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 op- posite signs and \(f^{\prime} \neq 0\) between \(a\) and \(b\) . Show that \(f(x)=0\) exactly once between \(a\) and \(b .\)

Short answer

By first applying the Intermediate Value Theorem to confirm the existence of at least one root and then the Mean Value Theorem to contradict the existence of more than one root, we conclude that the function \( f \) has exactly one root in \( (a, b) \)

Step-by-step solution

Verify the function has at least one root

By applying the Intermediate Value Theorem (IVT), due to the property that \( f(a) \) and \( f(b) \) have opposite signs, we know that there exists a \( c \) in the open interval \( (a, b) \) such that \( f(c) = 0 \). In other words, the IVT guarantees at least one root.

Prove the function has only one root

Now suppose that there were two roots, say \( c_1 \) and \( c_2 \), such that \( f(c_1)=f(c_2)=0 \) and \( c_1 < c_2 \). We can then apply the Mean Value Theorem (MVT), which states that there exists a \( c \) in \( (c_1, c_2) \) such that \( f'(c)= \frac{f(c_2)-f(c_1)}{c_2-c_1} \). However, since \( f(c_1) = f(c_2) = 0 \), the fraction becomes zero, which contradicts the assumption that \( f' \neq 0 \) in \( (a, b) \). Therefore, there can only be one root.

Key concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) comes into play when we're trying to pinpoint the existence of roots for a function within a closed interval. When discussing the IVT, imagine two points on a graph, linked by a path created by the continuous function over the interval. If those two points happen to have values with opposite signs, this theorem tells us with certainty that there must be at least one point along the path where the function crosses the horizontal axis (i.e., where the function's value is zero).

In practical terms, if you're playing a game of hot and cold, the IVT assures you that at some point, as you move continuously from hot to cold, you must pass through a 'just right' warm spot. This theorem doesn't tell us where that spot is or how many there are; it simply guarantees that at least one exists. This theorem is fundamental when we're trying to prove the existence of a unique solution to a function, as in the exercise provided.

Mean Value Theorem

Like a snapshot capturing the average speed of a car on a road trip, the Mean Value Theorem (MVT) provides a snapshot of a function's instantaneous rate of change at some point within an interval. The theorem requires two conditions: continuity on a closed interval and differentiability on the open interval between two points.

It states that there is at least one point where the derivative (the function's slope at that point) will be equal to the function's average rate of change over the entire interval. In simpler terms, no matter the twists and turns on your path from point A to B, there will be a moment where you're traveling exactly at your average pace. This concept was crucial in the solution to show that the function could not have more than one zero without violating the condition that its derivative is never zero within the interval.

Continuity of Functions

When we speak of a function being continuous, we're describing a smooth, unbroken curve with no jumps, gaps, or abrupt changes in direction. For a function to be continuous at a point, three conditions must be met: the function is defined at the point, it has a limit as it approaches the point, and the limit matches the function's value at that point.

Ensuring that a function is continuous over an interval is like assuring that a bridge has no gaps between its start and end; it allows us to traverse from one end to the other confidently. Continuity is a bedrock principle for both the Intermediate and Mean Value Theorems—without it, these theorems simply cannot be applied. In our example, the function's continuity on \[a, b\] is why we could use the IVT to assert the existence of at least one root.

Differentiability

A differentiable function is smooth in a more strict sense than a continuous function—it's not enough to be free of gaps and jumps; the function must also be without sharp turns or cusps at every point in its domain. Think of differentiability as the guarantee that a function has a well-defined slope, or derivative, at each point.

If a path is differentiable, we can safely skate along it, knowing we can predict its slope at any given moment, which is essential for understanding how it changes over time. In the context of the exercise, the function's differentiability on \(a, b\) prevents it from having a zero derivative—another factor ensuring that the function can cross the axis only once, and does not contradict the Mean Value Theorem.