Problem 156 - Calculus Volume 1

Question

Determine over what intervals (if any) the Mean Value Theorem applies. Justify your answer. $$ y=\ln (3 x-5) $$

Step-by-Step Solution

Answer

The MVT applies on any interval \([a, b]\) where \(a > \frac{5}{3}\) and \(b > \frac{5}{3}\).

Step 1: Confirm continuity on a closed interval

The Mean Value Theorem (MVT) applies if the function is continuous on a closed interval \([a, b]\). The function \(y = \ln(3x - 5)\) is continuous where its argument \(3x - 5 > 0\). Thus, \(x > \frac{5}{3}\). Choose any interval \( [a, b] \) where both \(a > \frac{5}{3}\) and \(b > \frac{5}{3}\) to ensure continuity.

Step 2: Confirm differentiability on an open interval

The function must be differentiable on the open interval \((a, b)\). The derivative \(y' = \frac{3}{3x - 5}\) exists and is well-defined as long as \(x > \frac{5}{3}\). Ensure the chosen interval does not include \(x = \frac{5}{3}\) to meet this condition.

Step 3: Identify valid intervals for the MVT

Considering Steps 1 and 2, the Mean Value Theorem applies on any interval \([a, b]\) where both endpoints satisfy \(a > \frac{5}{3}\) and \(b > \frac{5}{3}\). This ensures both continuity and differentiability conditions are met on \([a, b]\) and \((a, b)\), respectively.

Key Concepts

Continuity of Logarithmic Functions

Logarithmic functions, such as the natural logarithm function involved in the expression \(y = \ln(3x - 5)\), are known for their unique properties related to continuity. By definition, a logarithmic function is continuous wherever its argument is positive. This is because the logarithm is only defined for positive numbers, ensuring smooth and unbroken behavior in its output.

For the given function \(y = \ln(3x - 5)\), continuity is observed where \(3x - 5 > 0\). To find the interval, solve the inequality: \(3x > 5\), leading to \(x > \frac{5}{3}\). This means that the function is continuous for all \(x\) values greater than \(\frac{5}{3}\).

When applying the Mean Value Theorem, it's crucial to select a closed interval \([a, b]\) such that both \(a\) and \(b\) are greater than \(\frac{5}{3}\). This ensures the function is continuous throughout the entire interval, a necessary condition for the theorem.

Differentiability

Differentiability refers to the existence of a derivative at each point in an interval, confirming smoothness and no abrupt changes or corners. A function is differentiable in an open interval if you can calculate a derivative at every point within it.

For the logarithmic function \(y = \ln(3x - 5)\), its derivative is determined as \(y' = \frac{3}{3x - 5}\). The derivative is well-defined anywhere the logarithm is continuous, that is, when \(x > \frac{5}{3}\). This means the function is differentiable for any \(x\) greater than \(\frac{5}{3}\).

Hence, for the Mean Value Theorem to apply, choose an open interval \((a, b)\) where both \(a > \frac{5}{3}\) and \(b > \frac{5}{3}\). This ensures the function is differentiable over the entire interval, an essential requirement for employing the theorem.

Intervals of Applicability

The Mean Value Theorem applies to functions that are continuous on a closed interval and differentiable on the open interval contained within it. Understanding when it can be used is crucial for deriving meaningful results.

For \(y = \ln(3x - 5)\), the intervals where the Mean Value Theorem holds are determined by both its continuity and differentiability. From the information derived:

• The function is continuous for \(x > \frac{5}{3}\).
• It is also differentiable on this same condition.

Therefore, the Mean Value Theorem can be applied on any closed interval \([a, b]\) where \(a > \frac{5}{3}\) and \(b > \frac{5}{3}\). Here, continuity and differentiability conditions align perfectly, allowing proper application of the theorem. The chosen interval must not include points where \(x \leq \frac{5}{3}\) because the function would not meet the necessary conditions at those points.