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Chapter 2: Problem 69

Find all points where \(f\) fails to be differentiable. Justify your answer. \(\begin{array}{llll}{\text { (a) } f(x)=|3 x-2|} & {\text { (b) } f(x)=\left|x^{2}-4\right|} & {}\end{array}\)

Short Answer

Expert verified
Function (a) is not differentiable at \(x = \frac{2}{3}\). Function (b) is not differentiable at \(x = -2\) and \(x = 2\).

Step by step solution

01

Analyze function (a) for differentiability

The function given is \( f(x) = |3x - 2| \). The absolute value function can fail to be differentiable at points where the expression inside the absolute value equals zero, as this can create a cusp. We solve \( 3x - 2 = 0 \) to find any potential non-differentiable points.\( 3x = 2 \) leads to \( x = \frac{2}{3} \). We need to check if \( f(x) \) is differentiable at \( x = \frac{2}{3} \).
02

Check left and right derivatives for function (a)

To check if \( f \) is differentiable at \( x = \frac{2}{3} \), find the left-hand derivative \( \lim_{h \to 0^-} \frac{|3(x+h)-2| - |3x-2|}{h} \) and the right-hand derivative \( \lim_{h \to 0^+} \frac{|3(x+h)-2| - |3x-2|}{h} \). Calculating for \( x = \frac{2}{3} \), from the left side, \( \frac{|3(x+h)-2|} = -3 \) as \( 3x-2 \) changes sign, and from the right side, \( \frac{|3(x+h)-2|} = 3 \). Since the left and right derivatives are not equal to each other, the function is not differentiable at \( x = \frac{2}{3} \).
03

Analyze function (b) for differentiability

The function is \( f(x) = |x^2 - 4| \). Here, the expression inside the absolute value \( x^2 - 4 \) is equal to zero at points where the function may be non-differentiable. Solving \( x^2 - 4 = 0 \), we find the solutions \( x = 2 \) and \( x = -2 \). We need to investigate these points further.
04

Check derivatives for function (b) at potential points

To verify if differentiability fails at \( x = 2 \) and \( x = -2 \), calculate the left-hand and right-hand derivatives at these points similarly as earlier. For \( x = 2 \), as you approach from the left \( x < 2 \), the derivative is negative, while from the right \( x > 2 \), the derivative is positive; they don't match, so not differentiable at \( x = 2 \). Similarly for \( x = -2 \), the left derivative matches neither with the right derivative, hence \( f \) is not differentiable at \( x = -2 \).

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Key Concepts

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

Absolute Value Function
The absolute value function, denoted as \( |x| \), represents the distance of a number from zero on the number line. This function always returns a non-negative number, regardless of whether the input is positive or negative. Thus, for any real number \( x \), the absolute value is defined as \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).

When analyzing functions like \( f(x) = |3x - 2| \) or \( f(x) = |x^2 - 4| \), we look for points where the expression inside the absolute value becomes zero. These are spots where the function inside might change its behavior, often forming points known as cusps. Cusps are points where a function could fail to be smooth, hence not differentiable.
  • Find these critical points by setting the inside of the absolute value to zero and solving for \( x \).
  • For \( f(x) = |3x - 2| \), solve \( 3x - 2 = 0 \) to find \( x = \frac{2}{3} \).
  • For \( f(x) = |x^2 - 4| \), solve \( x^2 - 4 = 0 \) to find \( x = 2 \) and \( x = -2 \).
Left-Hand Derivative
Examining the left-hand derivative involves approaching the point of interest from the left, or from smaller values. This process helps us understand the behavior of a function approaching that critical point. If a function is differentiable at a point, the left-hand and right-hand derivatives must equal.

To calculate the left-hand derivative at point \( x = a \), we use the limit:\[\lim_{h \to 0^-} \frac{f(a + h) - f(a)}{h}\]

For the function \( f(x) = |3x - 2| \) at \( x = \frac{2}{3} \), compute this limit as approaching from values less than \( \frac{2}{3} \). Since the absolute value changes sign at this point:
  • Left-Hand Derivative outcome: negative value implies differing slope than approaching from the right.
If these calculations yield differing results than approaching from the right side of \( x=a \), it indicates non-differentiability at that point.
Right-Hand Derivative
The right-hand derivative allows us to see how a function behaves as we approach a specific point from the right (larger values). Similar to the left-hand derivative, the right-hand derivative is needed to determine if a function is smooth at a point.

To find the right-hand derivative at \( x = a \), use the limit:\[\lim_{h \to 0^+} \frac{f(a + h) - f(a)}{h}\]

Consider \( f(x) = |3x - 2| \) specifically at \( x = \frac{2}{3} \). Approaching from values greater than \( \frac{2}{3} \):
  • Right-Hand Derivative outcome: positive result signifies a shift in slope direction compared to coming from the left.
If the left and right derivatives do not match, the function shows a cusp at \( x=a \), confirming it is not differentiable at that specific point.

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Most popular questions from this chapter

Find \(f^{\prime}(x)\) $$ f(x)=\sqrt{\cos (5 x)} $$

Newton's Law of Universal Gravitation states that the magnitude \(F\) of the force exerted by a point with mass \(M\) on a point with mass \(m\) is $$ F=\frac{G m M}{r^{2}} $$ $$ \begin{array}{l}{\text { where } G \text { is a constant and } r \text { is the distance between the bod- }} \\ {\text { ies. Assuming that the points are moving, find a formula for }} \\ {\text { the instantaneous rate of change of } F \text { with respect to } r .}\end{array} $$

Determine whether the statement is true or false. Explain your answer. $$ \text { If } y=\sin ^{3}\left(3 x^{3}\right), \text { then } d y / d x=27 x^{2} \sin ^{2}\left(3 x^{3}\right) \cos \left(3 x^{3}\right) $$

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0} .\) You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section \(4.8) .\) Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0},\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow x_{0}} f^{\prime}(x) .\) $$ \begin{array}{l}{\text { Let } \quad f(x)=\left\\{\begin{array}{ll}{x^{2},} & {x \leq 1} \\ {\sqrt{x},} & {x>1}\end{array}\right.} \\ {\text { Determine whether } f \text { is differentiable at } x=1 . \text { If so, find }} \\\ {\text { the value of the derivative there. }}\end{array} $$

Suppose that the function \(f\) is differentiable everywhere and \(F(x)=x f(x)\) (a) Express \(F^{\prime \prime \prime}(x)\) in terms of \(x\) and derivatives of \(f\) (b) For \(n \geq 2\), conjecture a formula for \(F^{(n)}(x)\)
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