Problem 75 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Prove that if the function \(f(x)\) is continuous in the interval \((a, b)\) and \(x_{1}, x_{2}, \ldots \ldots \ldots, x_{n}\) are any values in this open interval, then we can always find a real number \(c\) in this open interval such that \(f(c)=\frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots \ldots .+f\left(x_{n}\right)}{n}\)

Step-by-Step Solution

Answer

We defined a new function \(F(x) = n \cdot f(x)\) which is continuous in the interval \((a, b)\). Using the extreme value theorem, we found maximum and minimum values for \(F(x)\) in this interval, denoted as \(M\) and \(m\). Then, we expressed the sum of function values \(S\) as a sum of \(F(x)\) values. The intermediate value theorem was applied to find a point \(c\) in the open interval \((a, b)\) such that \(F(c) = \frac{S}{n}\). Finally, we related \(F(c)\) with \(f(c)\) and obtained the desired result: \(f(c) = \frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots +f\left(x_{n}\right)}{n}\). This proves that there exists a real number \(c\) in the open interval \((a, b)\) that satisfies the given condition.

Step 1: Define a new function and its continuous interval

Define a new function \(F(x)\) as follows: \[F(x) = n \cdot f(x)\] Since \(f(x)\) is continuous in the interval \((a, b)\), it means that \(F(x)\) is also continuous in \((a, b)\).

Step 2: Find the maximum and minimum values of the new function

Since \(F(x)\) is continuous in the interval \((a, b)\), according to the extreme value theorem, \(F(x)\) has a maximum and minimum in the interval \((a, b)\). Let's denote them as \(M\) and \(m\) respectively: \[ m \le F(x) \le M \quad \forall x \in (a, b) \]

Step 3: Define the sum of function values with new function

With the defined function \(F(x)\), rewrite the sum of function values as follows: \[ S = F\left(x_{1}\right)+F\left(x_{2}\right)+\ldots +F\left(x_{n}\right) \]

Step 4: Use intermediate value theorem

Since \(F(x)\) is continuous, we can apply the intermediate value theorem. From Step 2, we know that \(m \le F(x) \le M\). Hence, when we sum all the values of \(F(x)\) to get the sum \(S\), we can see clearly that for any point in the interval \((a, b)\) there must exist a point \(c\) such that: \[ nm \le S \le nM \quad \Longrightarrow \quad m \le \frac{S}{n} \le M \] Using the intermediate value theorem, we conclude that because the function \(F(x)\) is continuous, there must exist a value \(c\) in the open interval \((a, b)\) such that: \[ F(c) = \frac{S}{n} \]

Step 5: Find the value of \(f(c)\)

In the last step, we need to find the value of \(f(c)\) by using the relationship between function \(F(x)\) and \(f(x)\). Since \(F(x) = n \cdot f(x)\), we have: \[ F(c) = n \cdot f(c) \] Now, replacing the value of \(F(c)\) from step 4, we get: \[ n \cdot f(c) = \frac{S}{n} \] So finally, we get the desired result: \[ f(c) = \frac{f\left(x_{1}\right)+f\left(x_{2}\right)+\ldots +f\left(x_{n}\right)}{n} \] Hence, we've proved that there exists a real number \(c\) in the open interval \((a, b)\) such that the given condition is satisfied.

Key Concepts

Extreme Value Theorem

The Extreme Value Theorem (EVT) is a crucial concept in calculus. It states that if a function is continuous over a closed interval \([a, b]\), then it must attain a maximum and a minimum value somewhere within that interval.
This means there are points where the function hits its highest and lowest values. Although the exercise deals with an open interval \(a, b\), knowing that any extensions of continuous functions to closed intervals \[a, b\] will also have maximums and minimums serves as a foundational understanding.

• This theorem provides assurance that no matter how a function "behaves" in between, its extreme values are guaranteed when considered on a closed interval.
• The concept helps in determining the bounds between which the continuous function's values lie, assisting further exploration, like using the Intermediate Value Theorem.

Understanding EVT helps grasp the continuous function's behavior deeply and is pivotal in proving many other mathematical theorems and problems.

Continuous Functions

A continuous function is one that has no breaks, holes, or sudden jumps. Every point in the function's domain flows smoothly to the next. Mathematically, for a function \(f(x)\) to be continuous at a point \(x = c\), the following must hold true:

• \(f(c)\) must be defined.
• \(\lim_{{x \to c}} f(x)\) must exist.
• \(\lim_{{x \to c}} f(x) = f(c)\).

Understanding continuous functions is crucial when applying the Extreme Value Theorem and Intermediate Value Theorem. In our exercise, knowing that \(f(x)\) is continuous in \(a, b\) meant it could smoothly transition to \(F(x) = n \cdot f(x)\), ensuring that \(F(x)\) remains continuous too.
Continuous functions are the backbone for many theorems in calculus because they ensure smooth transitions and predictability, which are essential for proving mathematical properties.

Mathematical Proofs

Mathematical proofs are logical arguments that demonstrate the truth of a mathematical statement. They are essential in verifying concepts and theories conclusively. Proofs follow a structured approach:

• Start with known information or axioms.
• Use logical reasoning to derive conclusions step by step.
• Transition smoothly from one step to another using well-established rules.

In our problem, the proof starts by defining a new function \(F(x) = n \cdot f(x)\). This helps manipulate the given setup to find \(c\) conveniently. The core steps involved checking continuity and applying known theorems like the EVT and Intermediate Value Theorem.
Each statement in the proof builds logically on the preceding ones, guiding us to the final conclusion. Mastering proofs not only aids in solving exercises but also develops critical thinking and problem-solving skills. They’re a key component in understanding deeper mathematical theories and verifying their validity.