Problem 47 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(f(x)=1+x, \quad 0 \leq x \leq 2\) \(=3-x, \quad 2

Step-by-Step Solution

Answer

The function \(f(f(x))\) is defined as: $$ f(f(x)) = \begin{cases} 2-x, & 0 \leq x \leq 2 \\ 4-x, & 2 < x \leq 3 \end{cases} $$ The function is continuous in the intervals \(0 \leq x \leq 2\) and \(2< x\leq3\), but not continuous at the point \(x=2\).

Step 1: Determine the function \(f(f(x))\) for the given domain:

To determine \(f(f(x))\), we need to compute the composition \(f(f(x))\) for the given domain of \(f(x)\). We know that \(f(x) = 1+x\) for \(0\leq x\leq 2\), and \(f(x)=3-x\) for \(2 2\), we have that \(0\le q 3-x\le 1\), thus the composition is of the form: $$ f(f(x)) = f(3-x) = 1+(3-x) = 4-x $$ Then, we have \(f(f(x))\) defined as: $$ f(f(x)) = \begin{cases} 2-x, & 0 \leq x \leq 2 \\ 4-x, & 2 < x \leq 3 \end{cases} $$

Step 2: Test continuity of \(f(f(x))\)

To test the continuity of \(f(f(x))\), we will first check continuity of each part of the function, and then continuity at the meeting point \(x=2\). For \(0 \leq x \leq 2\), \(f(f(x))=2-x\), which is a linear function and is continuous. For \(2 < x \leq 3\), \(f(f(x))=4-x\), which is also a linear function and is continuous. Now we need to check for continuity at \(x=2\). To check the continuity at the meeting point, we need to see if the functions match in value and therefore the limit as \(x\) approaches \(2\) exists. $$ \lim_{x\to 2^-} (2-x) = \lim_{x\to 2^+} (4-x) $$ Substituting \(x = 2\) in both expressions, we get: $$ 2-2 = 4-2 \Rightarrow 0 = 2 $$ Since these values are different, the function \(f(f(x))\) is not continuous at \(x = 2\). In conclusion, the function \(f(f(x))\) is continuous in the intervals \(0 \leq x \leq 2\) and \(2< x\leq3\), but not continuous at the point \(x=2\).

Key Concepts

Piecewise Functions

Piecewise functions are a type of mathematical function that are defined using different expressions based on different parts of their domain. In simpler terms, a piecewise function stitches together several functions, each part having its own rule, or expression, for its specific interval. This allows for versatility and the ability to represent complex situations more accurately.
Think of a piecewise function as a function with split personality. It acts one way under certain conditions and another way when those conditions change. The key takeaway is that each piece of the function applies over a specific range of numbers (called intervals), and the endpoints of these intervals are usually critical points where the behavior of the function might change.
In our exercise, the function \( f(x) \) is piecewise because it is defined by two different expressions depending on whether \( x \) is between 0 and 2, or between 2 and 3. These conditions are expressed clearly as intervals, making it easy to identify and work with each segment of the function.

Continuity Testing

Testing for continuity involves determining whether a function behaves in an unbroken or 'smooth' way across its entire domain. For a function to be continuous at a point, there should be no jumps or holes in its graph. More formally, a function \( g(x) \) is continuous at a point \( c \) if three conditions are met:

• \( g(c) \) is defined.
• The limit of \( g(x) \) as \( x \) approaches \( c \) exists.
• The limit of \( g(x) \) as \( x \) approaches \( c \) is equal to \( g(c) \).

With piecewise functions, it is particularly important to check continuity at the points where the pieces meet, called meeting points. In our problem, testing for continuity required checking whether the composed function \( f(f(x)) \) was continuous at the meeting point \( x = 2 \). While the individual parts were continuous, the meeting point revealed a discontinuity because the limits from the left and right at \( x = 2 \) did not match.

Limits at a Point

Limits are a fundamental concept in calculus that describe the behavior of a function as it approaches a particular point. When we talk about the limit of a function \( h(x) \) as \( x \) approaches some value \( c \), we are interested in what \( h(x) \)'s values are getting close to, not necessarily what happens exactly at \( c \).
In continuity testing, examining limits ensures that the function transitions smoothly at critical points, like boundaries in a piecewise function.
For the exercise's function \( f(f(x)) \), we checked the limit at \( x = 2 \) to confirm smoothness at this point. Calculating \( \lim_{x \to 2^-}(2-x) \) and \( \lim_{x \to 2^+}(4-x) \) showed that they were unequal, leading to our conclusion of discontinuity at this point. This mismatch is a clear signal that the function "jumps," confirming the function is not continuous at the meeting point.