Let's check if \(f(x) = \frac{x-4}{x-3}\) is continuous on \([0,4]\). The function has a potential discontinuity where the denominator is zero, which is at \(x = 3\). Therefore, \(f(x)\) is not continuous at \(x = 3\), and since \(3 \in [0,4]\), the function is not continuous on \([0,4]\).

Step 2: Check Differentiability

Even though we already found discontinuity, let's consider the differentiability aspect. The MVT also requires the function to be differentiable on the open interval \((a,b)\). For \(f(x)\) to be differentiable at all, it needs to be defined; however, it is undefined at \(x = 3\). It is already not continuous at \(x=3\), so it can't be differentiable there either. Therefore, \(f(x)\) is not differentiable on any interval containing \(x=3\).

Step 3: Conclusion

Since \(f(x)\) is neither continuous nor differentiable on \([0,4]\), the Mean Value Theorem cannot be applied to this function over the given interval. Therefore, we cannot find a value of \(c\) as required by the theorem.

Key Concepts

Continuity in Calculus

Continuity is a fundamental concept in calculus that describes how a function behaves as its input approaches a certain value. In simple terms, a function is continuous at a point if there are no abrupt changes or breaks in its graph at that point. This means you could draw the graph without lifting your pen. For the Mean Value Theorem (MVT) to be applied to a function, it is essential that the function is continuous on the closed interval \([a, b]\).
In our example, the function \(f(x) = \frac{x-4}{x-3}\) has a "hole" at \(x = 3\) because the denominator becomes zero, making the function undefined at this point. Since \(x = 3\) is included in the interval \([0, 4]\), the function is discontinuous on this interval. Consequently, the lack of continuity at \(x = 3\) excludes the possibility of using the MVT. Discontinuities can occur in various forms such as gaps, jumps, or asymptotes, and they are crucial in determining the applicability of calculus theorems like the MVT.

Differentiability in Calculus

Differentiability refers to the existence of a derivative at a certain point in a function's domain. If a function is differentiable at a point, its graph has a well-defined tangent line at that point, suggesting it has a consistent slope. A key requirement for the MVT is that the function should not only be continuous on a closed interval \([a, b]\) but also differentiable on the open interval \((a, b)\).
It should be noted that if a function is differentiable at a point, it automatically implies continuity at that point, but not vice versa.
In our specific case for \(f(x) = \frac{x-4}{x-3}\), we found that the function is not continuous at \(x = 3\). This discontinuity directly affects its differentiability since a function that is not continuous cannot be differentiable. Hence, \(f(x)\) is not differentiable over any interval containing \(x = 3\). Failing to meet this requirement further confirms that the MVT cannot be applied.

Rational Functions

Rational functions are expressions that can be represented as the ratio of two polynomials. They are defined wherever the denominator does not equal zero. These functions often exhibit interesting behavior such as vertical asymptotes, which occur where the denominator is zero.
In the example \(f(x) = \frac{x-4}{x-3}\), the function appears to resemble a line, but has notable behavior at \(x = 3\). Here, the denominator becomes zero, leading to a vertical asymptote, and thus the function is undefined at \(x = 3\).
Understanding rational functions is crucial when analyzing continuity and differentiability. It is particularly important to identify points where the function is undefined, as these will directly affect both the continuity and differentiability of the function. Rational functions, due to their nature, often require careful examination within the context of calculus, particularly when applying theorems like the Mean Value Theorem.