Problem 67 - Thomas Calculus in SI Units

Question

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

a) No, \(f'(2)\) does not exist. b) Local minimum at \(x=2\). c) No contradiction. d) Similar: \(f'(a)\) undefined, min at \(x=a\).

Step 1: Understanding the Function and Existence of Derivative

The function given is \( f(x) = (x-2)^{2/3} \). To determine whether \( f'(2) \) exists, we need to check the derivative at \( x=2 \). This involves differentiating \( f(x) \) and analyzing the behavior at \( x=2 \). Function \( f(x) \) can be rewritten as \( y = u^{2/3} \) where \( u = x-2 \). The derivative is \( f'(x) = \frac{d}{dx} (x-2)^{2/3} = \frac{2}{3}(x-2)^{-1/3} \). As \( x \to 2 \), \( (x-2)^{-1/3} \to \infty \), therefore \( f'(2) \) does not exist.

Step 2: Determining Local Extreme Values

To find local extreme values, analyze the first derivative \( f'(x) = \frac{2}{3}(x-2)^{-1/3} \). The derivative is undefined at \( x = 2 \) suggesting a potential extreme point. Check values around \( x = 2 \): \( f'(x) > 0 \) for \( x > 2 \) and \( f'(x) < 0 \) for \( x < 2 \), indicating a local minimum at \( x=2 \). Thus, the only local extreme value is a minimum at \( x = 2 \).

Step 3: Analyzing the Extreme Value Theorem

The Extreme Value Theorem (EVT) states that a continuous function on a closed interval has a maximum and minimum. \( f(x) = (x-2)^{2/3} \) is continuous on any chosen interval around \( x=2 \), but since \( f(x) \) is not defined over any closed interval around \( x = 2 \), the EVT does not apply. Thus, the result does not contradict EVT.

Step 4: Repeating Steps for General Function

For the function \( f(x) = (x-a)^{2/3} \), following similar steps as before: \( f'(x) = \frac{2}{3}(x-a)^{-1/3} \) which is undefined at \( x=a \). Analyzing \( f'(x) \) shows \( f'(x) > 0 \) when \( x > a \) and \( f'(x) < 0 \) when \( x < a \), indicating a local minimum at \( x=a \). Therefore, results are consistent.

Key Concepts

Derivative

In calculus, a derivative measures how a function's output changes as its input changes. For the function \( f(x) = (x-2)^{2/3} \), when we try to find the derivative \( f'(2) \), we check the behavior of the derivative formula itself. Here, \( f'(x) = \frac{2}{3}(x-2)^{-1/3} \).

As you approach \( x = 2 \), the term \((x-2)^{-1/3}\) tends towards infinity, meaning the slope becomes extremely steep. Because it blasts off to infinity, \( f'(2) \) does not exist. Thus, while you can compute derivatives at most points on a curve, at some special points, like sharp edges or cusps, derivatives can become undefined.

Local Extreme Value

Local extreme values are points where a function reaches a peak (maximum) or a valley (minimum) locally—meaning within a small neighborhood. To identify these points, you'd typically use the first derivative test.

For \( f(x) = (x-2)^{2/3} \), by observing \( f'(x) = \frac{2}{3}(x-2)^{-1/3} \), it's clear that \( x=2 \) is a crucial spot because the derivative is undefined. However, looking around this point:

• For \( x > 2 \), \( f'(x) > 0 \) which indicates the function is increasing.
• For \( x < 2 \), \( f'(x) < 0 \) suggesting the function is decreasing.

Hence, \( x=2 \) marks a local minimum since the derivative changes sign from negative to positive as it passes through \( x=2 \), effectively making it a 'bottom' point nearby.

Extreme Value Theorem

The Extreme Value Theorem (EVT) is like a guarantee that continuous functions on a closed interval will reach at least one highest and one lowest point.

For our function \( f(x) = (x-2)^{2/3} \), although it is continuous, we are dealing with an open interval around \( x=2 \). Since \( f(x) \) isn't defined on a fully closed interval that includes \( x=2 \), we don't get a playground for EVT to operate.

• The function doesn’t reach any closed boundary on the interval, making EVT non-applicable.
• Thus, the absence of contradicting results simply reflects on its domain and EVT's conditions not matching the scenario’s demands.

Continuity

Continuity of a function means you can draw its graph without lifting your pencil. It suggests smoothness, with no breaks or holes.

For the function \( f(x) = (x-2)^{2/3} \), the function is continuous at every point—even at \( x=2 \), where the derivative is undefined. Here’s why:

• Despite the steepness at \( x=2 \) caused by \( (x-2)^{-1/3} \) shooting towards infinity, the function itself is well-defined around and at \( x=2 \).
• The graph won't have jumps or interruptions, ensuring its continuity across its domain—except for the sharp edge at \( x=2 \).

The trick here is grasping that while derivatives rely on behavior, continuity is purely about uninterrupted existence.