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

The graphs of \(y = x^{-1/2}\) and \(y = x^{1/2}\) intersect at \(x = 1\). For \(0 < x \leq 1\), \(x^{-1/2} \geq x^{1/2}\), and for \(x \geq 1\), \(x^{-1/2} \leq x^{1/2}\).

Step 1: Understanding the Graphs

For the functions given, we first consider \(y = x^{-1/2}\) which is equivalent to \(y = \frac{1}{\sqrt{x}}\). This graph represents the reciprocal of the square root function, and it is defined for \(x > 0\). Second, we consider \(y = x^{1/2}\) which is simply the square root function \(y = \sqrt{x}\), defined for \(x \geq 0\). Both graphs will be plotted on a coordinate system where the x-axis represents the values of \(x\) and the y-axis the corresponding values of \(y\).

Step 2: Graphing the Functions

To graph these functions, remember:- For \(y = x^{-1/2}\), as \(x\) approaches 0 from the right, \(y\) approaches infinity, and as \(x\) increases, \(y\) decreases.- For \(y = x^{1/2}\), as \(x\) increases, \(y\) also increases.Now, plot these functions on the same coordinate system. You'll notice that these graphs intersect at \(x = 1\) where both functions equal 1.

Step 3: Show Algebraically for \(0<x\leq 1\)

For \(0 0\)):\[ 1 \geq x^2 \]This inequality is true for all \(0 < x \leq 1\) because \(x^2\) is less than or equal to \(x\) for values in this range.

Step 4: Show Algebraically for \(x \geq 1\)

For \(x \geq 1\), show \(x^{-1/2} \leq x^{1/2}\).Begin by assuming it is true and attempt to prove it:\[ x^{-1/2} \leq x^{1/2} \Rightarrow \frac{1}{\sqrt{x}} \leq \sqrt{x} \]Square both sides:\[ \frac{1}{x} \leq x \]Multiply through by \(x\):\[ 1 \leq x^2 \]This inequality holds for all \(x \geq 1\) because \(x^2\) is greater than or equal to 1 when \(x\) is 1 or larger.

Key Concepts

Graphing Functions

Graphing functions can seem daunting at first, but it becomes clearer when you break down the steps. We start with two functions: the reciprocal of the square root function, \(y = x^{-1/2}\), and the square root function, \(y = x^{1/2}\).

- For \(y = x^{-1/2}\), imagine flipping the square root function upside down over the x-axis. As \(x\) gets closer to zero (from the right side), \(y\) increases towards infinity. As \(x\) increases, \(y\) decreases. It never touches the axes but gets very close, forming a curve that dips towards the horizontal line. - For \(y = x^{1/2}\), it’s simpler; it starts at zero and increases steadily. The graph looks like the top right part of a sideways parabola. When you plot these two equations on the same coordinate system, they intersect at \(x = 1\), both equaling 1. From this point, \(y = x^{1/2}\) goes higher and \(y = x^{-1/2}\) descends lower as \(x\) continues to grow.

Inequalities

Understanding inequalities helps us determine how one expression relates to another. When you're asked to show if \(x^{-1/2}\) is greater than or equal to \(x^{1/2}\) for \(0 < x \leq 1\), you have to compare two functions over this interval.

We start by rewriting the original functions: \(x^{-1/2}\) becomes \(rac{1}{\sqrt{x}}\) and \(x^{1/2}\) stays as \(\sqrt{x}\).

- Square both sides to eliminate the square root, which gives \(rac{1}{x} \) and \(x\). Then, multiply each side by \(x\), assuming \(x\) is positive, to get \(1 \geq x^2\).- Between 0 and 1, \(x^2\) is naturally smaller than \(1\), confirming that \(x^{-1/2} \geq x^{1/2}\).When \(x \geq 1\), the comparison flips because both functions respond differently to larger values of \(x\). The same steps show how \(x^{-1/2} \leq x^{1/2}\), which is true for all numbers \(1\) and above, as \(x^2\) is always equal or greater than 1.

Reciprocal and Square Root Functions

The reciprocal and square root functions are fascinating because of their inverse relationships and visual differences. Let’s break them down individually:

- **Reciprocal Function (\(y = x^{-1/2}\)):** This function means you take the square root first and then find its reciprocal. As a fraction, it becomes \(rac{1}{\sqrt{x}}\). This function has a significant behavior shift around \(x = 1\), dropping from high to low as \(x\) increases.- **Square Root Function (\(y = x^{1/2}\)):** It takes any number \(x\) and finds what number, when multiplied by itself, gives \(x\). It creates a curve growing outwards from the origin. Over time, it plateaus, climbing slower as \(x\) increases.These functions complement each other: the reciprocal provides an inverse mechanism to the straightforward nature of the square root. Together, they demonstrate important calculus principles when graphed simultaneously, especially their intersection and who is 'winning' over different intervals of \(x\). Understanding these relationships is key to mastering more complex calculus problems.