Problem 18 - Essential Calculus

Question

Determine whether Rolle's Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If Rolle's Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=0\). $$ f(x)=x-2 \ln x,[1,3] $$

Step-by-Step Solution

Answer

Rolle's Theorem cannot be applied to this function on the interval [1, 3] and hence there is no value of \( c \) in the open interval (1, 3) such that the derivative of the function at point \( c \) is zero.

Step 1: Check the conditions of Rolle's theorem

The function \( f(x)=x-2 \ln x \) is continuous on [1, 3] because polynomials and the natural log function are continuous everywhere in their domains and the domain of this function includes the interval [1, 3]. The function is also differentiable on the open interval (1, 3) because it is made of differentiable functions. To check condition 3, we must calculate \( f(a) \) and \( f(b) \) and see if they are equal: \n \( f(1)=1-2 \ln 1=1 \) \n \( f(3)=3-2 \ln 3 \approx 3-2(1.1) \approx 0.8 \). The function does not meet the third condition so Rolle's theorem cannot be applied.

Step 2: Find all values of c

Since the conditions of Rolle's theorem are not met, there are no values of \( c \) on the open interval (1, 3) such that \( f'(c)=0 \).

Key Concepts

Continuous Functions

A function is continuous on a closed interval when it can be drawn without lifting the pencil from the paper. In simpler terms, there are no sudden jumps, breaks, or holes in the graph. A function is continuous if you can move from one end of the interval to the other smoothly. For example, polynomials and some logarithmic functions are continuous across their domains because they connect without interruption.

To determine if a function is continuous over a specific interval, you can check each type of function involved:

• Polynomial functions are continuous everywhere because they are simple curves.
• Logarithmic functions are continuous on any interval that doesn't hit zero or negative values, as \ln(x)\ is undefined for zero or negative numbers.

In the case of our function, \( f(x) = x - 2 \ln(x)\) over the interval [1, 3], both components,—the linear \(x\) and logarithmic \(\ln(x)\)—are continuous. Thus, the combined function is continuous over this interval. Checking continuity is crucial in Rolle's Theorem because it helps ensure there are no unexpected breaks that could affect the behavior of the function.

Differentiable Functions

For a function to be differentiable on an interval, it means the function has a derivative at every point within that interval. This essentially indicates smoothness, aligning with the idea that a derivative measures how steep a curve is at any point.

If a function is differentiable, you can calculate a derivative and predict the function's behavior effectively. Here's how you can tell if a function is differentiable:

• The function must not have any sharp corners or cusps; it should be smooth.
• There must be no vertical tangents; the slope cannot be infinitely steep.
• No discontinuities or jumps; the graph must transition smoothly.

The function \(f(x) = x - 2 \ln(x)\) involves both a linear component, \(x\), and a logarithmic part, \(-2 \ln(x)\). Both of these are differentiable over their domains. Especially, if you focus on the interval (1, 3), this combination guarantees that the function is differentiable throughout the interval. Differentiability is a core requirement of Rolle's Theorem because it helps to identify the behavior of a curve, ensuring the function can transition smoothly from one point to another without sudden direction changes.

Interval Analysis

Interval analysis, particularly when discussing Rolle's Theorem, refers to the scrutiny of a function's behavior over a particular span of numbers, also known as an interval. Rolle’s Theorem specifically requires examining a closed interval \([a, b]\), checking whether the function:

• Is continuous on the closed interval \([a, b]\).
• Is differentiable on the open interval \((a, b)\).
• Satisfies \(f(a) = f(b)\), meaning the function has equal values at the endpoints.

In our exercise involving \(f(x) = x - 2 \ln(x)\) on the interval \([1, 3]\), although the function meets the first two conditions, it doesn't satisfy \(f(a) = f(b)\) because:
- \(f(1) = 1\)- \(f(3) \approx 0.8\)

This difference means that there is no guarantee a horizontal tangent line (slope zero) will exist between the interval ends. Thus, even though the function seems smooth and has no breaks in this span, Rolle's Theorem is inapplicable due to the unequal endpoint values.