Problem 11 - Calculus Early Transcendentals

Question

Determine whether Rolle's Theorem applies to the following functions on the given interval. If so, find the point(s) that are guaranteed to exist by Rolle's Theorem. $$f(x)=1-x^{2 / 3} ;[-1,1]$$

Step-by-Step Solution

Answer

Answer: Rolle's Theorem does not apply to the function $$f(x) = 1-x^{2/3}$$ on the interval $$[-1,1]$$ because the function isn't differentiable at $$x=0$$, which is within the open interval $$(-1, 1)$$. While other conditions are satisfied, it is required for the function to be differentiable at all points within the open interval for Rolle's theorem to apply. This means that we cannot guarantee the existence of any point in this interval with a horizontal tangent.

Step 1: Verify if the function is continuous on the interval [-1,1]

The function $$f(x) = 1-x^{2/3}$$ is the difference of two continuous functions ($$1$$ and $$x^{2/3}$$) and thus is continuous on its entire domain. The domain of $$f(x)$$ is all real numbers, so we can conclude that $$f(x)$$ is continuous on the closed interval $$[-1,1]$$.

Step 2: Check if the function is differentiable on the open interval (-1,1)

We will find the first derivative of $$f(x)$$. The derivative of the function will give us information about when its tangent is horizontal, which is what Rolle's Theorem guarantees. $$f'(x) = -\frac{2}{3}x^{-1/3}$$ The derivative is defined for all real values of $$x$$ except $$x=0$$ because of the negative exponent. Thus, $$f(x)$$ is differentiable on the open interval $$(-1, 1)$$ except at the point $$x=0$$.

Step 3: Check if f(a) = f(b)

Rolle's Theorem requires that $$f(a) = f(b)$$. In this case, $$a = -1$$ and $$b = 1$$. Let's evaluate the function at these points: $$f(-1) = 1-\sqrt[3]{(-1)^{2}} = 1-1 = 0$$ $$f(1) = 1-\sqrt[3]{(1)^2} = 1-1 = 0$$ Since $$f(a) = f(b)$$, this condition is satisfied.

Step 4: Find the point(s) guaranteed by Rolle's Theorem

Rolle's Theorem guarantees there exists at least one point $$c$$ in the open interval $$(-1, 1)$$ for which $$f'(c) = 0$$. However, because the function isn't differentiable at $$x=0$$, we cannot apply Rolle's theorem to this problem. While the other conditions were met, it is necessary for the function to be differentiable at all points in the open interval of $$(-1,1)$$. In summary, Rolle's Theorem does not apply to the function $$f(x) = 1-x^{2/3}$$ on the interval $$[-1,1]$$, and thus we cannot guarantee the existence of any point satisfying the theorem's conditions.

Key Concepts

Continuous Functions

Understanding the concept of continuity in functions is crucial for grasping many theorems in calculus, including Rolle's Theorem. A function is said to be continuous at a point if the limit as you approach that point equals the function's value at that point. In simpler terms, you can draw the graph of a continuous function without lifting your pencil.

For a function to be continuous on a closed interval like \[[-1,1]\], it must be continuous at every point within that interval, and it must be defined at the end points. In the given exercise, the function \(f(x) = 1 - x^{2/3}\) is a difference of two continuous functions, which makes \(f(x)\) itself continuous over the entire domain of real numbers, including \[[-1,1]\].

Differentiability

Differentiability is another cornerstone concept in calculus. If a function is differentiable at a point, it means that it has a well-defined tangent at that point, or in other words, a unique slope can be determined. A function that's differentiable on an interval allows the calculation of its derivative anywhere within that interval, and the graph does not have any sharp turns or cusps.

For a function to be applicable to Rolle's Theorem, it must be differentiable on the open interval between the two points in question. The exercise presents a snag: \(f(x)\) is not differentiable at \(x=0\) because its derivative, \(f'(x) = -\frac{2}{3}x^{-1/3}\), is undefined here. Although \(f(x)\) is differentiable everywhere else within the interval, this single point of non-differentiability precludes the use of Rolle's Theorem.

First Derivative

The first derivative of a function represents the slope of the tangent line at any point of its graph, and essentially gives us the rate at which the function's value is changing at that point. It is often denoted as \(f'(x)\) or \(\frac{df}{dx}\).

In practical terms, the first derivative can inform us about the behavior of the function: where it's increasing, decreasing, and where it achieves local maxima or minima. In the context of our exercise, by finding the first derivative \(f'(x)\), we attempted to identify points at which the slope of the tangent would equal zero—these are points where the graph of the function might have a horizontal tangent line. Unfortunately, the non-existence of the derivative at \(x=0\) is enough to make Rolle's Theorem inapplicable.

Closed Interval

Lastly, let's identify what it means for an interval to be a closed interval. A closed interval, denoted by square brackets, includes its boundary points. For the interval \[a,b\], it encompasses all real numbers \(x\) such that \(a \leq x \leq b\). The continuity and behavior of a function at these boundary points are often subjects of examination in various theorems and analytical methods in calculus.

In our function \(f(x)\), not only does the closed interval \[[-1,1]\] specify the region we are interested in, but it also sets the conditions for Rolle's Theorem along with the requirement that \(f(a) = f(b)\), which was satisfied when \(f(-1) = f(1)\). The closed interval's end points are crucial in checking the applicability of the theorem even though it demands differentiability only in the interval's interior, that is, the open interval \((-1,1)\).