Problem 22 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, find the critical numbers and the open intervals on which the function is increasing or decreasing. Then use a graphing utility to graph the function. $$ g(x)=(x-1)^{2 / 3} $$

Step-by-Step Solution

Answer

The critical number is \(x = 1\). The function is increasing on \((-∞,1)\) and \((1,∞)\).

Step 1: Find the Derivative

To find the critical numbers, differentiate \(g(x)=(x-1)^{2 / 3}\) with respect to \(x\), it gives \(g'(x) = \frac{2(x-1)^{-1/3}}{3}\)

Step 2: Determine Critical Numbers

Set \(g'(x) = 0\) and solve for \(x\). However, there are no solutions since \((x-1)^{-1/3}\) is never zero. Next, find for what \(x\) values \(g'(x)\) is undefined. This happens when the denominator of \((x-1)^{-1/3}\) is zero, which gives \(x = 1\). Thus, the critical number is \(1\).

Step 3: Find Increasing and Decreasing Intervals

To find where the function is increasing or decreasing, one can test the sign of \(g'(x)\) in intervals determined by the critical numbers. For \(x<1\), \(g'(x) > 0\), so the function is increasing. For \(x>1\), \(g'(x)>0\), so the function is also increasing. Therefore, the function is increasing on \((-∞,1)\) and \((1,∞)\). The function is neither increasing nor decreasing at \(x = 1\).

Key Concepts

Increasing and Decreasing Intervals

When working with functions, it’s crucial to understand where they are increasing or decreasing as this tells us a lot about the behavior of the function. For a given function, the intervals of increase or decrease are determined by its derivative.Once we have the derivative, we check the sign of this derivative over various intervals. If the derivative is positive over an interval, the function is increasing there. Conversely, if the derivative is negative, the function is decreasing.In the case of our exercise with the function \(g(x) = (x-1)^{2/3}\), we found that the derivative \(g'(x)\) is always positive except at \(x=1\) where it's undefined. This indicates that the function is increasing on both \((-∞,1)\) and \((1,∞)\). It’s important to understand these intervals to fully grasp the behavior and shape of the function's graph.

Derivative

The derivative of a function gives us valuable information about the slope or rate of change of the function at any given point. In our exercise, we differentiated \(g(x) = (x-1)^{2/3}\) to find \(g'(x)\).The process of differentiation involves using rules like the power rule and chain rule to find the derivative of functions that might not be straightforward. The derivative we obtained was \(g'(x) = \frac{2(x-1)^{-1/3}}{3}\).This result is crucial because it helps us identify the critical numbers for the function where the slope is either zero or undefined, especially when determining the increasing and decreasing intervals. In this case, there was no point where \(g'(x) = 0\), meaning there were no slopes indicating a horizontal tangent. However, \(g'(x)\) was undefined at \(x = 1\), making \(x = 1\) a critical number.

Graphing Utility

A graphing utility is a helpful tool that aids in visualizing the behavior of functions across their domains. It provides an easy way to sketch and analyze the function's graph based on its equation and derivatives.In the problem at hand, using a graphing utility allowed us to visually confirm our analytical findings. It helped us see that the function \(g(x) = (x-1)^{2/3}\) is indeed increasing on both \((-∞,1)\) and \((1,∞)\), as predicted by the sign of the derivative.Graphing utilities are incredibly beneficial for looking at functions that might be complex to visualize manually. They allow students to develop deeper insights into how changes in the equation affect the overall graph, making understanding concepts like increasing and decreasing intervals much clearer.