Problem 6 - Calculus for Biology and Medicine

Question

Use the intermediate-value theorem to show that $$ \cos x=x $$ has a solution in \((0,1)\).

Step-by-Step Solution

Answer

The function \( f(x) = \cos x - x \) is continuous and changes sign between 0 and 1, so by the Intermediate-Value Theorem, a solution exists in \((0, 1)\).

Step 1: Understanding the Intermediate-Value Theorem

The Intermediate-Value Theorem states that if a function \( f \) is continuous on a closed interval \([a, b]\) and \(d\) is any number between \( f(a) \) and \( f(b) \), then there is at least one number \( c \) in \((a, b)\) such that \( f(c) = d \).

Step 2: Define the Function

Define a function \( f(x) = \cos x - x \). We need to find a value \( c \) in the interval \((0,1)\) where \( f(c) = 0 \).

Step 3: Checking Continuity

The function \( f(x) = \cos x - x \) is continuous everywhere because both \( \cos x \) and \( x \) are continuous functions. This implies \( f(x) \) is continuous on \((0, 1)\).

Step 4: Evaluate the Function at the Endpoints

Calculate \( f(0) = \cos(0) - 0 = 1 - 0 = 1 \) and \( f(1) = \cos(1) - 1 \). Since \( \cos(1) \approx 0.5403 \), \( f(1) \approx 0.5403 - 1 = -0.4597 \).

Step 5: Apply the Intermediate-Value Theorem

Since \( f(0) = 1 \) and \( f(1) \approx -0.4597 \), and since \( 0 \) is between \( f(0) \) and \( f(1) \), by the Intermediate-Value Theorem, there must exist a \( c \) in \((0, 1)\) such that \( f(c) = 0 \). Thus, \( \cos c = c \).

Key Concepts

Continuity of Functions

The concept of continuity is fundamental to understanding many principles in calculus, including the Intermediate-Value Theorem. A function is said to be continuous on a closed interval \([a, b]\) if there are no breaks, jumps, or holes in its graph in that interval. In simpler terms, you can imagine drawing the function without ever lifting your pen from the paper.

Continuity ensures that for any change in the input, there is a smooth and predictable change in the output. For instance, the function \(f(x) = \cos x - x\) is continuous because both component functions, \(\cos x\) and \(x\), are continuous functions. The subtraction of these two continuous functions retains the property of continuity. This smoothness across the domain allows us to use important theorems, such as the Intermediate-Value Theorem, to make conclusions about the existence of solutions to equations.

Trigonometric Functions

Trigonometric functions are a special class of functions in mathematics that relate the angles of a triangle to its side lengths. The function \(f(x) = \cos x\) is one of the basic trigonometric functions and is known for its wave-like pattern, oscillating between -1 and 1.

The cosine function is periodic, with a repeat every \(2\pi\), and it is continuous for all real numbers. It takes the value 1 when \(x = 0\). As \(x\) increases from 0 to \(\pi\), the cosine value decreases.

• The cosine function starts at 1 when \(x = 0\).
• It goes to 0 at \(x = \pi/2\).
• Returns to -1 at \(x = \pi\).

Understanding these properties of \(\cos x\) helps when examining functions like \(f(x) = \cos x - x\), as you can predict how the function behaves over specific intervals. This behavior is crucial when solving trigonometric equations, especially to see where the function \(f(x)\) can cross the x-axis.

Roots of Equations

Finding the roots, or solutions, of an equation involves determining the values of \(x\) which make the equation true. For the function \(f(x) = \cos x - x\), the root is the point where \(f(x) = 0\), meaning the graph of \(f(x)\) intersects the x-axis.

Using the Intermediate-Value Theorem, if a continuous function changes signs over an interval \(a\) to \(b\) (i.e., the function's values at these two points have opposite signs), there must be at least one root in that interval. In our original exercise, by evaluating \(f(x)\) at the boundaries, we found:

• At \(x = 0\), \(f(0) = 1\).
• At \(x = 1\), \(f(1) \approx -0.4597\).

Since 0 lies between \(f(0)\) and \(f(1)\), the theorem guarantees a solution exists between 0 and 1. This is a practical approach for finding roots when exact algebraic methods are challenging, especially for transcendental equations like \(\cos x = x\).