Problem 51 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

Functions intersect at (1,1); 1 \geq x for 0

Step 1: Graph the Functions

We graph the functions \( y = x^{-1/2} \) and \( y = x^{1/2} \) on the same coordinate system. Plot points for some values of \( x \). For \( y = x^{-1/2} \), when \( x = 0.25 \), \( y = 2 \); when \( x = 1 \), \( y = 1 \); when \( x = 4 \), \( y = 0.5 \). For \( y = x^{1/2} \), when \( x = 0 \), \( y = 0 \); when \( x = 1 \), \( y = 1 \); when \( x = 4 \), \( y = 2 \). The graphs show that these functions are reflections of each other across the line \( y = 1 \).

Step 2: Algebraic Proof for Part (b)

To show \( x^{-1/2} \geq x^{1/2} \) for \( 0

Step 3: Algebraic Proof for Part (c)

To show \( x^{-1/2} \leq x^{1/2} \) for \( x \geq 1 \), we start with \( x^{-1/2} \leq x^{1/2} \), which means \( \frac{1}{\sqrt{x}} \leq \sqrt{x} \). By multiplying both sides by \( \sqrt{x} \), we obtain \( 1 \leq x \). This inequality holds true for \( x \geq 1 \).

Key Concepts

Graphing Functions

Graphing functions is fundamental in calculus, providing a visual representation of algebraic equations. It helps in understanding the behavior of functions and their relationships. In this exercise, we are asked to graph two specific functions, \( y = x^{-1/2} \) and \( y = x^{1/2} \), on the same coordinate plane. Here’s how you can approach it:

• First, identify the domain of each function. For \( y = x^{-1/2} \), the domain is \( x > 0 \) because you cannot take the square root of a negative number or zero when it's in the denominator. For \( y = x^{1/2} \), the domain is \( x \geq 0 \).
• Next, choose strategic values of \( x \) to plot points for each function. For example, compute \( y \) for \( x = 0.25, 1, \) and \( 4 \). For \( y = x^{-1/2} \): at \( x = 0.25 \), \( y = 2 \); at \( x = 1 \), \( y = 1 \); at \( x = 4 \), \( y = 0.5 \). For \( y = x^{1/2} \): at \( x = 0 \), \( y = 0 \); at \( x = 1 \), \( y = 1 \); at \( x = 4 \), \( y = 2 \).
• Finally, plot these points and draw the curves, observing their behaviors especially as \( x \) approaches the boundaries of their domains.

Graphing these functions illustrates that they are reflections of each other across the line \( y = 1 \), showing an interesting symmetry in their graphs.

Inequality Proofs

Inequality proofs are a crucial skill in calculus, involving algebraic manipulation to establish the relationship between expressions over a specified range. For this problem, we have to rigorously prove two inequalities.Proof for \( x^{-1/2} \geq x^{1/2} \) when \( 0 < x \leq 1 \): To start, understand what the inequality means in numerical terms: \( \frac{1}{\sqrt{x}} \geq \sqrt{x} \). Multiplying both sides by \( \sqrt{x} \) simplifies the expression to \( 1 \geq x \), which holds true for \( 0 < x \leq 1 \). This proof shows that as \( x \) decreases from 1, \( x^{-1/2} \) becomes larger than \( x^{1/2} \).Proof for \( x^{-1/2} \leq x^{1/2} \) when \( x \geq 1 \): Similarly, start with \( \frac{1}{\sqrt{x}} \leq \sqrt{x} \). By multiplying both sides by \( \sqrt{x} \), we get \( 1 \leq x \). This proof shows that as \( x \) increases beyond 1, \( x^{1/2} \) grows faster than \( x^{-1/2} \), making the latter smaller.

Function Reflection

Function reflection reveals a fascinating aspect of symmetry about the graphs of certain functions. In this exercise, the functions \( y = x^{-1/2} \) and \( y = x^{1/2} \) reflect over the line \( y = 1 \).Here’s how reflection works with these functions:

• Imagine the line \( y = 1 \) on your graph. It serves as an axis of reflection. For every point \( (x, y) \) on one graph, there exists a corresponding point \( (x, 2-y) \) on the other.
• This reflects intuitively because if a point on one function is \( (x, y) \), its mirrored counterpart on the other function swaps how \( y \) is calculated by flipping over 1. For instance, at \( x = 4 \), \( y = 0.5 \) for \( y = x^{-1/2} \) and \( y = 2 \) for \( y = x^{1/2} \), maintaining a balance around \( y = 1 \).

Understanding this reflective behavior provides insights into more complex transformations and symmetries in calculus, aiding in recognizing patterns in different functions.

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Key Concepts

Graphing Functions

Graphing functions is fundamental in calculus, providing a visual representation of algebraic equations. It helps in understanding the behavior of functions and their relationships. In this exercise, we are asked to graph two specific functions, \( y = x^{-1/2} \) and \( y = x^{1/2} \), on the same coordinate plane. Here’s how you can approach it:

• First, identify the domain of each function. For \( y = x^{-1/2} \), the domain is \( x > 0 \) because you cannot take the square root of a negative number or zero when it's in the denominator. For \( y = x^{1/2} \), the domain is \( x \geq 0 \).
• Next, choose strategic values of \( x \) to plot points for each function. For example, compute \( y \) for \( x = 0.25, 1, \) and \( 4 \). For \( y = x^{-1/2} \): at \( x = 0.25 \), \( y = 2 \); at \( x = 1 \), \( y = 1 \); at \( x = 4 \), \( y = 0.5 \). For \( y = x^{1/2} \): at \( x = 0 \), \( y = 0 \); at \( x = 1 \), \( y = 1 \); at \( x = 4 \), \( y = 2 \).
• Finally, plot these points and draw the curves, observing their behaviors especially as \( x \) approaches the boundaries of their domains.

Graphing these functions illustrates that they are reflections of each other across the line \( y = 1 \), showing an interesting symmetry in their graphs.

Inequality Proofs

Inequality proofs are a crucial skill in calculus, involving algebraic manipulation to establish the relationship between expressions over a specified range. For this problem, we have to rigorously prove two inequalities.Proof for \( x^{-1/2} \geq x^{1/2} \) when \( 0 < x \leq 1 \): To start, understand what the inequality means in numerical terms: \( \frac{1}{\sqrt{x}} \geq \sqrt{x} \). Multiplying both sides by \( \sqrt{x} \) simplifies the expression to \( 1 \geq x \), which holds true for \( 0 < x \leq 1 \). This proof shows that as \( x \) decreases from 1, \( x^{-1/2} \) becomes larger than \( x^{1/2} \).Proof for \( x^{-1/2} \leq x^{1/2} \) when \( x \geq 1 \): Similarly, start with \( \frac{1}{\sqrt{x}} \leq \sqrt{x} \). By multiplying both sides by \( \sqrt{x} \), we get \( 1 \leq x \). This proof shows that as \( x \) increases beyond 1, \( x^{1/2} \) grows faster than \( x^{-1/2} \), making the latter smaller.

Function Reflection

Function reflection reveals a fascinating aspect of symmetry about the graphs of certain functions. In this exercise, the functions \( y = x^{-1/2} \) and \( y = x^{1/2} \) reflect over the line \( y = 1 \).Here’s how reflection works with these functions:

• Imagine the line \( y = 1 \) on your graph. It serves as an axis of reflection. For every point \( (x, y) \) on one graph, there exists a corresponding point \( (x, 2-y) \) on the other.
• This reflects intuitively because if a point on one function is \( (x, y) \), its mirrored counterpart on the other function swaps how \( y \) is calculated by flipping over 1. For instance, at \( x = 4 \), \( y = 0.5 \) for \( y = x^{-1/2} \) and \( y = 2 \) for \( y = x^{1/2} \), maintaining a balance around \( y = 1 \).

Understanding this reflective behavior provides insights into more complex transformations and symmetries in calculus, aiding in recognizing patterns in different functions.