Problem 175 - Calculus Volume 1

Question

Determine whether the Mean Value Theorem applies for the functions over the given interval \([a, b]\). Justify your answer. \(y=\frac{1}{|x+1|}\) over [0,3]

Step-by-Step Solution

Answer

Yes, the Mean Value Theorem applies because the function is continuous on [0, 3] and differentiable on (0, 3).

Step 1: Verify Continuity on the Interval

The Mean Value Theorem (MVT) requires that the function be continuous on the closed interval \([a, b]\). The function \(y = \frac{1}{|x+1|}\) is defined as \(y = \frac{1}{x+1}\) for \(x > -1\). Here, the interval is \([0, 3]\), so we need to check for continuity over this interval. Since neither 0 nor 3 makes the denominator zero, the function is continuous on \([0, 3]\).

Step 2: Verify Differentiability on the Open Interval

The MVT also requires that the function be differentiable on the open interval \((a, b)\). The function \(y = \frac{1}{|x+1|}\) is differentiable as \(y = \frac{1}{x+1}\) for \(x > -1\). There are no sharp turns or discontinuities in the derivative for \(x > 0\) in the interval \((0, 3)\), ensuring differentiability.

Step 3: Conclusion about Mean Value Theorem

Since the function \(y = \frac{1}{|x+1|}\) is continuous on \([0, 3]\) and differentiable on \((0, 3)\), both conditions for the Mean Value Theorem are satisfied. Therefore, the Mean Value Theorem applies to this function on the given interval.

Key Concepts

Understanding Continuity

In calculus, continuity of a function is fundamental for the application of the Mean Value Theorem (MVT). A function is continuous on a closed interval \([a, b]\) if there are no interruptions or breaks in its graph. It's like drawing the function's graph without lifting your pen. For the function \(y = \frac{1}{|x+1|}\), we check continuity over the interval \([0, 3]\):

• The absolute value \(|x+1|\) ensures the denominator does not equal zero within this range.
• Thus, the function is well-defined and continuous over the entire interval.

Continuity is a must for MVT, which guarantees that if a function is continuous over a closed interval, even if it changes direction several times, it will smoothly transition from point \(a\) to point \(b\). Always check for endpoints being defined to determine continuity in such problems.

Grasping Differentiability

Differentiability is a core requirement of the Mean Value Theorem on the open interval \(a, b\). A differentiable function at a point means it has a defined tangent and no sharp corners or edges at that point. For \(y = \frac{1}{|x+1|}\), differentiability can be confirmed:

• The expression \(y = \frac{1}{x+1}\) smooths out the edges caused by the absolute value for \(x > -1\).
• Within \((0, 3)\), the function remains smooth without disruptions, ensuring its differentiability.

Differentiability means being able to find the derivative across the interval and having no sudden jumps or vertical tangents. This property secures that the Mean Value Theorem's conditions are met, specifically outcomes that reflect predictable behavior, such as estimated average slopes matching instantaneous slopes.

Solving Calculus Problems with MVT

The Mean Value Theorem is a powerful tool in calculus for solving problems involving rates of change. Apply the MVT only if both continuity and differentiability are confirmed, as shown with the function \(y = \frac{1}{|x+1|}\):

• Verify the function is continuous on the closed interval \([a, b]\).
• Ensure it's differentiable on the open interval \((a, b)\).
• Look for a point \(c\) within \((a, b)\) where the instantaneous rate of change (slope of tangent) equals the average rate of change over the interval.

Successfully applying these steps means you can predict specific behaviors of the function within an interval. Though occasionally complex, mastering the theorem significantly bolsters your calculus toolkit for analyzing how functions behave beyond just finding derivatives and integrals.