Problem 240 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Draw the graph of the function \(y=|x-1|+|x-2|\) in the interval \([0,3]\) and discuss the continuity and differentiability of the function in this interval. \\{Ans. continuous, not differentiable at \(x=1 \& 2\\}\)

Step-by-Step Solution

Answer

The graph of the function \(y = |x-1| + |x-2|\) in the interval \([0, 3]\) consists of three line segments: \(y = 3-x\) for \(x<1\), \(y=1\) for \(1\le x<2\), and \(y=x-1\) for \(2\le x\). The function is continuous in the interval but not differentiable at \(x = 1\) and \(x = 2\), as there are sharp corners at these points.

Step 1: Understand the given function and domain

The given function is \(y = |x-1| + |x-2|\), with domain \([0, 3]\). We need to graph this function in the given domain and determine its continuity and differentiability.

Step 2: Simplify the function in each interval

First, we determine the intervals in which the absolute value expressions can be simplified. Absolute value functions change behavior at the points, where the expression inside them is equal to zero: - For \(x - 1 = 0\), \(x = 1\) - For \(x - 2 = 0\), \(x = 2\) According to these points, our function consists of three pieces (\(x < 1\), \(1 \le x < 2\), and \(2 \le x\)): 1. If \(x < 1\), then \(x-1<0\) and \(x-2<0\), so the function simplifies to \(y = -(x-1) -(x-2) = 3-x\). 2. If \(1 \le x < 2\), then \(x-1\geq0\) and \(x-2<0\), so the function simplifies to \(y = (x-1) -(x-2) = 1\). 3. If \(2 \le x\), then \(x-1\geq0\) and \(x-2\geq0\), so the function simplifies to \(y = (x-1) + (x-2) = x-1\). Now, we have simplified the function as follows: \(y(x) = \begin{cases} 3 - x, & x < 1 \\ 1, & 1 \le x < 2 \\ x - 1, & 2 \le x \end{cases} \)

Step 3: Sketch the graph of the function

Next, we sketch the graph of the simplified function, in the interval \([0, 3]\): 1. For \(0 \le x < 1\), the graph is a line with negative slope, starting from \((0, 3)\) and ending at \((1, 2)\). 2. For \(1 \le x < 2\), the graph is a horizontal line with the height of \(1\), starting from \((1, 1)\) and ending at \((2, 1)\). 3. For \(2 \le x \le 3\), the graph is a line with positive slope, starting from \((2, 1)\) and ending at \((3, 2)\).

Step 4: Determine continuity and differentiability

The function is continuous in the given interval \([0, 3]\) because each piece of the function connects smoothly to the adjacent pieces at the end points. However, the function is not differentiable at \(x = 1\) and \(x = 2\), because there are sharp turning points (corners) in the graph at these points. At such corners, the derivative of the function doesn't exist, as the slopes of the tangents to the graph from the left and right are different. In conclusion, the given function is continuous but not differentiable at \(x = 1\) and \(x = 2\) in the interval \([0, 3]\).

Key Concepts

Continuity

A function is said to be continuous if you can draw its graph without lifting your pen from the paper. This means that there are no breaks, holes, or gaps in the function's graph over its domain. For the function \(y = |x-1| + |x-2|\), continuity is assessed over the interval \([0, 3]\).

• The intervals are divided based on the critical points where the function changes form, specifically at \(x = 1\) and \(x = 2\).
• In this case, each piece of the function seamlessly transitions into the next at these points, ensuring there are no breaks in the graph.
• The graph has no interruptions, allowing us to say that the function is continuous over \([0, 3]\).

Thus, you can confidently say the function is continuous, as it shows no discontinuations within the given range.

Differentiability

Differentiability refers to the ability to determine a function's derivative at any given point. For a function to be differentiable, it must be smooth; that is, its graph can't have any sharp turns or corners. In the original exercise, the function \(y = |x-1| + |x-2|\) is not differentiable at \(x = 1\) and \(x = 2\). Here's why:

• At \(x = 1\), the graph transitions from a negative slope in the interval \(x < 1\) to a flat horizontal line in the interval \(1 \le x < 2\). This creates a corner, indicating a change in direction that is too abrupt for differentiability.
• Similarly, at \(x = 2\), the function again switches abruptly, this time from a horizontal line to a positive slope. This sudden change resembles another corner.
• At these points, the derivative does not exist because the limit approaching from the left does not equal the limit approaching from the right.

In summary, despite the continuity of the function, the presence of sharp corners at \(x = 1\) and \(x = 2\) make it non-differentiable at these points.

Piecewise Functions

Piecewise functions are functions defined by different expressions depending on the interval of the input value, \(x\). The complexity of their form often stems from the absolute value expressions they may contain and how these are resolved over specific intervals. Let's break down the function \(y = |x-1| + |x-2|\):

• The function is split into three intervals: \(x < 1\), \(1 \le x < 2\), and \(2 \le x\).
• For \(x < 1\), both \(x-1\) and \(x-2\) are negative, leading us to the expression \(3-x\).
• At \(1 \le x < 2\), \(x-1\) is non-negative while \(x-2\) is still negative, simplifying the function to a constant value, \(1\).
• For \(2 \le x\), both \(x-1\) and \(x-2\) are non-negative, resulting in the expression \(x-1\).

Each piece corresponds to a linear segment whose parameters are dictated by the behavior of the absolute value terms over the specified intervals. By understanding the breakdown into simple line equations, one can effectively interpret and graph the piecewise function.