Problem 52 - Calculus for Biology and Medicine

Question

(a) Graph \(y=x^{5 / 2}, x \geq 0\), and \(y=x^{1 / 2}, x \geq 0\), together, in one coordinate system. (b) Show algebraically that \(x^{5 / 2} \leq x^{1 / 2}\) for \(0 \leq x \leq 1\). (Hint: Show that \(x^{1 / 2} / x^{-1 / 2}=x \leq 1\) for \(0

Step-by-Step Solution

Answer

For \( 0 \leq x \leq 1 \), \( x^{5/2} \leq x^{1/2} \); for \( x \geq 1 \), \( x^{5/2} \geq x^{1/2} \).

Step 1: Understanding the Graphs

First, we need to graph the equations \( y = x^{5/2} \) and \( y = x^{1/2} \) on the same coordinate system. Both functions are defined for \( x \geq 0 \). Notice that \( y = x^{5/2} \) is increasing faster than \( y = x^{1/2} \) for larger values of \( x \). Near \( x = 0 \), both functions start at zero. Sketching these graphs helps to visually understand the relationship we will show algebraically next.

Step 2: Show \( x^{5/2} \leq x^{1/2} \) for \( 0 \leq x \leq 1 \)

To show this algebraically, we follow the hint. We consider the expression \( \frac{x^{1/2}}{x^{5/2}} = x^{-2} = \frac{1}{x^2} \). For \( 0 < x \leq 1 \), \( \frac{1}{x^2} \geq 1 \) since squaring a number between 0 and 1 gives a smaller number. Thus, this leads to \( x^{5/2} \leq x^{1/2} \) within the interval.

Step 3: Show \( x^{5/2} \geq x^{1/2} \) for \( x \geq 1 \)

For this, note the expression \( \frac{x^{5/2}}{x^{1/2}} = x^2 \). Since \( x \geq 1 \), \( x^2 \geq 1 \) because the square of any number greater than or equal to 1 is also greater than or equal to 1. Thus, in this region \( x^{5/2} \geq x^{1/2} \) because \( x^2 - 1 \geq 0 \). Accordingly, \( x^{5/2} \) is always greater than or equal to \( x^{1/2} \) for all \( x \geq 1 \).

Key Concepts

Graphing Functions

Graphing functions such as \( y = x^{5/2} \) and \( y = x^{1/2} \) requires a basic understanding of their behavior and shapes on a coordinate plane. Both functions only exist for \( x \geq 0 \), meaning we only consider the right side of the coordinate system. Start by marking known points like zero where both graphs meet. When graphing:

• \( y = x^{1/2} \), the square root function, grows fast but levels out as \( x \) increases.
• \( y = x^{5/2} \) grows slower near zero but accelerates faster for larger \( x \).

The visual intersection and separation of these graphs across different \( x \) values helps us interpret inequalities algebraically. For intervals \( 0 \leq x \leq 1 \), \( y = x^{5/2} \) is below \( y = x^{1/2} \). When \( x > 1 \), \( y = x^{5/2} \) overtakes \( y = x^{1/2} \). Recognizing this visual behavior underpins understanding why one power function is less or more than another in given ranges.

Inequality Proofs

Proving inequalities relies on solid algebraic reasoning. For the interval \( 0 \leq x \leq 1 \), the challenge is to prove \( x^{5/2} \leq x^{1/2} \). Use inequality transformation by considering:\[ \frac{x^{1/2}}{x^{5/2}} = x^{-2} = \frac{1}{x^2} \]Within this interval, anything squared results in a smaller fraction. Therefore, \( \frac{1}{x^2} \) flips it, making it greater than or equal to one, asserting \( x^{5/2} \leq x^{1/2} \).Similarly, for \( x \geq 1 \), we need to show that \( x^{5/2} \) is not less than \( x^{1/2} \). Simplify their relation:\[ \frac{x^{5/2}}{x^{1/2}} = x^2 \]Since \( x \geq 1 \), squaring enhances any positive number, proving \( x^2 \geq 1 \) holds, so \( x^{5/2} \geq x^{1/2} \). Mastery of inequality proofs is key for confident algebraic manipulation across various mathematical concepts.

Algebraic Manipulation

Algebraic manipulation is like rearranging elements to unveil truths about equations. It often involves simplifying, factoring, or restructuring expressions to make them easily comparable. In comparing powers such as \( x^{5/2} \) and \( x^{1/2} \), simplifying helps:Begin with the hint: use the division of like bases:

• For \( 0 < x \leq 1 \), consider \( \frac{x^{1/2}}{x^{5/2}} = \frac{1}{x^2} \)—the higher the denominator’s power, the greater the result, supporting \( x^{5/2} \leq x^{1/2} \).
• For \( x \geq 1 \), \( \frac{x^{5/2}}{x^{1/2}} = x^2 \)—here, exponents intensify the base’s impact due to \( x \) being greater than or equal to one, confirming \( x^{5/2} \geq x^{1/2} \).

Efficiently moving terms, recognizing patterns, and applying rules of exponents are core skills in algebraic manipulation, essential for solving equations and inequalities.