Problem 53 - Thomas Calculus

Question

In Exercises 53 and \(54,\) give reasons for your answers. Let \(f(x)=(x-2)^{2 / 3}\) a. Does \(f^{\prime}(2)\) exist? b. Show that the only local extreme value of \(f\) occurs at \(x=2\) . c. Does the result in part (b) contradict the Extreme Value Theorem? d. Repeat parts (a) and (b) for \(f(x)=(x-a)^{2 / 3}\) , replacing 2 by a.

Step-by-Step Solution

Answer

f'(2) and f'(a) do not exist, local extremes at x=2 and x=a, respectively, do not contradict the theorem.

Step 1: Assess Derivative Existence at x=2

To determine if \( f'(2) \) exists, we first find the derivative of \( f(x) = (x-2)^{2/3} \). The function is \( f(x) = u^{2/3} \) with \( u = x-2 \). The derivative of \( u^{2/3} \) is \( \frac{2}{3}u^{-1/3} \), so \( f'(x) = \frac{2}{3}(x-2)^{-1/3} \). At \( x = 2 \), the term \( (x-2)^{-1/3} \) becomes undefined. Hence, \( f'(2) \) does not exist.

Step 2: Identify Local Extreme Value at x=2

To find local extreme values, usually, we check where \( f'(x) = 0 \). Since \( f'(x) \) doesn't exist at \( x=2 \), and \( (x-2)^{2/3} \) has a cusp (critical point) at \( x=2 \), it can be considered a local extreme. The function is increasing on both sides of \( x=2 \), confirming that the only local extreme occurs at \( x=2 \).

Step 3: Check for Contradiction with Extreme Value Theorem

The Extreme Value Theorem asserts that a continuous function on a closed interval must have a maximum and minimum. However, \( f(x) = (x-2)^{2/3} \) isn't bounded on a closed interval traditionally, so the theorem isn't applied directly here. As a result, there is no contradiction since the function itself on any typical closed interval including such a cusp doesn't fully apply the theorem.

Step 4: Generalize to f(x) = (x-a)^{2/3}

Repeating steps for \( f(x) = (x-a)^{2/3} \), we see \( f'(x) = \frac{2}{3}(x-a)^{-1/3} \), which is undefined at \( x=a \). Hence, \( f'(a) \) does not exist, just like \( f'(2) \). The local extreme value occurs at \( x=a \), as the reasoning doesn't change due to translation from 2 to \( a \).

Key Concepts

Extreme Value Theorem

The Extreme Value Theorem (EVT) is an essential principle in calculus. It states that if a function is continuous on a closed interval \[a, b\], then the function must have both a maximum and minimum value on that interval. However, it's crucial to understand that EVT only applies when a function is truly continuous on a bounded, closed interval.
This theorem doesn't apply directly in the case of functions like \((x-2)^{2/3}\). Even though it is continuous, for EVT to hold, the domain should be a closed interval, and the function needs to be differentiable at all but a finite number of points within the interval. A function like \((x-2)^{2/3}\) defined on an open interval \(-\infty, \infty\) can still give you points where EVT might not apply, particularly when such points are critical points that don't yield a typical extremum behavior.

Derivative Existence

To determine if the derivative of a function exists at a particular point, you need to look at the behavior around that point. If we take the function \(f(x) = (x-2)^{2/3}\), we first differentiate it to get \(f'(x) = \frac{2}{3}(x-2)^{-1/3}\).
At \(x = 2\), evaluating \(f'(2)\) gives an undefined expression since \(x-2\) becomes zero, leading to an undefined division in the expression \(\frac{2}{3}(x-2)^{-1/3}\).
Whenever the derivative involves dividing by zero or results in an infinite value at any point, the derivative does not exist at that point. This concept is crucial as it indicates where the function might have a cusp or vertical tangent, signifying a critical point.

Local Extreme Values

Local extreme values of a function are points where the function reaches a local maximum or minimum. These occur where the function's derivative equals zero or does not exist. In tackling the function \(f(x) = (x-2)^{2/3}\), we examine points where \(f'(x)\) is zero or undefined to identify local extremes.
Since \(f'(x) = \frac{2}{3}(x-2)^{-1/3}\) is undefined at \(x = 2\), we inspect \(x=2\) for any local extreme value. Here, the behavior of the function \(f(x)\) suggests a cusp, a type of extreme point. The function increases on both sides of the cusp, confirming that the local extreme value occurs at \(x = 2\).
This is different from typical smooth transitions like a gentle peak or valley because of the undefined derivative, yet it is crucial to recognize such points in the study of calculus to fully grasp the nature of local extremas.

Critical Points

Critical points of a function are places where the function's derivative is either zero or doesn't exist. These points often indicate potential locations for local maximums or minimums. Examining the function \(f(x) = (x-2)^{2/3}\), the critical point arises at \(x=2\), where the derivative \(f'(x) = \frac{2}{3}(x-2)^{-1/3}\) does not exist.
This lack of a derivative can signal features like a sharp turn or a cusp, which is evident in this case. Cusp points, specific types of critical points, don't provide clear maxima or minima in the traditional sense but are still essential for understanding a function's overall behavior.
In general, identifying critical points helps to analyze and predict the behavior of functions across intervals, enabling more profound insights into their structures and dynamics.