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Chapter 2: Problem 146

Explain why $$ \sqrt{x}<\sqrt[3]{x} \quad \text { if } \quad 0\sqrt[3]{x} \quad \text { if } \quad x>1 $$ Sketch the graphs of the functions \(\sqrt{x}\) and \(\sqrt[3]{x}\) on the interval [0,4] .

Short Answer

Expert verified
The inequality \(\sqrt{x}<\sqrt[3]{x}\) holds for \(0\sqrt[3]{x}\) holds for \(x>1\) because, for numbers greater than 1, a larger exponent results in a larger value. The graphs of \(\sqrt{x}\) and \(\sqrt[3]{x}\) on the interval [0, 4] confirm this algebraically derived result.

Step by step solution

01

Convert the inequality into a single expression

To compare the behaviors of \(\sqrt{x}\) and \(\sqrt[3]{x}\), we'll convert the inequality into a single expression by taking the difference between the two functions: $$ f(x) = \sqrt[3]{x} - \sqrt{x} $$ Now we'll analyze the behavior of this function for different values of x.
02

Factor the expression

To analyze the function, we'll factor the expression: $$ f(x) = x^{\frac{1}{3}} - x^{\frac{1}{2}} $$ Step 2: Analyze the behavior of the function for different intervals
03

Analyze the interval 0 < x < 1

Let's analyze the function for 0 < x < 1. When x is in this range, both \(\sqrt{x}\) and \(\sqrt[3]{x}\) are positive. Using the properties of positive exponents, we know that if \(0 < a < b\), then \(a^n < a^m\) for \(n > m\). Applying this property to our function, we find: $$ f(x) > 0 \quad \text { if } \quad 0 \sqrt{x}\) for 0 < x < 1.
04

Analyze the interval x > 1

Now let's analyze the function for x > 1. In this range, both \(\sqrt{x}\) and \(\sqrt[3]{x}\) are positive and greater than 1. Because the properties of exponents reverse for numbers greater than 1, we have that if \(a > b > 1\), then \(a^n < a^m\) for \(n < m\). Applying this property to our function, we find: $$ f(x) < 0 \quad \text { if } \quad x>1 $$ This means that \(\sqrt[3]{x} < \sqrt{x}\) for x > 1. Step 3: Sketch the graphs of the functions
05

Plot the functions

To visualize the behavior of these functions, sketch the graphs of \(\sqrt{x}\) and \(\sqrt[3]{x}\) on the same axes, focusing on the interval [0, 4] (which includes the regions 0 < x < 1 and x > 1). You should see that the graph of \(\sqrt{x}\) lies below the graph of \(\sqrt[3]{x}\) for 0 < x < 1 (as we proved algebraically) and above the graph of \(\sqrt[3]{x}\) for x > 1 (also as we proved algebraically). You can use graphing software or graph the functions by hand. In conclusion, we have proven algebraically that \(\sqrt{x}<\sqrt[3]{x}\) for 0 < x < 1 and \(\sqrt{x}>\sqrt[3]{x}\) for x > 1. By sketching the functions on the interval [0, 4], we have also visually confirmed this result.

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