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Chapter 4: Problem 74

Check the differentiability of the function \(f(x)=\sin x+\sin |x|\) in \([-2 \pi, 2 \pi] .\)

Short Answer

Expert verified
The function \(f(x) = \sin x + \sin |x|\) is differentiable in the interval \([-2\pi, 2\pi]\).

Step by step solution

01

Identify function within the given interval

Firstly, split the interval into three regions: \(x < 0\), \(x = 0\), and \(x > 0\) which correspond to \(-2\pi \leq x < 0\), \(x = 0\), and \(0 < x \leq 2\pi\) respectively due to the modulus function.
02

Simplify the function

Since \(|x|\) equals \(x\) for \(x > 0\) and \(-x\) for \(x < 0\), the function simplifies to \(f(x) = \sin x + \sin x\) for \(x > 0\) and \(f(x) = \sin x - \sin x\) for \(x < 0\). At \(x = 0\), \(|x| = 0\), thus the function takes the form \(f(x) = \sin x + \sin 0\).
03

Find the derivative function

The derivative of \(f(x) = 2\sin x\) for \(x > 0\) and \(f(x) = 0\) for \(x < 0\). At \(x = 0\) the function is continuous, the left hand derivative equals the right hand derivative, so it is differentiable here.
04

Conclude differentiability

Since the function is differentiable at all points in the interval \([-2\pi, 2\pi]\), including at \(x = 0\), it is differentiable within the entire interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Modulus Function
The modulus function, symbolized as \(|x|\), is an essential concept in calculus. It represents the absolute value of \(x\). This means it tells you how far \(x\) is from zero, regardless of the direction on the number line. Here's what that looks like:
  • For any positive value \(x > 0\), \(|x| = x\).
  • For any negative value \(x < 0\), \(|x| = -x\).
This characteristic causes the function to behave differently on either side of zero, which is why it's important to examine both sides when determining differentiability. In the given exercise, for \(x < 0\), the modulus changes \(\sin |x|\) to \(\sin (-x)\), effectively flipping the sign. This behavior influences how we analyze derivatives of functions involving the modulus.
Exploring the Left-Hand Derivative
When examining a function's differentiability at a specific point, the left-hand derivative is crucial. This type of derivative considers the slope of the function as it approaches from the left side of a specific point. For the function in the exercise, here's how it works:
  • Calculate the derivative of the function for values of \(x\) just below the target point.
  • If we target \(x = 0\), for \(x < 0\), the function simplifies to \(f(x) = \sin x - \sin x = 0\). Hence, the left-hand derivative at \(x = 0\) is 0.
The role of the left-hand derivative is to ensure that the function does not have a sudden change in slope or direction as it approaches the point from the left side. It provides a clear picture of the function's behavior as we come from negative values toward zero.
Understanding the Right-Hand Derivative
The right-hand derivative, like its counterpart, assesses the slope of a function as it approaches a certain point, but this time from the right. This derivative checks the behavior on the positive side of the point.
  • To find it, calculate the derivative for values of \(x\) slightly greater than the point of interest.
  • For \(x = 0\) in the exercise, with \(x > 0\), the function simplifies to \(f(x) = 2\sin x\). Thus, the right-hand derivative at \(x = 0\) is the derivative of \(2\sin x\), which equals 2 at \(x = 0\).
The right-hand derivative confirms how the function extends from the zero point towards positive values and ensures no abrupt slope change. For overall differentiability, both the left and right hand derivatives should match at the point of interest, indicating smooth transit across that point.

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Most popular questions from this chapter

Find the number of points of discontinuity of the function \(f(f(f(x)))\), where \(f(x)=1 / 1-x\).

Let \(f(x)=\left|x^{3}\right| .\) Examine whether the function is twice differentiable or not.

The left hand derivative of \(f(x)=[x] \sin (\pi x)\) at \(x=k\), where \(k\) is an integer, is (a) \((-1)^{k}(k-1) \pi\) (b) \((-1)^{k-1}(k-1) \pi\) (c) \((-1)^{k} k \pi\) (d) \((-1)^{k-1} k \pi\).

Let \(f(x)=x^{3}-x^{2}+x+1\) and \(g(x)= \begin{cases}\max \cdot \mid f(t): 0 \leq t \leq x\\} & : 0 \leq x \leq 1 \\ 3-x & : 1

If \(f(x)=\left\\{\begin{array}{lll}-1: & x<0 \\ 0 & : & x=0 \text { and } g(x)=x\left(1-x^{2}\right) \\ 1 & : & x>0\end{array}\right.\) then discuss the continuity of the function \(h(x)\), where \(h(x)=f(g(x))\).
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