Problem 41 - Calculus Early Transcendentals

Question

If \(f\) is increasing on an interval \([0, b),\) then it follows from Definition 4.1 .1 that \(f(0)< f(x)\) for each \(x\) in the interval \((0, b) .\) Use this result in these exercises. Show that \(\sqrt[3]{1+x}<1+\frac{1}{3} x\) if \(x>0,\) and confirm the in equality with a graphing utility. [Hint: Show that the function \(\left.f(x)=1+\frac{1}{3} x-\sqrt[3]{1+x} \text { is increasing on }[0,+\infty) .\right]\)

Step-by-Step Solution

Answer

The inequality \( \sqrt[3]{1+x} < 1 + \frac{1}{3}x \) holds for \( x > 0 \).

Step 1: Define the Function

First, we define the function \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \). Our goal is to show this function is increasing for \( x > 0 \).

Step 2: Find the Derivative

To determine if \( f(x) \) is increasing, we need to find its derivative. Derivatives show the rate of change. We compute \( f'(x) = \frac{d}{dx} \left( 1 + \frac{1}{3}x - \sqrt[3]{1+x} \right) \). By differentiating, we get \( f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \).

Step 3: Analyze the Derivative

Analyze \( f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \). We need to show that this derivative is positive for \( x > 0 \) to prove that \( f(x) \) is increasing:\[ f'(x) = \frac{1}{3} \left( 1 - (1+x)^{-2/3} \right). \] For \( x > 0, \ 1+x > 1 \), thus \( (1+x)^{-2/3} < 1 \), which means \( f'(x) > 0 \).

Step 4: Conclude the Inequality

Since \( f'(x) > 0 \) for \( x > 0 \), \( f(x) \) is increasing over this interval. From the property of increasing functions, \( f(0) < f(x) \) for \( x > 0 \). Therefore, \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} > 0 \) implies \( \sqrt[3]{1+x} < 1 + \frac{1}{3}x \).

Step 5: Confirm with a Graph

To visually confirm, you can use a graphing utility to plot the functions \( y = \sqrt[3]{1+x} \) and \( y = 1 + \frac{1}{3}x \). Observe that \( y = 1 + \frac{1}{3}x \) lies above \( y = \sqrt[3]{1+x} \) for \( x > 0 \). This confirms our analytical solution.

Key Concepts

Increasing Functions

When we say a function is increasing, it simply means that as you move to the right on the graph, the function keeps getting larger. Mathematically, a function \( f(x) \) is increasing on an interval if for any two numbers \( a \) and \( b \) in that interval with \( a < b \), it follows that \( f(a) < f(b) \). In this context, increasing doesn't mean the function always goes upwards without ever dipping; it just means it's consistently getting larger over the interval in question.
Determining that a function is increasing is important because it helps us understand the behavior of the function on that interval. If we know the function \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \) is increasing for \( x > 0 \), we can say that \( f(0) < f(x) \) for any \( x > 0 \). This increasing nature helps us establish inequalities such as \( \sqrt[3]{1+x} < 1 + \frac{1}{3}x \) in the problem we're examining.

Derivative Analysis

Finding a derivative helps us analyze how a function behaves. The derivative of a function gives us the rate at which it changes. If the derivative is positive over an interval, the function is increasing in that interval. For the function \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \), we determined its derivative:

• \( f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \).

To prove the function is increasing, we need to show that \( f'(x) > 0 \) for \( x > 0 \). The expression
inside the derivative can be simplified to \[ \frac{1}{3} \left( 1 - (1+x)^{-2/3} \right) \]. Since \( (1+x) > 1 \) for \( x > 0 \), it ensures \( (1+x)^{-2/3} < 1 \). Therefore, \( f'(x) > 0 \).
The positive derivative confirms that the original function \( f(x) \) is indeed increasing over the interval \( x > 0 \).

Graphical Confirmation

Visual confirmation through graphing can solidify our understanding of inequalities. When you graph the two functions \( y = \sqrt[3]{1+x} \) and \( y = 1 + \frac{1}{3}x \) on the same set of axes, you can directly observe the behaviors of these functions for \( x > 0 \).
On the graph, \( y = 1 + \frac{1}{3}x \) should always lie above \( y = \sqrt[3]{1+x} \), confirming the inequality discovered analytically. This visual aid can clarify instances where analytical proof alone might seem abstract, by giving intuition through tangible evidence.
Graphs provide a practical way to verify solutions, offering a second layer of understanding that pairs nicely with the more abstract calculus results.

Real-Valued Functions

A real-valued function is one that has real numbers as its range. This means for every real input \( x \), the function outputs a real number. In calculus, most functions we analyze, including \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \), are real-valued.
Understanding real-valued functions is crucial as it constitutes the basis of calculus and analysis. They allow us to model real-world phenomena where inputs and outputs need to be real numbers, such as calculating distances, areas, and speeds.
Real-valued functions are generally more intuitive to work with and are the primary objects of study in many areas of mathematics and the sciences.