Problem 8 - 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. $$ F(t)=\frac{1}{t-1} ;[0,2] $$

Step-by-Step Solution

Answer

Mean Value Theorem does not apply because the function is not continuous on \([0, 2]\).

Step 1: Check Continuity

The function \( F(t) = \frac{1}{t-1} \) is defined for all \( t eq 1 \). We must check its continuity on the interval \([0, 2]\). The function is not defined at \( t = 1 \), which is within our interval. Thus, \( F(t) \) is not continuous on \([0, 2]\).

Step 2: Conclusion on Application of Mean Value Theorem

The Mean Value Theorem requires the function to be continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\). Since \( F(t) \) is not continuous at \( t = 1 \), which lies in the interval \([0, 2]\), the Mean Value Theorem does not apply to this function on the given interval.

Key Concepts

Continuity of Functions

A function is said to be continuous on a particular interval if, for every point within this interval, the function remains unbroken and smooth. In simpler terms, you should be able to draw the function on this segment without lifting your pencil!
For the Mean Value Theorem to apply, the function must be continuous on the closed interval \([a, b]\). For example, if we examine the provided function \(F(t) = \frac{1}{t-1}\), we note that the function becomes undefined at \(t = 1\).
Due to this point within our interval \([0, 2]\), the function is not continuous. Thus, the Mean Value Theorem cannot be used. Always remember that any gap, jump, or vertical asymptote in a function on an interval renders it non-continuous for that segment.

Differentiability

Differentiability refers to a function’s ability to have a derivative at every point in a given interval. Simply put, if you zoom in really close at any point on the function's graph, it should look like a straight line. This is important in calculus, as it helps in finding slopes and rates of change.
For the Mean Value Theorem, the function must be differentiable over the open interval, \((a, b)\). If we consider \(F(t) = \frac{1}{t-1}\), it needs to be differentiable on \((0, 2)\).
However, due to not being defined at \(t = 1\), \(F(t)\) is not differentiable at this point. Without differentiability, calculating a derivative at \(t = 1\) is impossible, reinforcing why the theorem doesn’t apply.

Closed Interval

A closed interval is expressed as \([a, b]\) and includes all points between \(a\) and \(b\), as well as the endpoints themselves. Think of it like a closed bracket that encompasses every value in between. These intervals are crucial when applying the Mean Value Theorem because it requires the function to be continuous on this very interval.
In our exercise, the closed interval is \([0, 2]\). The function \(F(t) = \frac{1}{t-1}\) must fulfill the theorem's conditions within this range.
However, because \(t = 1\) is not a point where the function is defined (resulting in discontinuity and lack of differentiability), it means the precondition is not met, showing the importance of a continuous function over the closed interval.

Calculus Problem Solving

Solving problems in calculus, especially those involving the Mean Value Theorem, requires a structured approach. First, ensure the function is continuous over the closed interval, \([a, b]\), and differentiable over the open interval, \((a, b)\).
Using \(F(t) = \frac{1}{t-1}\), we saw neither condition was met in \([0, 2]\), meaning we couldn’t apply the theorem. When tackling such problems:

• Verify continuity on given intervals.
• Check for differentiability by identifying any undefined or non-smooth points.
• Consider graphing the function to visually spot any issues.

These steps help dissect complex calculus problems, ensuring correct applications of theorems like the Mean Value Theorem.