Problem 52 - Calculus: Early Transcendentals

Question

Suppose \( f \) is continuous on \( [1, 5] \) and the only solutions of the equation \( f(x) = 6 \) are \( x = 1 \) and \( x = 4 \). If \( f(2) = 8\), explain why \( f(3) > 6 \).

Step-by-Step Solution

Answer

Since \( f(x) = 6 \) only at specific points and it is known \( f(2) = 8 \), for continuity, \( f(3) \) must also be greater than 6.

Step 1: Understanding the Problem

The problem is asking us to use the properties of continuous functions to determine why \( f(3) > 6 \), given that \( f(x) = 6 \) only has solutions at \( x = 1 \) and \( x = 4 \), and \( f(2) = 8 \).

Step 2: Analyze Function Behavior around Known Points

We know \( f \) is continuous on \([1, 5]\). This means it should not 'jump' or have 'gaps', considering the Intermediate Value Theorem (IVT) applies. Also, \( f(x) = 6 \) at only \( x=1 \) and \( x=4 \).

Step 3: Apply the Intermediate Value Theorem

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

Step 4: Consider Values at Interval Edges

Since \( f(2) = 8 \) and \( f(1) = 6 \), the value at \( x = 2 \) is greater than 6. Since \( f(x) = 6 \) has no other solutions in \((1, 5)\) besides 1 and 4, \( f(x) \) cannot again take the value 6 in the interval \((1, 4)\).

Step 5: Confirm Behavior Between Known Values

Continuity at these known points suggests that \( f(x) \) cannot dip back down to 6 before dipping below 6 and then going back up to 8 at \( x=2 \) without breaking the condition \( f(x)=6 \) only at \( x=1 \) and \( x=4 \). Thus \( f(3) > 6 \).

Key Concepts

Continuous Functions

A continuous function is one that has no breaks, jumps, or gaps across its domain. Imagine drawing the graph of such a function without lifting your pencil from the paper. This ability to draw the function smoothly is the essence of continuity. In mathematical terms, a function \( f(x) \) is continuous over an interval \([a, b]\) if, for every point \( c \) in the interval, the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \). This indicates a seamless progression of the function's values as \( x \) changes.Continuous functions are crucial as they ensure predictable behavior, where changes in input lead to gradual changes in output, following predictably without random spikes or drops. This is particularly helpful when applying mathematical theorems like the Intermediate Value Theorem (IVT), which relies on this property.

Function Behavior

Analyzing a function's behavior involves understanding how its values change over its domain. For example, in our problem, we know specific values: \( f(1) = 6 \), \( f(2) = 8 \), and \( f(4) = 6 \). By examining these known values, we can infer how the function behaves between these points.The critical part to focus on here is observing how \( f(x) \) transitions between these values, maintaining continuity. Given that \( f(x) = 6 \) only at \( x = 1 \) and \( x = 4 \), the function cannot drop to 6 between these points without violating this condition. Thus, as \( f(2) \) is greater than 6, and \( f \) is continuous, \( f(x) \) must remain greater than 6 as it moves from \( x = 1 \) through \( x = 3 \). This understanding of function behavior underpins the conclusion regarding \( f(3) \).

Solutions of Equations

Finding solutions of equations often involves identifying values of \( x \) that satisfy a given function equation. Here, we investigate solutions for \( f(x) = 6 \). The problem states that \( x = 1 \) and \( x = 4 \) are the only solutions, meaning these are the only points where the function equals 6.The Intermediate Value Theorem (IVT) plays a vital role here. According to the IVT, if a function is continuous on \([a, b]\) and \( f(a) \leq N \leq f(b) \), then there is some \( c \) in \((a, b)\) such that \( f(c) = N \). In our scenario, considering \([1, 4]\), if \(f(2) = 8\) and the only solutions for \( f(x) = 6 \) are at \( x = 1 \) and \( x = 4 \), the function cannot again equal 6 between \( x = 1 \) and \( x = 4 \). Hence, \( f(3) \) must be greater than 6 as it doesn't dip down to 6 without another solution appearing, which is a contradiction.