Problem 11 - 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. $$ h(t)=t^{2 / 3} ;[0,2] $$

Step-by-Step Solution

Answer

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

Step 1: Verify Continuity on the Closed Interval

First, we need to check if the function \( h(t) = t^{2/3} \) is continuous on the closed interval \([0, 2]\). The function \( t^{2/3} \) is continuous for all \( t \geq 0 \) because it is a basic power function. Thus, it is continuous on \([0, 2]\).

Step 2: Verify Differentiability on the Open Interval

Next, check if \( h(t) = t^{2/3} \) is differentiable on the open interval \((0, 2)\). The derivative is \( h'(t) = \frac{2}{3}t^{-1/3} \). This derivative exists for all \( t > 0 \). However, at \( t = 0 \), the derivative does not exist because it would require division by zero. Therefore, \( h(t) \) is not differentiable on the entire open interval \((0, 2)\).

Step 3: Conclusion on Applying the Mean Value Theorem

For the Mean Value Theorem to apply, the function must be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Since \( h(t) \) is not differentiable at \( t = 0 \), the Mean Value Theorem does not apply to \( h(t) \) on the interval \([0, 2]\).

Step 4: Sketch the Graph

Sketch the graph of \( h(t) = t^{2/3} \) over the interval \([0, 2]\). The graph is a curve starting at \( (0,0) \) and rising smoothly to \( (2, \, 2^{2/3}) \). There is a vertical tangent line at \( t = 0 \), indicating non-differentiability at that point.

Key Concepts

Continuity

Continuity of a function is a core concept in calculus. It means the graph of the function can be drawn without lifting your pencil. A function is continuous over an interval if it doesn't have any breaks, jumps, or holes in that interval.
For the Mean Value Theorem (MVT) to apply, continuity on the closed interval is crucial. In the original exercise, the function \( h(t) = t^{2/3} \) is continuous on the interval \([0, 2]\).
This specific function has no discontinuities since it is a basic power function, allowing it to remain unbroken over the interval.

Differentiability

Differentiability is a measure of how smoothly a function behaves and whether it has a defined tangent at each point in its domain.
For a function to be differentiable at a point, it must have a derivative at that point.
This concept is important for employing the Mean Value Theorem, which requires differentiability on an open interval.

• In the step-by-step solution, the derivative of \( h(t) = t^{2/3} \) is \( h'(t) = \frac{2}{3}t^{-1/3} \).
• However, this derivative doesn’t exist at \( t = 0 \), making \( h(t) \) non-differentiable over the entire interval \((0, 2)\).

The breakdown at this point makes the function unsuitable for the Mean Value Theorem.

Closed Interval

A closed interval, denoted as \([a, b]\), includes all the points \(x\) where \(a \leq x \leq b\). It incorporates both its endpoints \(a\) and \(b\).
This interval is essential for the continuity requirement in the Mean Value Theorem.
In the given exercise, the closed interval is \([0, 2]\).
The function \( h(t) \) is continuous over this full span, verifying the first condition needed for the Mean Value Theorem.

Open Interval

An open interval, represented by \((a, b)\), consists of all points \(x\) where \(a < x < b\). Importantly, it does not include the endpoints \(a\) and \(b\).
The role of the open interval is pivotal in determining differentiability.
In our example, \( h(t) = t^{2/3} \) needed to be differentiable over \((0, 2)\) to satisfy the Mean Value Theorem's criteria.
However, because its derivative becomes undefined at \( t = 0 \), the function can’t meet the Mean Value Theorem requirements on this open interval.