Problem 114 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The set of points where \(f(x)=\frac{x}{1+|x|}\) is differentiable is (A) \((-\infty, 0) \cup(0, \infty)\) (B) \((-\infty,-1) \cup(-1, \infty)\) (C) \((-\infty, \infty)\) (D) \((0, \infty)\)

Step-by-Step Solution

Answer

The set is (A) \((-\infty, 0) \cup (0, \infty)\).

Step 1: Identify the Function's Behavior

The function given is \( f(x) = \frac{x}{1 + |x|} \). The absolute value \(|x|\) affects the differentiability at \( x=0 \), so we will examine the function on the intervals \( (-\infty, 0) \) and \( (0, \infty) \).

Step 2: Consider the Left-side Derivative

For \( x < 0 \), the absolute value function becomes \( |x| = -x \). Thus, \( f(x) = \frac{x}{1 - x} \). This function is differentiable wherever \( 1 - x eq 0 \), i.e., at all points \( x eq 1 \). Since \( x < 0 \), the function is differentiable on \( (-\infty, 0) \).

Step 3: Consider the Right-side Derivative

For \( x > 0 \), the absolute value function becomes \( |x| = x \). Thus, \( f(x) = \frac{x}{1 + x} \). This function is differentiable wherever \( 1 + x eq 0 \), i.e., at all points \( x eq -1 \). Since \( x > 0 \), the function is differentiable on \( (0, \infty) \).

Step 4: Confirm Differentiability at x=0

To check differentiability at \( x=0 \), calculate the derivative from both sides. Left derivative: \( \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h}{1 - h^2} \). Right derivative: \( \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h}{1 + h^2} \). Both limits yield different results, confirming non-differentiability at \( x = 0 \).

Step 5: Determine the Differentiable Set

Since the function is not differentiable at \( x = 0 \) (Step 4), the set of points where it is differentiable is \( (-\infty, 0) \cup (0, \infty) \).

Key Concepts

Absolute Value

The absolute value function, denoted as \(|x|\), is a mathematical function that gives the non-negative value of \(x\). It is defined as \(|x| = x\) when \(x \geq 0\) and \(|x| = -x\) when \(x < 0\). This function is crucial because it changes the behavior of the numerator or denominator in expressions, affecting differentiability.
For example, in the function \((f(x) = \frac{x}{1+|x|})\), the absolute value \(|x|\) alters the expression based on whether \(x\) is positive or negative:

• When \(x < 0\), \(|x| = -x\), translating the function to \(f(x) = \frac{x}{1-x}\).
• When \(x \geq 0\), \(|x| = x\), making \(f(x) = \frac{x}{1+x}\).

This change must be considered when evaluating where a function is continuous or differentiable, especially at the points where \(|x|\) switches values.

Piecewise Functions

Piecewise functions are those defined by different expressions over different intervals of the domain. In the exercise, the function \((f(x) = \frac{x}{1+|x|})\) operates differently depending on whether \(x\) is positive or negative. Thus, it's evaluated as a piecewise function:

• For \(x < 0,\) the function is \(f(x) = \frac{x}{1-x}.\)
• For \(x \geq 0,\) it changes to \(f(x) = \frac{x}{1+x}.\)

Piecewise functions are instrumental in mathematics because they can model real-world scenarios where behavior changes across different conditions. The differentiability of a piecewise function can vary from one piece to another; hence, each piece must be assessed separately.
In this example, evaluating each part separately allowed us to determine that the function is not differentiable at \(x=0\), where these pieces converge but differ in slope. This illustrates a key challenge with piecewise functions: ensuring transitions between pieces are smooth to achieve differentiability.

Left and Right Derivatives

Left and right derivatives help assess whether a function is differentiable at a certain point. A function is said to be differentiable at a point if both the left and right derivatives at that point exist and are equal.
When dealing with the function \((f(x) = \frac{x}{1+|x|})\), it's important to evaluate it from both sides:

• Left derivative: For \(h \to 0^-\), the approach is from the left side. Calculate the limit as \((\frac{f(h) - f(0)}{h})\) approaches there.
• Right derivative: For \(h \to 0^+\), the approach is from the right side. Similarly, compute the limit of \((\frac{f(h) - f(0)}{h})\) as \(h\) goes to zero from the right.

In our exercise, these derivatives yielded different values, meaning the function is not differentiable at \(x=0\). This discrepancy is a classic signal in calculus that a function lacks smoothness or continuity at a point, highlighting the importance of checking both left and right limits when verifying differentiability.