Problem 43 - Calculus of a Single Variable

Question

Determine whether the Mean Value Theorem can be applied to \(f\) on the closed interval \([a, b] .\) If the Mean Value Theorem can be applied, find all values of \(c\) in the open interval \((a, b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). If the Mean Value Theorem cannot be applied, explain why not. \(f(x)=|2 x+1|, \quad[-1,3]\)

Step-by-Step Solution

Answer

The Mean Value Theorem cannot be applied to the function \(f(x) = |2x + 1|\) on the interval [-1, 3] because the function is not differentiable on the entire open interval which is an important prerequisite of the Mean Value Theorem.

Step 1: Identify the 'corners' of the function

Solve for the expression inside the absolute value function to be equal to zero, \(2x + 1 = 0\). This gives \(x = -\frac{1}{2}\) as the 'corner' or point of non-differentiability.

Step 2: Check for differentiability

Since the 'corner' \(x = -\frac{1}{2}\) lies in the open interval (-1, 3), the function is not differentiable on the entire open interval. Thus, the Mean Value Theorem cannot be applied.

Step 3: Mean Value Theorem applicability

The conclusion drawn is that because the function is not differentiable on the entire open interval, it doesn't meet the conditions of the Mean Value Theorem.

Key Concepts

Differentiability

Understanding differentiability is essential in calculus as it relates to the smoothness of functions and their ability to be represented by a tangent line at a point. A function is said to be differentiable at a point if its derivative – the rate at which it changes – exists at that point. More formally, if the limit of the slope of the secant lines as they approach the point from either side is the same, then the function is differentiable there.

To visualize this, imagine a curve on a graph. If you can draw a single, straight tangent line touching the curve at a given point without any ambiguity or sharp turns, then the function is differentiable at that point. In contrast, functions with 'corners' or 'cusps', just like the absolute value function at its vertex, do not have a unique tangent and are therefore not differentiable at those points. Differentiability is a stronger condition than continuity; a function must be continuous to be differentiable, but not all continuous functions are differentiable.

Absolute Value Function

The absolute value function, often denoted as |x|, is a piecewise function defined as follows: it is equal to x if x is positive or zero, and -x if x is negative. Visually, it produces a 'V' shaped graph which is symmetric across the y-axis. This sharp point where the graph changes direction is the 'cusp' of the function.

The important thing to note with absolute value functions is the presence of these corner or cusp points where the function switches from one piece to the next, creating a point of non-differentiability. In the case of the function |2x + 1|, we calculate the zero point to identify where this cusp occurs. This aspect of the absolute value function is particularly relevant when discussing the applicability of certain calculus theorems such as the Mean Value Theorem.

Continuity of Functions

A function is considered continuous on an interval if you can draw its graph without lifting your pen. Formally, a function is continuous at a point if three conditions are met: the function is defined at the point, the limit of the function as it approaches that point exists, and the limit equals the function's value at that point. Continuity ensures that there are no gaps, jumps, or holes in the function within the interval in question.

Continuity is often a prerequisite for applying various calculus theorems. Although an absolute value function is continuous everywhere since it does not have any asymptotes, jumps, or holes, the corner points pose a problem for differentiability, as seen in exercises involving the Mean Value Theorem.

Calculus Theorems

There are several key theorems in calculus that underpin the relationships between functions, derivatives, and integrals. The Mean Value Theorem is one such theorem with high significance. It states that if a function is continuous on a closed interval \[a, b\] and differentiable on the open interval (a, b), then there is at least one point c in the open interval where the derivative of the function equals the average rate of change over (a, b). Symbolically, this is represented as \( f'(c) = \frac{f(b) - f(a)}{b - a} \).

The applicability of this theorem depends on both the continuity and differentiability of the function in question over the specified interval. When a function, like the absolute value function, fails to be differentiable throughout the interval due to corners, the Mean Value Theorem cannot be applied, as seen in the original exercise. These calculus theorems are integral tools for analyzing and understanding the behaviors of different functions in various contexts.