Problem 31 - Essential Calculus

Question

In Exercises 31-36, 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 \(f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\). $$ f(x)=x^{2 / 3},[0,1] $$

Step-by-Step Solution

Answer

The Mean Value Theorem cannot be applied to the function \(f(x)=x^{2 / 3}\) on the interval [0,1]. Therefore, there are no values of \(c\) that satisfy the theorem.

Step 1: Check Continuity and Differentiability

The first step is to ensure that function is continuous on [0, 1] and differentiable on (0, 1). The function \(f(x)=x^{2 / 3}\) is continuous on [0, 1] but it's not differentiable at x= 0 and as per Mean Value Theorem, function should be differentiable on the open interval. Hence, the Mean Value Theorem can't be applied.

Step 2: No Solution

Since the function doesn't fulfill the requirements of the Mean Value Theorem, there is no point in looking for a value of \(c\) in the open interval (0,1), as it doesn't exist.

Key Concepts

Continuity and Differentiability

Understanding the concept of continuity and differentiability is fundamental when dealing with calculus problems, especially when applying the Mean Value Theorem. A function is considered continuous at a point if there is no sudden break or jump at that point. To visualize this, imagine drawing the graph of the function without lifting your pencil off the paper.

More formally, a function f is continuous at a point c if the limit as x approaches c of f(x) is equal to f(c). For an interval, if the function is continuous at every point within that interval, we say the function is continuous on that interval.

On the other hand, differentiability is related to the existence of a derivative. If a function is differentiable at a point, it means that it has a specific slope at that point, which is determined by its derivative. A function can be continuous without being differentiable, as in the case of the function f(x) = x^{2/3} at x=0. Here, while it doesn't have a sudden break, the function isn't smooth; it has a sharp turn or cusp, which prevents us from defining a slope at x=0.

Closed Interval

A closed interval is a range of numbers that includes its endpoints. In the notation \[a, b\], the brackets denote that both a and b are part of the interval. Closed intervals are particularly noteworthy in calculus because many theorems, like the Mean Value Theorem, require that the functions under consideration are continuous on a closed interval.

This inclusion of endpoints ensures that there's a 'beginning' and an 'end' value to the range under inspection, and it is essential because certain properties of functions can be guaranteed on this interval. For instance, by the Extreme Value Theorem, if a function is continuous on a closed interval, then it must attain a maximum and minimum value on that interval, which is not necessarily true on an open interval.

Open Interval

Contrastingly, an open interval, denoted as \(a, b\), is a set of real numbers that lies between a and b but does not include the endpoints a and b themselves. In the context of the Mean Value Theorem, the function in question must be differentiable on the open interval—that is, the derivative must exist at all points between a and b.

The reason for requiring differentiability on an open interval is to avoid the complications that can arise at the endpoints of a closed interval, like sudden changes in the direction of the graph or vertical tangents. If the function fails to be differentiable even at a single point within this open interval, as happened with our function f(x) = x^{2/3} at x=0, then the Mean Value Theorem cannot be applied.