Problem 55 - Thomas Calculus

Question

Unique solution Assume that \(f\) is continuous on \([a, b]\) and differentiable on \((a, b) .\) Also assume that \(f(a)\) and \(f(b)\) have opposite signs and that \(f^{\prime} \neq 0\) between \(a\) and \(b\) . Show that \(f(x)=0\) exactly once between \(a\) and \(b\) .

Step-by-Step Solution

Answer

The function \( f(x) \) has exactly one root in \((a, b)\).

Step 1: Understanding the Problem

We need to prove that the function \( f(x) \) has exactly one real root in the interval \((a, b)\). Given conditions: \( f \) is continuous on \([a, b]\) and differentiable on \((a, b)\), \( f(a) \) and \( f(b) \) have opposite signs, and \( f'(x) eq 0 \) for all \( x \) in \((a, b)\).

Step 2: Using the Intermediate Value Theorem

Since \( f \) is continuous on \([a, b]\) and \( f(a) \) and \( f(b) \) have opposite signs, the Intermediate Value Theorem guarantees at least one \( c \) in \( (a, b) \) such that \( f(c) = 0 \). Thus, there is at least one root between \( a \) and \( b \).

Step 3: Proving Uniqueness with Derivative Information

Since \( f'(x) eq 0 \) for all \( x \) in \((a, b)\), \( f \) is either strictly increasing or strictly decreasing on \((a, b)\). This property of strictly monotonic functions ensures that \( f(x) = 0 \) can occur at most once. Thus, there can only be one root in \((a, b)\).

Key Concepts

Intermediate Value Theorem

In the world of calculus, the Intermediate Value Theorem (IVT) is a fundamental concept. It revolves around the behavior of continuous functions. Simply put, if a function is continuous on a closed interval \([a, b]\), the function "takes on" every value between \((f(a))\) and \((f(b))\). This is quite intuitive if you think about drawing a curve without lifting a pencil from the paper.
For example, if initially the function is below the x-axis (negative) at point \((a)\) and above the x-axis (positive) at point \((b)\), then the curve must cross the x-axis somewhere between \((a)\) and \((b)\). This crossing point is where the function equals zero.
The Intermediate Value Theorem was used in the exercise to show the existence of at least one zero (or root) in the interval. This is crucial in proving that there's a solution to an equation within a particular range.

Monotonic Functions

Monotonic functions are those that consistently move in one direction. This means they are either solely increasing or solely decreasing over an interval. The key feature of a function being monotonic is its derivative.
When a function has a derivative that is nonzero over an interval – as in \(f'(x) eq 0\) – the function behaves in a monotonic manner:

• If \(f'(x) > 0\), the function is strictly increasing.
• If \(f'(x) < 0\), the function is strictly decreasing.

In the original problem, the function was shown to have a nonzero derivative across the interval \((a, b)\). This means it is strictly monotonic, hence continuously moving up or down with no reversals.
This property of monotonicity assures us that if the function crosses the x-axis once, it cannot return to do so again in the interval, ensuring uniqueness of the zero found using the Intermediate Value Theorem.

Continuity and Differentiability

Continuity and differentiability are two key properties of functions that play a crucial role in calculus. A function is continuous on an interval if there are no breaks or gaps over that range. Imagine drawing the graph of the function without lifting your pencil.

• Continuity ensures the application of the Intermediate Value Theorem, guaranteeing a root in the interval.

Differentiability, on the other hand, means the function has a derivative at each point in the interval. This implies that the slope of the tangent line to the function is well-defined everywhere in the interval.

• This is stronger than continuity because every point on the function's curve is smooth, meaning there are no sharp turns or cusps.

In the given problem, the condition that \(f(x)\) is continuous on \[a, b \] and differentiable on \((a, b)\) not only allows the use of the Intermediate Value Theorem but also supports the argument of monotonicity by providing the derivative information needed to ascertain that the function can cross the x-axis at most once.