Problem 52 - Single Variable Calculus

Question

In Exercises \(47-52\), determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. If \(f\) is continuous on \([1,3]\), differentiable on \((1,3)\), and satisfies \(f(1)=2, f(3)=5\), then there exists a number \(c\) satisfying \(1

Step-by-Step Solution

Answer

The statement is true. By applying the Mean Value Theorem to the function \(f\) which is continuous on the interval \([1, 3]\) and differentiable on \((1, 3)\), we find that there exists a number \(c\) in \((1, 3)\) such that \(f'(c) = \frac{3}{2}\).

Step 1: Mean Value Theorem (MVT)

The Mean Value Theorem states that if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a number \(c\) in \((a, b)\) such that \[f'(c) = \frac{f(b) - f(a)}{b - a}\]

Step 2: Apply the MVT to the function \(f\)

In this exercise, we have a function \(f\) continuous on the interval \([1, 3]\) and differentiable on \((1, 3)\). Therefore, we can apply the Mean Value Theorem to this function. Let \(a = 1\) and \(b = 3\), we get: \[f'(c) = \frac{f(3) - f(1)}{3 - 1}\]

Step 3: Find f'(c)

Substitute the given values of f(1) and f(3) in the equation: \[f'(c) = \frac{5 - 2}{3 - 1} = \frac{3}{2}\]

Step 4: Conclusion

Since the requirements of the MVT are satisfied and we have found \(f'(c) = \frac{3}{2}\), we can conclude that there exists a number c in the interval (1, 3) such that \(f'(c) = \frac{3}{2}\). Thus, the statement is true.

Key Concepts

Continuity

Continuity is a fundamental concept in mathematics, especially when dealing with functions. A function is said to be continuous on an interval if, intuitively, you can draw the function without lifting your pen from the paper. This means there are no breaks, jumps, or holes in the graph of the function.
For a function \( f \) to be continuous on a closed interval \([a, b]\), it must be continuous at every point within that interval. Specifically, it should satisfy:

• The function \( f \) is defined for all \( x \) in \([a, b]\).
• The limit \( \lim_{{x \to c}} f(x) = f(c) \) exists for all \( c \) in \([a, b]\).
• The left and right limits at the endpoints \( a \) and \( b \) should agree with the function's values at those points.

In the given exercise, we know that \( f \) is continuous on \([1, 3]\), which is crucial for applying the Mean Value Theorem (MVT). Without continuity, the theorem doesn't hold.

Differentiability

Differentiability is about the existence of a derivative. A function is differentiable at a point if it has a defined tangent line at that point, indicating a specific rate of change.
To be differentiable on an open interval \((a, b)\), \( f \) must have a derivative at every point in that interval. Differentiability implies continuity, but not vice versa.

• If \( f \) is differentiable at a point \( c \), then \( \lim_{{h \to 0}} \frac{f(c+h)-f(c)}{h} \) exists.
• Any sharp corners or cusps imply non-differentiability at those points.

In this exercise, \( f \) is stated to be differentiable on \((1, 3)\). This property allows us to use the MVT, as differentiability is a crucial requirement.

Interval Analysis

Interval analysis is the study of a function's behavior over a specific interval. It involves examining various properties such as continuity and differentiability within that range.
When applying the Mean Value Theorem, we focus on closed and open intervals:

• \([a, b]\): This is a closed interval where the function must be continuous.
• \((a, b)\): This is an open interval where the function must be differentiable.

For the case of \([1, 3]\) and \((1, 3)\), we are assured that the necessary conditions are met, allowing us to find a specific \( c \) where the derivative equals the average rate of change from \( a \) to \( b \). This analysis is why the statement in the exercise is true.

Function Behavior

Understanding a function's behavior involves looking at how it changes over its domain. Critical insights come from examining its continuity, differentiability, and rates of change.
The function's behavior on a specific interval helps us predict and understand key aspects, like finding a \( c \) such that \( f'(c) = \frac{3}{2} \).

• The difference \( f(b) - f(a) \) provides the overall change on the interval.
• The derivative \( f'(c) \) tells us the instantaneous rate of change at point \( c \).

This behavior is encapsulated by the MVT, linking the average rate of change over an interval to the function's instantaneous rate of change at some point within that interval.