Problem 26 - Calculus for Biology and Medicine

Question

Show algebraically that if \(n \geq m\), $$ x^{n} \leq x^{m} \quad \text { for } 0 \leq x \leq 1 $$ and $$ x^{n} \geq x^{m} \quad \text { for } x \geq 1 $$

Step-by-Step Solution

Answer

The conditions are satisfied; \( x^n \leq x^m \) for \( 0 \leq x \leq 1 \) and \( x^n \geq x^m \) for \( x \geq 1 \).

Step 1: Understand the Given Conditions

You are given that \( n \geq m \). You need to show two conditions for \( x \):1. \( x^n \leq x^m \) when \( 0 \leq x \leq 1 \).2. \( x^n \geq x^m \) when \( x \geq 1 \).

Step 2: Prove for \( 0 \leq x \leq 1 \)

Consider \( x^n \leq x^m \). Factor out the common term \( x^m \), giving \((x^{n-m}) \leq 1 \) since \( n-m \geq 0 \). As \( 0 \leq x \leq 1 \), raising \( x \) to a higher power results in a smaller or equal value. Therefore, \( x^{n-m} \leq x^{0} = 1 \), making \( x^n \leq x^m \) true for \( 0 \leq x \leq 1 \).

Step 3: Prove for \( x \geq 1 \)

Consider \( x^n \geq x^m \). Similarly, factor out the common term \( x^m \), giving \( (x^{n-m}) \geq 1 \) because \( n-m \geq 0 \). Since \( x \geq 1 \), raising \( x \) to a higher power results in a larger or equal value. Therefore, \( x^{n-m} \geq 1 \), making \( x^n \geq x^m \) true for \( x \geq 1 \).

Key Concepts

Algebraic Proof

Algebraic proof is a fundamental method in mathematics used to demonstrate the truth of a statement using symbolic manipulation and logical reasoning. In our exercise, we need to show that certain inequalities hold true for different intervals of the variable \( x \), given that \( n \geq m \). The beauty of algebraic proof lies in its ability to provide a solid and universal conclusion without relying on specific numerical examples. Let's see how it applies here.
The exercise requires us to prove two conditions:

• For \( 0 \leq x \leq 1 \): if \( n \geq m \), then \( x^n \leq x^m \).
• For \( x \geq 1 \): if \( n \geq m \), then \( x^n \geq x^m \).

To prove these situations algebraically, we derived inequalities by considering the nature of powers and how they change with varying \( x \). By strategically factoring terms and leveraging known properties, we constructed each proof step-by-step to solidly establish these conditions. This approach highlights how algebra provides a systematic way to tackle complex questions.

Exponents and Powers

Exponents and powers are crucial in this problem because they define the relationship between different expressions of \( x \). They help us determine whether one expression is larger or smaller than the other under given conditions or intervals.
Specifically, in the inequality \( x^n \leq x^m \) for \( 0 \leq x \leq 1 \), we see how increasing powers of \( x \) actually result in smaller values. This happens because when \( x \) is a fraction (between 0 and 1), raising it to a higher power contracts its magnitude further. Imagine as squeezing a number closer to zero the more you multiply it by itself. On the flip side, for \( x \geq 1 \), as in \( x^n \geq x^m \), increasing powers of \( x \) start expanding its size; the greater \( x \), the more it balloons as powers rise.

• For \(0 \leq x \leq 1\), higher exponents decrease the value.
• For \(x \geq 1\), higher exponents increase the value.

These observations concerning how exponents behave depending on whether \( x \) is less than or greater than 1 form the core part of solving the exercise.

Conditions on Intervals

Understanding the conditions on intervals is key to solving this exercise because it dictates how \( x \) behaves in each case, and subsequently how our exponents act. The problem specifies two distinct intervals for \( x \):

• Interval 1: \( 0 \leq x \leq 1 \)
• Interval 2: \( x \geq 1 \)

In Interval 1, \( x \'s \) behavior changes because squashing a fraction smaller than one through exponentiation results in a value that is always less than or equal to itself if given higher exponents. It's like what happens when you try to squash a soft sponge; it can shrink but not inflate beyond its shape unless more material is added.
On the other hand, in Interval 2, where \( x \geq 1 \), every multiplication by \( x \) amplifies its magnitude, making it easier for powers of \( x \) to surpass earlier ones if the exponents themselves are greater. This is equivalent to stacking blocks; with each new block making the tower taller.
Conditions on these intervals guide us because they capitalize on the inherent behavior of \( x \) under exponentiation, reinforcing why our inequalities are true.