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Chapter 4: Problem 75

Check the differentiability of the function \(f(x)=\min \\{|x+1|,|x|,|x-1|\\}\) in \([-4,4]\)

Short Answer

Expert verified
The function \(f(x)=\min \{|x+1|,|x|,|x-1|\}\) is continuous in \([-4,4]\), but it's not differentiable at \(x=-1\) and \(x=1\).

Step by step solution

01

Define the given function

The function is given as \(f(x)=\min \{|x+1|,|x|,|x-1|\}\) and needs to be checked for differentiability in the interval \([-4,4]\). This function has three separate parts defined by the absolute value functions.
02

Define the ranges for each function

Since the absolute value functions compare the distances from \(x=0\), \(x=1\) and \(x=-1\) respectively, each function will be 'dominant' in different parts of the \([-4,4]\) interval: |x+1| for \(x < -1\), |x| for \(-1 \leq x \leq 1\), and |x-1| for \(x > 1\).
03

Differentiate the functions

Each function can be differentiated as follows: For |x+1|, the derivative is -1 for \(x < -1\); for |x|, the derivative is 1 for \(-1 < x < 1\), and undefined for \(x=-1\) and \(x=1\); for |x-1|, the derivative is 1 for \(x > 1\).
04

Check for Discontinuities

The given function is continuous for all \(x\) as the min function always yields a value at a given point. However, it's not differentiable at \(x=-1\) or \(x=1\) because as we have seen, the derivative of |x| is undefined at those points.

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