Problem 16 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Prove that } f(x)=|\ln x| \text { is continuous but not differentiable at } x=1 \text { . } $$

Step-by-Step Solution

Answer

The function \(f(x)=|\ln x|\) is continuous at \(x=1\) because \(\lim_{x \rightarrow 1} f(x) = f(1) = 0\). However, it is not differentiable at this point. For \(x > 1\), the derivative is \(f'(x) = \frac{1}{x}\), while for \(x < 1\), the derivative is \(f'(x) = -\frac{1}{x}\). Since the left and right derivatives at \(x = 1\) do not match (\(f'(1^-) = -1\) and \(f'(1^+) = 1\)), the function is not differentiable at \(x=1\).

Step 1: First, let's understand what the given function represents. It is \(f(x) = |\ln x|\), which is the absolute value of the natural logarithm of \(x\). The absolute value ensures that the function is always non-negative. #Step 2: Proving continuity at x = 1#

In order to show that \(f(x)\) is continuous at \(x = 1\), we need to show that \(\lim_{x \rightarrow 1} f(x) = f(1)\). Let's calculate \(f(1)\): \[f(1) = |\ln(1)| = |0| = 0\] Now let's find \(\lim_{x \rightarrow 1} f(x)\): \[\lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} |\ln x|\] Since \(|\ln x|\) is always non-negative, the value of the function approaches 0 as \(x\) approaches 1: \[\lim_{x \rightarrow 1} f(x) = 0\] Therefore, since \(\lim_{x \rightarrow 1} f(x) = f(1)\), the function is continuous at \(x = 1\). #Step 3: Finding the derivative of the function#

Step 2: To check the differentiability at \(x = 1\), we need to find the derivative of \(f(x)\). As \(f(x) = |\ln x|\), we need to consider two cases: when \(\ln x\) is positive, and when \(\ln x\) is negative. Case 1: \(\ln x > 0\) \[f'(x) = \frac{d(\ln x)}{dx}\] Case 2: \(\ln x < 0\) \[f'(x) = \frac{d(-\ln x)}{dx}\] Now let's differentiate \(\ln x\) and \(-\ln x\): Case 1: \[f'(x) = \frac{1}{x}\] Case 2: \[f'(x) = -\frac{1}{x}\] #Step 4: Analyzing the derivative at x = 1#

Now, let's determine if the derivative exists at \(x = 1\). In the first case, as \(x\) approaches 1 from the right (\(x > 1\)), \(\ln x > 0\) and thus: \[f'(x) = \frac{1}{x}\] When \(x = 1\), the first case gives: \[f'(1) = 1\] In the second case, as \(x\) approaches 1 from the left (\(x < 1\)), \(\ln x < 0\) and thus: \[f'(x) = -\frac{1}{x}\] When \(x = 1\), the second case gives: \[f'(1) = -1\] Since the value of the derivative from the left does not equal the value of the derivative from the right, the derivative does not exist at \(x = 1\). #Conclusion# The given function \(f(x) = |\ln x|\) is continuous, as shown in step 2, but not differentiable at \(x = 1\), as shown in step 4.

Key Concepts

Absolute Value Function

An absolute value function takes a real number and returns its non-negative magnitude. For example, the absolute value of both 3 and -3 is 3. This property ensures that the output of an absolute value function is always zero or positive. Given a function of the form \( f(x) = |g(x)| \), it computes the absolute value of another function \( g(x) \).

Key characteristics of the absolute value function \( |g(x)| \):

• If \( g(x) > 0 \), then \( |g(x)| = g(x) \).
• If \( g(x) < 0 \), then \( |g(x)| = -g(x) \).
• If \( g(x) = 0 \), then \( |g(x)| = 0 \).

The function in our problem, \( f(x) = |\ln x| \), uses this property to ensure that its output is never negative. At \( x = 1 \), since \( \ln 1 = 0 \), \( |\ln x| \) simply equals zero.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is the power to which the base of the natural logarithms, \( e \), must be raised to obtain the value \( x \).

It is a critical function in mathematics with several noteworthy properties:

• \( \ln 1 = 0 \), since \( e^0 = 1 \).
• The function \( \ln x \) is undefined for \( x \leq 0 \).
• It is a monotonically increasing function, meaning it rises steadily as \( x \) increases.

In our function \( f(x) = |\ln x| \), \( \ln x \) changes its behavior around 1. Exactly at \( x = 1 \), \( \ln 1 = 0 \). To the left of 1, \( \ln x \) is negative, making \( f(x) \) take the negative of \( \ln x \). On the right, \( \ln x \) is positive, and \( f(x) \) equals \( \ln x \) directly.

Differentiability

Differentiability of a function at a point means the function has a well-defined derivative at that point. Derivatives represent the rate of change or the slope of a function's graph.

To check a function's differentiability, consider:

• The function must be continuous at the point.
• The left-hand derivative must equal the right-hand derivative.

For \( f(x) = |\ln x| \), differentiability was assessed at \( x = 1 \). To the right of \( x = 1 \) (\( x > 1 \)), the derivative is \( f'(x) = \frac{1}{x} \). To the left (\( x < 1 \)), it's \( f'(x) = -\frac{1}{x} \).

Because the derivatives approaching from the left and the right are not equal, \( f(x) \) is not differentiable at \( x = 1 \). This type of behavior, a sharp point or cusp on a graph, is where functions often fail to be differentiable.

Continuity

Continuity at a point means that as you approach that point from either direction, the function's value approaches a single, finite value. For a function \( f(x) \) to be continuous at \( x = a \), the following must hold:

• \( f(a) \) should be defined.
• \( \lim_{{x \to a}} f(x) \) should exist.
• \( \lim_{{x \to a}} f(x) = f(a) \).

For our function \( f(x) = |\ln x| \), at \( x = 1 \), it was shown:

• \( f(1) = |\ln(1)| = 0 \).
• \( \lim_{{x \to 1}} |\ln x| = 0 \), from both directions.

Thus, \( f(x) \) is continuous at \( x = 1 \) even though it's not differentiable there. This is a common situation with functions involving absolute values or abrupt changes.