Problem 29 - Essential Calculus

Question

Find the critical numbers of \(f\) (if any). Find the open intervals on which the function is increasing or decreasing and locate all relative extrema. Use a graphing utility to confirm your results. $$ f(x)=(x-1)^{2 / 3} $$

Step-by-Step Solution

Answer

The critical number of the function \(f(x) = (x-1)^{2 / 3}\) is \(x = 1\). The function is decreasing on the interval \((-∞, 1)\) and increasing on the interval \((1,∞)\). The function has a relative minimum at \(x = 1\).

Step 1: Derivation of the function

Start by calculating the derivative of \(f(x) = (x-1)^{2 / 3}\) using the chain rule. \[f'(x) = \frac{2}{3}(x-1)^{-1 / 3}\].

Step 2: Find the critical numbers

The critical numbers of a function are where the derivative is zero or undefined. From the derivative \(f'(x) = \frac{2}{3}(x-1)^{-1 / 3}\), it's clear that the derivative is undefined at \(x = 1\). There are no values of x that make the derivative zero. Thus, the only critical number is \(x = 1\).

Step 3: Find the intervals and relative extrema

To determine where \(f\) is increasing or decreasing, test a number from each interval in the first derivative. The intervals are \((-∞, 1)\) and \((1, ∞)\). Choose \(x = 0\) for the first interval and \(x = 2\) for the second interval. Substituting these into the derivative \(f'(x)\), we get \(f'(0) = -2/3 < 0\) and \(f'(2) = 2/3 > 0\). Therefore, \(f\) is decreasing on the interval \((-∞, 1)\) and increasing on the interval \((1,∞)\). Given that the function changes from decreasing to increasing at \(x = 1\), \(x = 1\) is a relative minimum.

Step 4: Confirmation with a graphing tool

Plotting \(f(x) = (x-1)^{2 / 3}\) using a graphing utility confirms the conclusions. It shows that the function decreases from \(-∞\) to \(1\), and then increases from \(1\) to \(∞\). Also, the relative minimum at \(x = 1\) is clearly visible.

Key Concepts

Chain Rule

The chain rule is an essential calculus technique used to find the derivative of composite functions. Essentially, when you have a function nested inside another, the chain rule allows you to differentiate it easily. It is based on the idea that if you have two functions, say, u(x) and v(u), and you want to find the derivative of the composite function f(x) = v(u(x)), you can do so by multiplying the derivative of the outer function v(u) with respect to u by the derivative of the inner function u(x) with respect to x. The formula for the chain rule is:
\[ \frac{d}{dx}f(x) = \frac{dv}{du} \cdot \frac{du}{dx} \]
In our exercise, we applied the chain rule to the function f(x) = (x-1)^{2/3}. We treated (x-1) as the inner function u(x) and u^{2/3} as the outer function v(u). By applying the chain rule, we found the derivative f'(x) of the function.

Relative Extrema

Relative extrema refer to the points on a function where the function value is at a peak or trough relative to its nearby points. These include relative maximums, where the function has a peak, and relative minimums, where it has a trough. Finding relative extrema involves looking for points on the function where the derivative either changes sign or fails to exist. Such points could indicate relative extrema, but additional tests, such as the First or Second Derivative Test, are usually employed to confirm their nature.
For our given function f(x) = (x-1)^{2/3}, we used the derivative to find that the function had a relative minimum at x=1. This is because the derivative changes from negative to positive as x increases through 1, indicating a transition from decreasing to increasing values of the function.

Increasing and Decreasing Intervals

Identifying increasing and decreasing intervals of a function is a process that adds context to the shape and behavior of the function. Intervals where the function is increasing are where the output of the function grows as the input gets larger. Conversely, intervals where the function is decreasing are where the output diminishes as the input increases. You can determine these intervals by analyzing the sign of the first derivative: a positive derivative indicates an increasing interval, while a negative derivative implies a decreasing interval.
In the exercise, by substituting a number from the intervals (-∞, 1) and (1, ∞) into the derivative f'(x), we observed that the function f(x) decreased and then increased. Specifically, the function was decreasing for x < 1 and increasing for x > 1. These intervals are critical for understanding the overall trend of the function.

Derivative of a Function

The derivative of a function at a point provides the rate at which the function's output value is changing at that point. More formally, if \( y = f(x) \), the derivative of \( f \) at \( x \) is the limit of the average rate of change of \( f \) as the change in \( x \) approaches zero. The notation for the derivative of a function \( f \) is often \( f'(x) \) or \( \frac{df}{dx} \). Derivatives are foundational to calculus, as they are used to examine the instantaneous rate of change and they play a crucial role in optimization and motion analysis.
In the context of our exercise, the derivative of the function \( f(x) = (x-1)^{2/3} \) was found to be \( f'(x) = \frac{2}{3}(x-1)^{-1/3} \) using the chain rule. This derivative was then used to analyze the function's behavior, such as its critical numbers and intervals of increase and decrease.