Problem 241 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Let \(f(x)\) be defined in the interval \([-2,2]\) such that \(f(x)=-1, \quad-2 \leq x \leq 0\) \(=x-1, \quad 0

Step-by-Step Solution

Answer

\(g(x)\) is differentiable in the interval \((-2,2)\) since its derivative, \(g'(x)\), is continuous at all points in this interval. For \(-2 \leq x \leq 0\), \(g'(x) = 0\). For \(0 < x \leq 1\), \(g'(x) = 0\), and for \(1 < x \leq 2\), \(g'(x) = 2\).

Step 1: Find the derivative of \(f(x)\) on its defined intervals

: As \(f(x)\) is defined piecewise for two intervals, we must find its derivative for both of these intervals: 1. For the interval \(-2 \leq x \leq 0\), we have \(f(x) = -1\). The derivative of a constant function is always zero, so \(f'(x) = 0\) in this interval. 2. For the interval \(0 < x \leq 2\), we have \(f(x) = x - 1\). Taking the derivative, we have \(f'(x) = 1\) in this interval.

Step 2: Define \(g(x)\) with respect to \(x\)

: We are given that \(g(x) = f(|x|) + |f(x)|\). We will now define \(g(x)\) with respect to \(x\). Since we already have derivatives for \(f(x)\) on both intervals, we need to define \(g(x)\) in terms of these derivatives: 1. For the interval \(-2 \leq x \leq 0\), we have \(f(x) = -1\) and \(f'(x) = 0\). Therefore, \(g(x) = f(|x|) + |f(x)| = -1 + 1 = 0\). 2. For the interval \(0 < x \leq 2\), we have \(f(x) = x - 1\) and \(f'(x) = 1\). Therefore, \(g(x) = f(|x|) + |f(x)| = x - 1 + |x - 1|\).

Step 3: Find the derivative of \(g(x)\) on its defined intervals

: 1. For the interval \(-2 \leq x \leq 0\), as \(g(x) = 0\), the derivative \(g'(x) = 0\). 2. For the interval \(0 < x \leq 2\), we have \(g(x) = x - 1 + |x - 1|\). To find the derivative of \(g(x)\) in this interval, we will have to analyze the absolute value term. We have two cases. - When \(x > 1\), we have \(|x-1| = x-1\). Thus, \(g(x) = x - 1 + x - 1 = 2x - 2\). Taking the derivative, we have \(g'(x) = 2\). - When \(0 < x \leq 1\), we have \(|x-1| = -(x-1)\). Thus, \(g(x) = x - 1 - (x - 1) = 0\). So, \(g'(x) = 0\).

Step 4: Check for differentiability of \(g(x)\)

: To check for differentiability, we need to make sure \(g'(x)\) is continuous. Recall that \(g'(x) = 0\) when \(-2 \leq x \leq 0\) and has different values when \(0 < x \leq 2\). The only point where \(g'(x)\) may not be continuous is when \(x=0\). To check for continuity, we check the value of \(g'(x)\) as it approaches \(0\) from left and right: - \(\lim_{x \to 0^-} g'(x) = 0\) - \(\lim_{x \to 0^+} g'(x) = 0\) Since the two limits are equal, \(g'(x)\) is continuous at \(x=0\). So, \(g(x)\) is differentiable over the interval \((-2,2)\).

Key Concepts

Piecewise functions

Piecewise functions are a great way to define functions that have different expressions for different parts of their domain. Essentially, it is like creating a function using multiple "pieces." In this exercise, the function \(f(x)\) is piecewise, which means it is composed of two parts:

• From \(-2 \leq x \leq 0\), \(f(x) = -1\)
• From \(0 < x \leq 2\), \(f(x) = x - 1\)

By using separate expressions for different intervals, piecewise functions provide flexibility to model diverse scenarios, like financial calculations, based on conditions. Working with these functions often requires analyzing each piece separately. This includes tasks like taking derivatives or evaluating continuity and differentiability at boundary points. By understanding and managing these segments, one can see the comprehensive behavior of the entire function.

Absolute value function

The absolute value function, denoted by \(|x|\), is a simple yet powerful tool in mathematics. It transforms any negative input into a positive output, while leaving positive numbers unchanged. In mathematical terms, the absolute value of a number is its distance from zero on the number line.

• If \(x \geq 0\), then \(|x| = x\)
• If \(x < 0\), then \(|x| = -x\)

In this exercise, the absolute value function affects the function \(f(x)\) by ensuring that any negative values it produces become positive in \(g(x) = f(|x|) + |f(x)|\). When dealing with absolute values, especially in piecewise functions, one must consider the influence of switching signs on different intervals. This often requires careful examination of the function's behavior at key points, like wherever the algebraic expression inside the absolute value equals zero.

Continuity and differentiability

A function is considered continuous if it can be drawn without lifting the pencil from the paper. Differentiability, on the other hand, is a step further and means that the function's graph not only is smooth but has a well-defined tangent line at every point within its domain.Continuity is a prerequisite for differentiability. If a function is not continuous at a point, it cannot be differentiable there. For differentiability, the function's derivative – the rate at which the function's value changes – needs to exist and be continuous.In our exercise, \(g(x)\) needs to be checked for differentiability. This involves ensuring its derivative is continuous over the interval (-2, 2). We find discontinuities by looking for jumps, spikes, or vertical tangents in the derivative. For this specific problem, after finding derivatives for each piece, we need to ensure they align smoothly at transition points, particularly at \(x = 0\). When these derivatives match from both sides of a point, it verifies their continuity and thus the differentiability of the function.

Interval notation

Interval notation is a simple and concise way to describe the set of all real numbers between two endpoints. It uses brackets and parentheses to indicate whether endpoints are included or not.There are different symbols used to represent intervals:

• \([a, b]\) means all numbers from \(a\) to \(b\) including both \(a\) and \(b\).
• \((a, b)\) means all numbers strictly between \(a\) and \(b\), excluding \(a\) and \(b\).
• \([a, b)\) or \((a, b]\) mix inclusion and exclusion at one endpoint.

In this exercise, the function \(f(x)\) is defined using the interval notation \([-2, 0]\) and\( (0, 2]\), illustrating where the different pieces of the piecewise definition apply. Using interval notation helps clarify which sections of a function's domain are affected by different expressions, making it a vital tool in calculus for ensuring precision, especially when determining domains and ranges.