Problem 3 - Calculus for Biology and Medicine

Question

Let $$ f(x)=\sqrt{x^{2}+2}, \quad 1 \leq x \leq 2 $$ (a) Graph \(y=f(x)\) for \(1 \leq x \leq 2\). (b) Use the intermediate-value theorem to conclude that $$ \sqrt{x^{2}+2}=2 $$ has a solution in \((1,2)\).

Step-by-Step Solution

Answer

Yes, the equation has a solution in (1,2) by the Intermediate Value Theorem.

Step 1: Understanding the Function

The given function is \( f(x) = \sqrt{x^2 + 2} \). This function represents the mathematical operation of taking the square root of \( x^2 + 2 \). We will explore this function for the domain \( 1 \leq x \leq 2 \).

Step 2: Graphing the Function

To graph \( y = \sqrt{x^2 + 2} \) for \( 1 \leq x \leq 2 \), consider evaluating a few points. For instance, at \( x = 1 \), \( f(1) = \sqrt{1^2 + 2} = \sqrt{3} \approx 1.73 \). At \( x = 2 \), \( f(2) = \sqrt{2^2 + 2} = \sqrt{6} \approx 2.45 \). The graph will be a smooth curve rising from approximately 1.73 to 2.45.

Step 3: Intermediate Value Theorem (IVT) Introduction

The Intermediate Value Theorem suggests that if a function \( f \) is continuous over an interval \([a, b]\) and \( f(a) \leq N \leq f(b)\), then there exists \( c \) in the interval \((a, b)\) such that \( f(c) = N \). We need to show this applies for \( f(x) = 2 \).

Step 4: Applying IVT to Find the Solution

First, ensure \( f(x) \) is continuous on the interval \([1, 2]\). This is true since \( \sqrt{x^2 + 2} \) is continuous as long as the expression inside the square root, \( x^2 + 2 \), is continuous and non-negative, which it is for all \( x \). Evaluate \( f \) at the endpoints: \( f(1) = \sqrt{3} \approx 1.73 \) and \( f(2) = \sqrt{6} \approx 2.45 \). Since 2 lies between \( 1.73 \) and \( 2.45 \), by the IVT, there is a value \( c \) in \( (1,2) \) such that \( f(c) = 2 \).

Key Concepts

Continuous Function

A fundamental concept in calculus is that of a continuous function. A function is continuous if you can draw its graph without lifting your pencil from the paper. In more technical terms, a function is continuous over an interval if there are no breaks, jumps, or holes in the graph of the function for that interval. For the given problem, our function is \(f(x) = \sqrt{x^2 + 2}\),which involves the square root of a polynomial function, meaning it is continuous for all real numbers since it simplifies into smooth operations with no sudden changes. For our task, we specifically look at the interval \([1, 2]\).Because \(x^2 + 2\) is always positive for all values of \(x\), the square root function is well defined, ensuring continuity over the entire real line including our interval of interest. This property is crucial for applying the Intermediate Value Theorem, which hinges on the continuity of the function within the specified bounds.

Graphing Functions

Graphing is a powerful tool in mathematics that helps visualize the behavior of functions across their domains. In this problem, we focus on graphing the function \(y = \sqrt{x^2 + 2}\) for the domain \(1 \leq x \leq 2\). Here's how to approach it:

• Start by evaluating the function at key points within the interval, such as \(x = 1\) and \(x = 2\). For \(x = 1\), \(f(1) = \sqrt{3} \approx 1.73\), and for \(x = 2\), \(f(2) = \sqrt{6} \approx 2.45\).
• These points help to determine where the curve of the graph will start and end within the specified interval.
• Between these points, the function will form a smooth upward curve, because \(\sqrt{x^2 + 2}\) is increasing on \([1, 2]\).

Using these evaluations, sketching a graph gives a visual representation of the range and helps see where the function achieves particular values.

Calculus Problem Solving

Tackling calculus problems often involves understanding key theorems and properties like the Intermediate Value Theorem (IVT). The IVT is useful for determining the existence of solutions within given intervals. In problem-solving:

• Start by confirming the function is continuous within the desired range. For \(f(x) = \sqrt{x^2 + 2}\), as established, continuity is a given from \([1, 2]\).
• Identify two y-values encompassing your target value. Here, \(f(1) = \sqrt{3} \approx 1.73\) and \(f(2) = \sqrt{6} \approx 2.45\) encompass \(2\).
• The IVT guarantees at least one solution \(c\) exists within \((1,2)\) where \(f(c) = 2\).

This systematic approach of breaking down a problem into smaller steps makes it easier to analyze and find solutions, particularly when graphing or evaluating a function is involved.

Function Evaluation

Evaluating a function at specific points is a fundamental task that helps in understanding its behavior and is particularly important in solving calculus problems. For the function \(f(x) = \sqrt{x^2 + 2}\),we perform evaluations at points in the domain \(1 \leq x \leq 2\) to build an understanding. Here’s how to do it:

• Choose strategic points within the domain, like \(x = 1\) and \(x = 2\) in this scenario.
• Calculate the function's value at these points. For \(x=1\), we find \(f(1) = \sqrt{3} \approx 1.73\). For \(x=2\), \(f(2) = \sqrt{6} \approx 2.45\).
• These values provide insight into the function's output over the interval and illustrate how the function transitions between these points.

Function evaluation is crucial for graphical analysis and solving equations involving the function—like determining the presence of solutions within an interval by applying the Intermediate Value Theorem, as is done in this exercise.