Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

In each of the Problems 1-21, 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. $$ h(t)=t^{2 / 3} ;[0,2] $$

Step-by-Step Solution

Answer

The Mean Value Theorem does not apply because the function is not differentiable at \( t = 0 \).

Step 1: Verify Conditions for Mean Value Theorem

First, we check if the function \( h(t)=t^{2/3} \) is continuous on the closed interval \([0, 2]\). Since \( h(t) \) is a power function and its base \( t \) is raised to a fractional power, it is continuous everywhere in its domain, which includes the interval \([0, 2]\).

Step 2: Check Differentiability Inside the Interval

Next, we need to determine if \( h(t) \) is differentiable on the open interval (0, 2). The derivative is given by \( h'(t) = \frac{2}{3}t^{-1/3} \). This derivative is not defined at \( t = 0 \), meaning \(h(t)\) is not differentiable over the entire open interval \((0, 2)\).

Step 3: Conclusion on Mean Value Theorem Applicability

Since \( h(t) \) is not differentiable at \( t = 0 \), one of the conditions of the Mean Value Theorem is violated, and therefore the theorem cannot be applied to \( h(t) \) over \([0, 2]\). Hence, there are no values of \( c \) that satisfy the theorem in the given interval.

Key Concepts

Function Continuity

When deciding if the Mean Value Theorem (MVT) can be applied, one must first assess if the function is continuous on the closed interval given. Continuity means that the function doesn't have any jumps, breaks, or holes in the specified interval. For the function \( h(t) = t^{2/3} \), it is continuous on the interval [0, 2] because

• It is constructed from a power function,\( t^{2/3} \), where any positive real number for \( t \) keeps the function smooth without interruptions.
• There are no undefined points, leaps or holes within this interval, making it perfectly continuous from \( t = 0 \) to \( t = 2 \).

Ensuring continuity is crucial because the MVT requires the function to be continuous over a closed interval as one of its primary conditions. If this condition is met, which is the case here, then we can proceed to the next step in analyzing differentiability.

Differentiability

After confirming continuity, the next step is to check differentiability of the function on the open interval. Differentiability means that the derivative of the function exists at each point in the interval. For \( h(t) = t^{2/3} \), the derivative \( h'(t) = \frac{2}{3}t^{-1/3} \) exists, but not for the whole range (0, 2).

• This derivative becomes undefined at \( t = 0 \) because \( t^{-1/3} \) leads to an infinite slope when \( t = 0 \).
• A function failing differentiability even at one point in the open interval means the MVT cannot be applied. Differentiability over the entire interval is a must for using the theorem.

Thus, although the function is continuous on [0, 2], it is not differentiable throughout (0, 2), namely at \( t = 0 \). Because differentiability fails, the MVT cannot be used in this case.

Calculus Problem Solving

Solving calculus problems often involves determining when essential theorems like the Mean Value Theorem can be applied. For MVT in particular, it's vital to verify whether a function meets both continuity on the closed interval and differentiability on the open interval.
To solve calculus problems:

• Ensure the function is continuous on the interval. This often means checking for points where the function might be undefined or have abrupt changes.
• Verify differentiability. Calculate the derivative. The function should have a defined derivative for every point in the open interval.
• If both conditions fail, then the theorem doesn't apply, and solving the problem may require alternative methods or the conclusion that no solution exists in that context.

In the case of \( h(t) = t^{2/3} \), since differentiability at \( t = 0 \) fails, calculus problem-solving directs us to conclude that no \( c \) can satisfy the conditions of the MVT on [0, 2]. Applying these systematic checks helps in navigating more complex calculus problems.