Chapter 4: Problem 56
Check the differentiability of the function \(f(x)=|x-2|\) at \(x=2\).
Short Answer
Expert verified
The function \(f(x)=|x-2|\) is not differentiable at \(x=2\), because the left-hand derivative -1 is not equal to the right-hand derivative +1.
Step by step solution
01
Understand the function
Firstly, note that \(f(x)=|x-2|\) is a modulus function, which is absolute value function in case of real numbers. It measures the 'distance' of \(x\) from 2. Therefore, it will have a sharp turn or 'cusp' at \(x = 2\).
02
Calculate the left-hand derivative
The left-hand derivative at a point, is the derivative of the function as we approach that point from the left (smaller \(x\) values). Since the function is \(f(x)=|x-2|\), for \(x < 2\), |x-2| is equal to -(x-2) as the inside part of the absolute function equals a negative value. Therefore, we compute \(f'(x)\) for the part of the function where \(x < 2\), which is \(f'(x) = -1\).
03
Calculate the right-hand derivative
The right-hand derivative at a point is the derivative of the function as we approach that point from the right (larger \(x\) values). On this side of the cusp, for \(x > 2\), |x-2| is equal to \(x-2\), because the inside part of the absolute function equals a positive value. So the right-hand derivative of the function where \(x > 2\) is \(f'(x) = 1\).
04
Compare the left-hand and right-hand derivatives
The function will be differentiable at \(x = 2\) if the left-hand and right-hand derivatives are equal. However, as computed in step 2 and step 3, the left-hand derivative is -1 and the right-hand derivative is +1. Therefore, the left-hand derivative is not equal to the right-hand derivative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Modulus Function
A modulus function, also known as an absolute value function, is a fundamental concept in calculus and algebra. It is denoted by the symbol \(|x|\), and, in essence, it extracts the positive magnitude of a real number. For instance, the modulus of both -3 and 3 is 3. This function transforms negative inputs into positive outputs while keeping positive inputs unchanged. The equation of a modulus function could be written in the form \(|x-c|\) where \(c\) is some constant. Here, it measures the distance of \(x\) from \(c\) on a number line. In the exercise, we dealt with \(f(x) = |x - 2|\), which exhibits a distinct property: a sharp point or 'cusp', indicating potential non-differentiability.
Left-hand Derivative
To understand differentiability at a point, one needs to examine the concept of left-hand derivative. It represents the rate of change of the function as \(x\) approaches a particular point from the left. Mathematically, the left-hand derivative of a function \(f(x)\) at a point \(x = a\) is given by:
- \[ f'_-(a) = \lim_{{h \to 0^-}} \frac{{f(a+h) - f(a)}}{h} \]
Right-hand Derivative
The right-hand derivative, analogous to its left-hand counterpart, gauges how a function behaves as it approaches a designated point from the right. It's defined by the formula:
- \[ f'_+(a) = \lim_{{h \to 0^+}} \frac{{f(a+h) - f(a)}}{h} \]
Cusp
A 'cusp' emerges in a function where there is an abrupt directional transition, typified by a sharp point or corner in the graph. At these points, the function often fails to be differentiable, as seen with \(f(x) = |x-2|\) at \(x=2\). Instead of a smooth curve, the graph displays a distinct V-shape at the cusp, where both the left-hand and right-hand derivatives yield different values (specifically, -1 and +1). This disparity accentuates the non-differentiability at cusps because for a function to be differentiable at a particular point, its left-hand and right-hand derivatives must coincide, which is evidently not the case here. Such unique characteristics make cusps an intriguing yet challenging feature in calculus.