Problem 41 - Calculus

Question

\(f(x)\) is any continuous function with \(f(0)=f(1) .\) (a) Draw a typical \(f(x)\). Mark where \(f(x)=f\left(x+\frac{1}{2}\right)\). (b) Explain why \(g(x)=f\left(x+\frac{1}{2}\right)-f(x)\) has \(g\left(\frac{1}{2}\right)=-g(0)\). (c) Deduce from (b) that (a) is always possible: There must be a point where \(g(x)=0\) and \(f(x)=f\left(x+\frac{1}{2}\right)\)

Step-by-Step Solution

Answer

There is a point \(c\) where \(f(c) = f(c+\frac{1}{2})\).

Step 1: Understanding the Problem

We are given a continuous function \(f(x)\) with the property \(f(0) = f(1)\). We need to visualize where \(f(x) = f(x+\frac{1}{2})\) on the graph, analyze the function \(g(x) = f(x+\frac{1}{2}) - f(x)\), and prove there exists a point where \(g(x)=0\).

Step 2: Drawing a Typical Function

Consider a typical continuous function that satisfies \(f(0) = f(1)\). Draw it smoothly over the interval \([0, 1]\), ensuring that it starts and ends at the same point. Highlight sections where the function appears to repeat, because these may give clues about where \(f(x) = f(x + \frac{1}{2})\).

Step 3: Analyzing the Function g(x)

Define \(g(x) = f(x + \frac{1}{2}) - f(x)\). The problem states we must show that \(g(\frac{1}{2}) = -g(0)\). This implies the difference between function values at points \(x\) and \(x+\frac{1}{2}\) reverses from \(x = 0\) to \(x = \frac{1}{2}\).

Step 4: Evaluating g(x) at Key Points

Calculate \(g(0) = f(\frac{1}{2}) - f(0)\) and \(g(\frac{1}{2}) = f(1) - f(\frac{1}{2})\). Since \(f(1) = f(0)\), we have \(g(\frac{1}{2}) = f(0) - f(\frac{1}{2}) = -g(0)\), thus verifying the given identity.

Step 5: Applying Intermediate Value Theorem

Because \(g(x)\) is continuous (given \(f(x)\) is continuous) and we go from \(g(0)\) to \(g(\frac{1}{2}) = -g(0)\), there must exist some \(c\) in \([0, \frac{1}{2}]\) where \(g(c) = 0\) due to the Intermediate Value Theorem, implying \(f(c) = f(c + \frac{1}{2})\).

Step 6: Conclusion: Final Deduction

From our exploration, it becomes evident that in a continuous transition from \(g(0)\) to \(-g(0)\), \(g(x)\) must be zero at some point, confirming that \(f(x) = f(x + \frac{1}{2})\) is always possible within \([0, \frac{1}{2}]\).

Key Concepts

Continuous Function

A continuous function is one that can be drawn without lifting your pen off the paper. This property is essential when analyzing functions like in our exercise. The function starts and ends smoothly, which means there are no breaks or jumps anywhere along its path. For the function given in the exercise, the continuity is crucial because it ensures there are no sudden changes in value between any two points.

• Continuous functions are smooth and unbroken.
• This nature helps in applying the Intermediate Value Theorem, which we use in this exercise.

Whenever you have a continuous function, transitions between points happen gradually. This is why we can confidently reason about the function's behavior between two particular values.

Function Analysis

Function analysis involves breaking down a function's behavior and understanding its characteristics at different intervals. In our exercise, we're analyzing a function where we know the starting and ending points are equal. This offers a special symmetry that we leverage.

• Consider how the function compares at different points, such as its midpoint.
• Analysis helps identify properties like symmetry or periodicity.

By thoroughly examining these aspects, you gain insights into the function's behavior, enabling predictions of what the function does at any given section. For instance, the comparison of values at intervals of \[x\] and \[x + \frac{1}{2} \] reveals symmetrical properties we can prove using algebraic manipulation.

Graphing Functions

Graphing functions is an intuitive way to understand their behavior visually. When tackling our exercise, we graph the continuous function over the interval \([0, 1]\). This visual representation allows us to identify repeating sections, satisfying \(f(x) = f(x + \frac{1}{2})\).

• Graphs provide a clear picture of how a function behaves.
• Identifying repeated patterns is easier visually than algebraically.

Marking points where patterns arise or where transitions like the ones at \(f(0) = f(1)\) occur, graphing turns an abstract idea into something tangible and easier to work with.

Function Properties

To understand our function, we delve into its properties, like continuity and periodic behavior, in this exercise. These properties are what fundamentally allow application of the Intermediate Value Theorem.

• The theorem leverages these properties, stating that a continuous function that changes from one value to another must pass through every intermediate value.
• This applies when determining where \(g(x) = 0\).

Given that \(g(x)\) is formed from a continuous \(f(x)\), it inherits this property. The theorem assures us that between any two points of opposite value, there's a point where the value is zero. This underpins the proof that such a point always exists in our exercise within the stated interval.