Problem 35 - Calculus A New Horizon

Question

If \(f\) is increasing on an interval \([0, b),\) then it follows from Definition 5.1.1 that \(f(0)0,\) and confirm the inequality 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

We are given the function \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \). We need to show that \( f(x) \) is increasing on the interval \([0, +\infty)\).

Step 2: Find the Derivative of the Function

To determine if \( f(x) \) is increasing, we calculate its derivative. The derivative \( f'(x) \) is found using the power rule and chain rule and it's given by:\[ f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \]

Step 3: Analyze the Derivative

Check whether \( f'(x) \) is positive on \([0, +\infty)\). Since \( f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \), observe that \( (1+x)^{-2/3} \) is always non-negative and less than or equal to one, meaning \( \frac{1}{3} > \frac{1}{3}(1+x)^{-2/3} \) for \( x > 0 \). Thus, \( f'(x) > 0 \) on \([0, +\infty)\).

Step 4: Conclusion from the Analysis

Since \( f'(x) > 0 \) for \( x > 0 \), the function \( f(x) \) is increasing on \( [0, +\infty) \), implying \( f(0) < f(x) \) for \( x > 0 \). Accordingly, \( 1 + \frac{1}{3}x > \sqrt[3]{1+x} \), proving the required inequality.

Step 5: Confirm with a Graphing Utility

Using a graphing utility, plot \( \sqrt[3]{1+x} \) and \( 1+\frac{1}{3}x \). Notice that the graph of \( \sqrt[3]{1+x} \) lies below the graph of \( 1+\frac{1}{3}x \) for \( x > 0 \), verifying the inequality visually.

Key Concepts

Derivative Analysis

To determine if a function is increasing on a given interval, it is crucial to analyze its derivative. For the function \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \), its derivative gives us insights into whether the function is increasing or decreasing within the interval \([0, +\infty)\). The derivative, found using the power rule and chain rule, is \( f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \).

If the derivative \( f'(x) \) is positive on the interval, it indicates that the function \( f(x) \) is increasing. Analyzing \( f'(x) = \frac{1}{3} - \frac{1}{3}(1+x)^{-2/3} \), we note that \( (1+x)^{-2/3} \) remains non-negative and is always less than or equal to one. Hence, \( \frac{1}{3} > \frac{1}{3}(1+x)^{-2/3} \) for \( x > 0 \).

Therefore, \( f'(x) > 0 \) on \([0, +\infty)\), confirming that the function is indeed increasing in this interval, which supports our inequality \( 1 + \frac{1}{3}x > \sqrt[3]{1+x} \) for \( x > 0 \).

Graphing Utility

A graphing utility can be a valuable tool in visually confirming mathematical conjectures or inequalities. In our case, plotting \( \sqrt[3]{1+x} \) and \( 1+\frac{1}{3}x \) allows us to visually verify the inequality \( \sqrt[3]{1+x} < 1+\frac{1}{3}x \) for \( x > 0 \).

By inputting these equations into a graphing calculator or software, you will observe that the curve for \( \sqrt[3]{1+x} \) consistently remains below the line for \( 1+\frac{1}{3}x \). This visual check reinforces the derivative analysis that the function \( f(x) \) is increasing in the interval \([0, +\infty)\).

A graphing utility doesn't replace the mathematical proof but complements it by providing a clear image of function behavior that can be easier to interpret and understand, especially for those who are more visually oriented.

Inequality Proof

In the context of this exercise, we needed to prove the inequality \( \sqrt[3]{1+x} < 1+\frac{1}{3} x \) for \( x > 0 \). Inequations often require establishing that a related function is increasing or decreasing, which can directly demonstrate one expression's supremacy over another.

Here, we define \( f(x) = 1 + \frac{1}{3}x - \sqrt[3]{1+x} \), and we have shown that \( f(x) \) is increasing for \( x > 0 \) through derivative analysis. If a function is increasing, by definition, any value of the function at a point \( x > 0 \) will be greater than its initial value at \( x = 0 \).

Therefore, since \( f(x) \) is increasing on \([0, +\infty)\), it follows that \( f(0) = 0 \) is less than \( f(x) \) for \( x > 0 \). This proves that \( 1 + \frac{1}{3}x \) is indeed greater than \( \sqrt[3]{1+x} \), effectively proving our inequality.