Question

Prove: Let \(f\) be continuous on \([a, b]\) and differentiable on \((a, b)\). If \(f(a)\) and \(f(b)\) have opposite signs and if \(f^{\prime}(x) \neq 0\) for all \(x\) in \((a, b)\), then the equation \(f(x)=0\) has one and only one solution between \(a\) and \(b .\) Hint: Use the Intermediate Value Theorem and Rolle's Theorem (Problem 22).

Short answer

There is exactly one solution in \((a, b)\) for \(f(x)=0\).

Step-by-step solution

Introduction to the Problem and Conditions

We need to prove that there is exactly one solution for the equation \(f(x)=0\) in the interval \((a, b)\) given certain conditions. These include that \(f\) is continuous on \([a, b]\), differentiable on \((a, b)\), \(f(a)\) and \(f(b)\) have opposite signs, and \(f'(x) eq 0\) for all \(x\) in \((a, b)\).

Applying the Intermediate Value Theorem

By the Intermediate Value Theorem (IVT), since \(f(a)\) and \(f(b)\) have opposite signs and \(f\) is continuous on \([a, b]\), there exists at least one value \(c\in (a, b)\) such that \(f(c)=0\). This establishes the existence of a solution.

Using Rolle's Theorem to Prove Uniqueness

Assume there are two solutions \(c_1\) and \(c_2\) such that \(f(c_1)=0\) and \(f(c_2)=0\) with \(c_1 < c_2\). By Rolle's Theorem, since \(f\) is continuous on \([c_1, c_2]\) and differentiable on \((c_1, c_2)\), there must exist some \(c\in (c_1, c_2)\) such that \(f'(c)=0\). This contradicts the given condition that \(f'(x) eq 0\) for any \(x\) in \((a, b)\).

Conclusion

Since assuming two solutions leads to a contradiction, there must be exactly one solution \(c\) in \((a, b)\) such that \(f(c)=0\). Thus, the function \(f\) crosses the x-axis only once in the interval due to the continuous change between the opposite signs and the non-zero derivative condition.

Key concepts

Rolle's Theorem

Rolle's Theorem is a mathematical theorem used in calculus to make conclusions about a differentiable function based on the behavior of its derivative. It's a special case of the Mean Value Theorem and can be quite handy in proving the uniqueness of solutions for certain types of equations.

Here’s how it works: if you have a function, say \( f \), that is continuous on a closed interval \([a, b]\) and differentiable on an open interval \((a, b)\), and \( f(a) = f(b) \), then there is at least one point \( c \) in \((a, b)\) where the derivative \( f'(c) = 0 \).

This is like saying that if you start and stop at the same elevation while walking on a continuous, smooth trail, there must be at least one spot in between where you are walking perfectly flat. In our exercise, we use Rolle's Theorem to demonstrate that if there were two solutions, it would mean there's a point where \( f' \) must be zero, which contradicts our conditions. Thus, proving the solution's uniqueness.

Continuity

Continuity is a fundamental concept in calculus and real analysis, focusing on how functions behave in terms of consistency and predictability. A function \( f \) is continuous on an interval \([a, b]\) if you can draw its graph on that interval without lifting your pen from the paper. Mathematically, this means for every \( x \) in \([a, b]\), the limit of \( f \) as you approach \( x \) from any direction equals \( f(x) \).

In our context, the Intermediate Value Theorem (IVT) relies heavily on continuity. This theorem tells us that if a function is continuous over an interval and takes on different signs at the ends of that interval, it must cross the x-axis somewhere in between. This crossing ensures that there's at least one solution in the interval where \( f(x) = 0 \).

Continuity, thus, helps in establishing the existence of solutions, sealing the first part of solving the exercise.

Differentiability

Differentiability adds another layer to understanding how smoothly a function behaves. If a function \( f \) is differentiable on an interval \((a, b)\), it means \( f \) not only changes smoothly but has a defined slope (or derivative) at every point in \((a, b)\).

In simple terms, differentiability implies there are no "sharp" corners or discontinuities in \( f \). The function can be stretched or compressed, but it keeps a consistent slope. For instance, the graph of \( f \) can be tangentially touched at every point in \((a, b)\) without interruption.

In practice, differentiability is essential in applying Rolle's Theorem. It lets us reason about conditions where \( f'(c) = 0 \), which usually corresponds to peaks, valleys, or flattening points on a graph. However, in our exercise, the requirement that \( f'(x) eq 0 \) clearly contradicts multiple solutions, enforcing the idea of the uniqueness when paired with the other conditions.

Unique Solution

A unique solution in mathematical problems indicates that there's precisely one solution meeting given criteria or constraints. This notion of uniqueness ensures clarity and certainty: no ambiguities or multiple outcomes are present under specified conditions.

In the exercise, achieving a unique solution for \( f(x) = 0 \) within \((a, b)\) is contingent on several criteria:
- The continuity of \( f \), confirmed by the Intermediate Value Theorem, presents the existence of a crossing point for the x-axis.
- The differentiability across the interval paired with the nonzero derivative requirement (\( f'(x) eq 0 \)) disallows any flat spots or repeated solutions.

By ensuring these conditions, one essentially "locks in" the solution. The process invalidates any possibility of two solutions simultaneously satisfying \( f(x) = 0 \). Thus, the exercise guarantees that f crosses the axis only once, underscoring the importance of both continuity and the absence of a zero derivative.