Problem 7 - Advanced Calculus

Question

Let \(f\) and \(g\) be continuous on \([0,1]\) and suppose that \(f(0)g(1)\). Prove that there is a point \(x, 0

Step-by-Step Solution

Answer

By applying the Intermediate Value Theorem to the newly defined function \(h\), it is demonstrated that there exists a point \(c\), in the interval (0,1), such that \(f(c) = g(c)\). Therefore, affirmative, there is a point \(x\) with \(0

Step 1: Define a New Function

Define a new function \(h(x) = f(x) - g(x)\). The function \(h\) is continuous because it is the difference of the two continuous functions \(f\) and \(g\).

Step 2: Evaluate the New Function at Endpoints

Evaluate the function \(h\) at the endpoints of the interval [0, 1]. We have \(h(0) = f(0) - g(0) < 0\) and \(h(1) = f(1) - g(1) > 0\).

Step 3: Apply the Intermediate Value Theorem

Apply the Intermediate Value Theorem for the function \(h\) over the interval [0, 1]. Since \(h(0) < 0\) and \(h(1) > 0\), this implies that there is some point \(c\) in the interval (0, 1) such that \(h(c) = 0\).

Step 4: Equate with Origin and Solve.

From \(h(c) = 0\), by substitute \(h(x) = f(x) - g(x)\), we get: \(f(c) = g(c)\). So there is at least one point c which lies between 0 and 1 such that \(f(c) = g(c)\).

Key Concepts

Continuous Functions

In calculus, a continuous function is one that has no interruptions or breaks in its graph. This property means for every value of the function's domain, there is a corresponding point on the curve that is not disconnected from any other point. The definition ties directly to limits: a function \( f \) is continuous at a point \( a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to the function's value at that point, \( f(a) \).
Continuous functions are powerful in calculus due to their predictable nature. When a function is continuous over a closed interval \([a, b]\), it's guaranteed to take on every value between \( f(a) \) and \( f(b) \) at some point within that interval. This trait is instrumental in solving various equation types and in applying important theorems like the Intermediate Value Theorem.

Function Properties

Functions have various properties that can be analyzed, such as continuity, differentiability, and integrability, each giving us different insights into the function's behavior. Evaluating a function's properties allows us to classify and understand its structure, predict future values, and identify points of interest like maxima, minima, or roots.

In the context of continuous functions, we can apply the Intermediate Value Theorem when dealing with continuous functions over an interval. If two points \((a, b)\) satisfy \(f(a) < f(b)\) or \(f(a) > f(b)\), the theorem guarantees that for any value between \(f(a)\) and \(f(b)\), there exists a point \(c\) such that \(f(c)\) equals this value. This concept is pivotal in demonstrating how functions "behave" over intervals, proving the presence of certain function values within those gaps.

Calculus Proofs

Calculus proofs are logical arguments that establish the truth behind calculus principles and theorems like the Intermediate Value Theorem. In these proofs, mathematicians use properties of functions, often employing limits and continuity, to systematically demonstrate why certain conditions or conclusions hold true.

For example, in the original exercise, the proof involved showing that the function \( h(x) = f(x) - g(x) \) is continuous because it results from the subtraction of two continuous functions, \( f \) and \( g \). Next, by assessing its value at boundary points \( h(0) \) and \( h(1) \), and observing a change from negative to positive, the Intermediate Value Theorem ensures a point exists where \( h(x) \) equals zero within that interval. Hence, at this point \( c \), we have \( f(c) = g(c) \). This transparent logical flow demonstrates how each property and theorem builds upon the last, creating a cohesive and reliable proof.

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Problem 7

Next

Problem 8

Other exercises in this chapter

Problem 7

Let \(D\) be a bounded set and let \(f\) be uniformly continuous on \(D \subset \mathbf{R}^{n} .\) Prove that \(f\) is bounded on \(D\).

View solution

Problem 7

Show that a continuous function \(f\) cannot map the interval \([0,1]\) onto itself exactly 2 -to-1.

View solution

Problem 8

Use the example \(f(x)=x^{2} /\left(1+x^{2}\right)\) to show that a continuous function does not always have to map a closed set onto a closed set.

View solution

Problem 8

Let \(f\) be a function defined on a set \(E\) which is such that it can be uniformly approximated within \(\varepsilon\) on \(E\) by functions \(F\) that are u

View solution

Key Concepts

Continuous Functions

In calculus, a continuous function is one that has no interruptions or breaks in its graph. This property means for every value of the function's domain, there is a corresponding point on the curve that is not disconnected from any other point. The definition ties directly to limits: a function \( f \) is continuous at a point \( a \) if the limit of \( f(x) \) as \( x \) approaches \( a \) is equal to the function's value at that point, \( f(a) \).
Continuous functions are powerful in calculus due to their predictable nature. When a function is continuous over a closed interval \([a, b]\), it's guaranteed to take on every value between \( f(a) \) and \( f(b) \) at some point within that interval. This trait is instrumental in solving various equation types and in applying important theorems like the Intermediate Value Theorem.

Function Properties

Functions have various properties that can be analyzed, such as continuity, differentiability, and integrability, each giving us different insights into the function's behavior. Evaluating a function's properties allows us to classify and understand its structure, predict future values, and identify points of interest like maxima, minima, or roots.

In the context of continuous functions, we can apply the Intermediate Value Theorem when dealing with continuous functions over an interval. If two points \((a, b)\) satisfy \(f(a) < f(b)\) or \(f(a) > f(b)\), the theorem guarantees that for any value between \(f(a)\) and \(f(b)\), there exists a point \(c\) such that \(f(c)\) equals this value. This concept is pivotal in demonstrating how functions "behave" over intervals, proving the presence of certain function values within those gaps.

Calculus Proofs

Calculus proofs are logical arguments that establish the truth behind calculus principles and theorems like the Intermediate Value Theorem. In these proofs, mathematicians use properties of functions, often employing limits and continuity, to systematically demonstrate why certain conditions or conclusions hold true.

For example, in the original exercise, the proof involved showing that the function \( h(x) = f(x) - g(x) \) is continuous because it results from the subtraction of two continuous functions, \( f \) and \( g \). Next, by assessing its value at boundary points \( h(0) \) and \( h(1) \), and observing a change from negative to positive, the Intermediate Value Theorem ensures a point exists where \( h(x) \) equals zero within that interval. Hence, at this point \( c \), we have \( f(c) = g(c) \). This transparent logical flow demonstrates how each property and theorem builds upon the last, creating a cohesive and reliable proof.