Problem 4 - Calculus for Biology and Medicine

Question

Let \(f(x)=\sin x-x, \quad-1 \leq x \leq 1\). (a) Graph \(y=f(x)\) for \(-1 \leq x \leq 1\). (b) Use the Intermediate Value Theorem to conclude that \(\sin x=x\) has a solution in \((-1,1)\)

Step-by-Step Solution

Answer

The graph of \(f(x) = \sin x - x\) crosses the x-axis in \((-1, 1)\), implying \(\sin x = x\) has a solution in \((-1, 1)\).

Step 1: Understanding the Function

The function given is \( f(x) = \sin x - x \). We need to analyze it within the interval \(-1 \leq x \leq 1\). This function is the difference between the sine function and the line \(y = x\). We will graph this to see where it crosses the x-axis.

Step 2: Graph of Function

To graph \(y = f(x)\), plot points where \(x\) ranges from -1 to 1. Use specific values such as \(-1, -0.5, 0, 0.5, 1\) to compute \(f(x)\). Note that \(\sin(-1) \approx -0.8415\), \(\sin(0) = 0\), and \(\sin(1) \approx 0.8415\). The graph will intersect the x-axis where \(\sin x = x\).

Step 3: Intermediate Value Theorem Explanation

The Intermediate Value Theorem states if a continuous function \(f\) takes on values \(f(a)\) and \(f(b)\) at each end of an interval \([a, b]\), and if \(f(a)\) and \(f(b)\) have opposite signs, there must be at least one value \(c\) in the interval where \(f(c) = 0\).

Step 4: Applying Intermediate Value Theorem

Calculate \(f(-1) = \sin(-1) + 1 \approx -0.8415 + 1 = 0.1585\) and \(f(1)=\sin(1) - 1 \approx 0.8415 - 1 = -0.1585\). Since \(f(-1) > 0\) and \(f(1) < 0\), the Intermediate Value Theorem guarantees there is at least one \(c\) in the interval \((-1, 1)\) where \(f(c) = 0\). This implies there is a solution \(\sin c = c\) in \((-1,1)\).

Key Concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a fundamental concept in calculus. It helps us understand scenarios where a continuous function crosses the x-axis. Imagine driving a car from point A to point B without stopping. If point A is below sea level and point B is above it, you surely must have passed sea level at some point. The IVT formalizes this idea for functions. If you have two values on the y-axis that are on opposite sides of zero, the function's graph must cross the x-axis between those points.

To apply the IVT, two things must hold:

• The function must be continuous over the interval.
• It must have different signs at the two endpoints of the interval.

Using it, you can determine there's at least one x-value (let's call it c) where the function equals zero. In our exercise, we have the function \( f(x) = \sin x - x \), which is continuous over \( [-1, 1] \). We checked:

• \( f(-1) > 0 \)
• \( f(1) < 0 \)

Since these conditions are met, the Intermediate Value Theorem confirms there is a solution where \( \sin x = x \) within \( (-1, 1) \).

Continuous Functions

In calculus, continuous functions have no holes, jumps, or gaps. They smoothly connect points without any breaks. Think of drawing a line without lifting your pen from the paper. That's continuity.

Continuous functions have precise mathematical requirements. For a function to be continuous at a point \( c \), three conditions must be satisfied:

• The function must be defined at \( c \).
• \( \lim_{x \to c} f(x) \) must exist.
• The limit of \( f(x) \) as \( x \) approaches \( c \) must equal \( f(c) \).

In our case, the function \( f(x) = \sin x - x \) is composed of the sine function, which is continuous everywhere, and the linear function \( x \), also continuous everywhere. As a result, \( f(x) = \sin x - x \) maintains continuity over the interval \( [-1, 1] \). This ensures we can confidently apply the Intermediate Value Theorem.

Graphing Functions

Graphing functions is a way to visually explore mathematical relationships. You plot points and draw a curve or line that best describes those points. For our function \( f(x) = \sin x - x \), it's like combining the wave of the sine curve with the slant of a line, subtracting one from the other.

To create a graph:

• Identify key points. We used specific \( x \) values like \(-1, -0.5, 0, 0.5, 1\).
• Compute the corresponding \( f(x) \) values using \( \sin(x) \).
• Plot these points on the coordinate system.
• Draw a smooth curve through these points.

With \( f(x) = \sin x - x \), you'll notice that the curve intersects the x-axis where \( \sin x = x \). This x-intersection is where the function value becomes zero, indicating a solution to our equation. Through graphing, we not only determine this intersection visually but enable a clearer understanding of the behavior of such functions.