Problem 177 - Calculus Volume 1

Question

Determine whether the Mean Value Theorem applies for the functions over the given interval \([a, b] .\) Justify your answer. $$ y=\frac{x^{2}+3 x+2}{x} \text { over }[-1,1] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because the function is not continuous on \([-1, 1]\).

Step 1: Verify Function Continuity

The first condition for the Mean Value Theorem is that the function must be continuous on the closed interval \([-1, 1]\). \[ y = \frac{x^{2} + 3x + 2}{x} = x + 3 + \frac{2}{x} \]The function has a discontinuity at \(x = 0\) because \(\frac{2}{x}\) is not defined at \(x = 0\). Since the interval includes this point, the function is not continuous on \([-1, 1]\).

Step 2: Examine Differentiability

The second condition requires the function to be differentiable on the open interval \((-1, 1)\). For differentiability, the function must be continuous, which as previously established, it is not at \(x = 0\). Therefore, the function is not differentiable over the entire \((-1, 1)\).

Step 3: Conclusion on Mean Value Theorem Applicability

Since the function \(y = \frac{x^{2} + 3x + 2}{x}\) is neither continuous nor differentiable over the interval \([-1, 1]\), the Mean Value Theorem does not apply. The discontinuity at \(x = 0\) disqualifies the function from meeting the necessary conditions for the theorem.

Key Concepts

Function Continuity

Function continuity is an essential component when considering the application of the Mean Value Theorem (MVT). For a function to be continuous on a closed interval, it must be defined without any interruptions within that range. In simpler terms, you should be able to draw its graph from the starting point to the endpoint without lifting your pencil.

Consider the function given in the problem: \[ y = \frac{x^{2} + 3x + 2}{x}\]This function can be rewritten as \[ y = x + 3 + \frac{2}{x}\]Analyzing this, you notice immediately a hint of a problem with the term \(\frac{2}{x}\). It is not defined at \(x = 0\), creating a gap or a sudden jump in the graph at this point.

In the given interval of \([-1, 1]\), since \( x = 0 \) falls inside this range and causes an undefined region, the function is not continuous. The Mean Value Theorem necessitates a continuous function over the whole interval, rendering it inapplicable here.

Function Differentiability

Function differentiability is closely related to continuity. For a function to be differentiable at a point, it must also be continuous there. However, differentiability is a stricter condition and refers to the smoothness of the function at a point. This means it should not have sharp turns or corners.

In mathematical terms, a function \(f(x)\) is differentiable at \(x = c\) if the derivative \(f'(x)\) exists at that point. In the case study problem, the function \[y = \frac{x^{2} + 3x + 2}{x}\]has already been determined as having a discontinuity at \(x = 0\), meaning it cannot smoothly be extended around that interval. Hence, the second condition of MVT - differentiability on the open interval \((-1, 1)\) - is not satisfied.

The presence of the term \(\frac{2}{x}\), which leads to a division by zero at a specific point, confirms that the function is not differentiable there. Therefore, the smoothness needed for the function isn’t maintained, violating the Mean Value Theorem condition.

Interval Notation

Interval notation provides a simplified way to express a set of numbers grouped together on the number line. This method efficiently communicates the start and end points of an interval while indicating whether these points are included or excluded.

In the original exercise, we have two notations to discuss: \([-1, 1]\) and \((-1, 1)\). The brackets \([a, b]\) indicate a closed interval, which includes both endpoints \(-1\) and \(1\). This suggests that values within this range, including the endpoints, are being considered.

• A closed interval using brackets reflects an inclusive range. Both \(-1\) and \(1\) belong to the set.
• An open interval, expressed as parentheses like \((-1, 1)\), is exclusive, meaning the endpoints are not part of the set.

In the problem, the function is checked for continuity on \([-1, 1]\) but differentiability on \((-1, 1)\). These notations clarify what each condition of the theorem requires: continuous throughout including endpoints for closed intervals, and smooth (differentiable) between the endpoints without touching them for open intervals. Remaining mindful of these notations helps greatly in analyzing and applying various calculus theorems.