Problem 72 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Let the function \(f(x)\) be continuous in the interval \([a, b]\) and \(a \leq f(x) \leq b \forall x \in[a, b]\). Prove that in this close interval there exists at least one point at which \(f(x)=x\). Explain this geometrically.

Step-by-Step Solution

Answer

As a short answer to the question: To prove that there exists at least one point in the interval \([a, b]\) where \(f(x)=x\), we define a new function \(g(x) = f(x) - x\), which is also continuous in the given interval. Check the boundary values: \(g(a) \geq 0\) and \(g(b) \leq 0\). Since the function values have opposite signs at the endpoints, we can use the Intermediate Value Theorem (IVT), which ensures that there is at least one point \(c \in [a, b]\) where \(g(c) = 0\), which implies that \(f(c) = c\). Geometrically, this means that the function \(f(x)\) intersects the line \(y=x\) at least one point in the interval \([a, b]\).

Step 1: Defining the problem

We have the function \(f(x)\) that is continuous in the interval \([a, b]\), and \(a \leq f(x) \leq b \forall x \in[a, b]\). We want to prove the existence of at least one point \(c \in [a, b]\) such that \(f(c)=c\).

Step 2: Define a new function to apply the Intermediate Value Theorem

Define a new function \(g(x) = f(x) - x\). The function \(g(x)\) is continuous because \(f(x)\) is continuous, and a sum/subtraction of continuous functions is continuous. Now, we have: \(g(x) = f(x) - x\) Our goal is to find a root for this function in the interval \([a, b]\). In other words, we are looking for a point \(c\) such that \(g(c) = 0\), which would consequently give us \(f(c) - c = 0 \Rightarrow f(c)=c\).

Step 3: Check boundary values

Check the boundary values of the new function: \(g(a) = f(a) - a \geq a - a = 0\) (since \(f(a) \geq a\)) \(g(b) = f(b) - b \leq b - b = 0\) (since \(f(b) \leq b\))

Step 4: Apply the Intermediate Value Theorem (IVT)

Since \(g(x)\) is continuous on the interval \([a, b]\), and \(g(a) \leq 0 \leq g(b)\), we can apply the Intermediate Value Theorem. The IVT states that there exists at least one value \(c \in [a, b]\) such that \(g(c) = 0\).

Step 5: Conclusion

By the IVT, there is a point \(c\) in the interval \([a, b]\) at which \(g(c) = 0\). Recalling the definition of the function \(g(x) = f(x) - x\), we obtain: \(g(c) = f(c) - c = 0 \Rightarrow f(c) = c\) Hence, there exists at least one point \(c \in [a, b]\) such that \(f(c) = c\).

Step 6: Geometrical Interpretation

Geometrically, the graph of the function \(f(x)\) intersects the graph of the line \(y=x\) at least one point in the interval \([a, b]\). This intersection point has the coordinates \((c, f(c))\), where \(f(c)=c\). This shows that the function \(f(x)\) equals its argument at a point \(c \in [a, b]\).

Key Concepts

Continuous Functions

When we speak about continuous functions, we're referring to a property of a function that ensures it does not have any interruptions or breaks. In essence, you could sketch the graph of a continuous function without lifting your pencil from the paper. This property is not only visually pleasing but has important implications in mathematics. For example, if a function is continuous on a closed interval \[a, b\], it will pass through every value between its maximum and minimum on that interval.

Continuous functions are essential when applying the Intermediate Value Theorem, because one of the requirements for using this theorem is that the function must be continuous on the interval being considered. In the context of our exercise, the function \(f(x)\) is given as continuous within the interval \[a, b\], which sets the stage for further analysis and application of the theorem to find roots.

Roots of Equations

The roots of an equation are the solutions that satisfy the equation, typically where the function's output is zero. In the context of the question, we're interested in finding where the function \(f(x)\) is equal to \(x\), or where the equation \(f(x) - x = 0\) holds true.

When we find a value \(c\) that satisfies \(f(c) - c = 0\), we have essentially found a 'root' of the equation \(g(x) = f(x) - x\). Uncovering the roots is a common goal in algebra and calculus, as it offers insight into the behavior of functions and has practical applications across various fields including physics, engineering, and economics.

Function Graphs and Intersection

The geometric interpretation of functions often involves analyzing their graphs. In algebra and calculus, an important aspect is determining where two function graphs intersect, which is the solution to the equation obtained by setting the functions equal to each other. In our textbook problem, we consider the graph of \(f(x)\) and the line \(y=x\).

The intersection points of these graphs represent the x-values where \(f(x)\) is equal to \(x\). If we plot both the function \(f(x)\) and the line \(y = x\) on the same coordinate plane, the x-coordinate of their intersection point(s) will give us the value(s) we're after. The Intermediate Value Theorem assures us at least one such intersection exists within the interval \[a, b\], given the conditions of the problem, reinforcing the deep connection between algebraic solutions and geometric representation.