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Chapter 3: Problem 66

A caution about centered difference quotients (Continuation of Exercise \(65 . )\) The quotient $$ \frac{f(x+h)-f(x-h)}{2 h} $$ may have a limit as \(h \rightarrow 0\) when \(f\) has no derivative at \(x .\) As a case in point, take \(f(x)=|x|\) and calculate $$ \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} $$ As you will see, the limit exists even though \(f(x)=|x|\) has no derivative at \(x=0 .\) Moral: Before using a centered difference quotient, be sure the derivative exists.

Short Answer

Expert verified
The limit exists and is equal to 1.

Step by step solution

01

Understanding the Expression

The centered difference quotient given is \( \frac{f(x+h)-f(x-h)}{2 h} \). For this specific problem, we're asked to evaluate \( \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} \). Here, \( f(x) = |x| \).
02

Simplifying the Expression for \(|0+h| - |0-h|\)

Since we're evaluating at \( x=0 \), and we need to find \( \lim _{h \rightarrow 0} \), the expression becomes \(|h| - |-h|\). Given the properties of absolute value, we have \(|h| = h\) when \( h > 0 \) and \(|h| = -h\) when \( h < 0 \). Thus, \(|0+h| = |h|\) and \(|0-h| = |h|\).
03

Evaluating the Expression \(|h| - |h|\)

For all \( h eq 0 \), the expression simplifies as follows: \(|h| - |h| = h - (-h)\). Therefore, \(|0+h|-|0-h| = h + h = 2h\).
04

Substitute Back into The Fraction

Substitute \(2h\) back into the expression \( \frac{|0+h|-|0-h|}{2h} = \frac{2h}{2h} \). This simplifies to \(1\).
05

Take the Limit as \(h \rightarrow 0\)

Since the expression simplifies to \(1\) regardless of \(h\) approaching zero from either the negative or positive direction, the limit \( \lim _{h \rightarrow 0} \frac{|0+h|-|0-h|}{2 h} \) is \(1\).

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Key Concepts

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

Centered Difference Quotient
The centered difference quotient is a way to estimate the derivative of a function at a given point. It is defined as the expression \( \frac{f(x+h)-f(x-h)}{2 h} \). This method uses two points that are equally spaced around a particular point \( x \), instead of the traditional one-sided approaches. This can sometimes provide a more balanced estimate.
  • In essence, it takes the average rate of change around the point, giving a potentially clearer picture of behavior there.
  • However, caution is advised as the method may suggest the existence of a derivative where none actually exists.
A notable characteristic of the centered difference quotient, as highlighted by our original exercise, is its ability to produce a limit even when the actual derivative does not exist at a point. This illustrates its strength in approximation but also signals a need for careful application, ensuring that any conclusions about differentiability are appropriate.
Absolute Value Function
The absolute value function, denoted by \(|x|\), measures the distance of a number from zero on the real number line, disregarding its sign. Its main feature causes some unique behaviors in calculus when compared to other standard functions.
  • The function is defined as \( |x| = x \) if \( x \geq 0 \) and \( |x| = -x \) if \( x < 0 \).
  • This piecewise definition can lead to non-smooth points where the function changes behavior, such as \( x = 0 \).
In the context of the original exercise, we explored how this piecewise nature of the absolute value function influences calculations with the centered difference quotient. The limit calculation illustrated the symmetrical nature of \(|x|\) at \( x = 0 \), leading to a result even though there is a cusp at the origin.This behavior emphasizes why the centered difference quotient might offer a limit when traditional differentiability fails.
Derivative Existence at a Point
A derivative at a point exists if the function is differentiable at that location. For a function \( f(x) \), the derivative at \( x \) is defined as the limit:\[\lim_{{h \rightarrow 0}} \frac{f(x+h) - f(x)}{h}\]This limit must exist and provide a finite number for the derivative to be defined at that point. However, certain functions, such as the absolute value function at \( x = 0 \), lack this differentiability due to sudden directional changes or cusps.
  • For the absolute value function, \( |x| \), at \( x = 0 \), the left-hand and right-hand derivatives differ, with one being \(-1\) and the other \(1\).
  • Thus, the derivative does not exist at this point, as the limit defining the derivative does not yield a single, consistent value.
The original exercise used this example to show that despite the centered difference quotient providing a limit, this does not confirm actual differentiability at \( x = 0 \). This serves as a reminder to check derivative existence with care and not purely rely on quotient limits.

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Most popular questions from this chapter

A growing sand pile Sand falls from a conveyor belt at the rate of 10 \(\mathrm{m}^{3} / \mathrm{min}\) onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 \(\mathrm{m}\) high? Answer in centimeters per minute.

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Use a CAS to perform the following steps for the functions \begin{equation} \begin{array}{l}{\text { a. Plot } y=f(x) \text { to see that function's global behavior. }} \\ {\text { b. Define the difference quotient } q \text { at a general point } x, \text { with }} \\ {\text { general step size } h .} \\\ {\text { c. Take the limit as } h \rightarrow 0 . \text { What formula does this give? }} \\ {\text { d. Substitute the value } x=x_{0} \text { and plot the function } y=f(x)} \\ \quad {\text { together with its tangent line at that point. }} \\ {\text { e. Substitute various values for } x \text { larger and smaller than } x_{0} \text { into }} \\ \quad {\text { the formula obtained in part (c). Do the numbers make sense }} \\ \quad {\text { with your picture? }} \\ {\text { f. Graph the formula obtained in part (c). What does it mean }} \\ \quad {\text { when its values are negative? Zero? Positive? Does this make }} \\ \quad {\text { sense with your plot from part (a)? Give reasons for your }} \\ \quad {\text { answer. }}\end{array} \end{equation} $$f(x)=\frac{x-1}{3 x^{2}+1}, \quad x_{0}=-1$$

Temperatures in Fairbanks, Alaska The graph in the accompanying figure shows the average Fahrenheit temperature in Fairbanks, Alaska, during a typical 365 -day year. The equation that approximates the temperature on day \(x\) is $$y=37 \sin \left[\frac{2 \pi}{365}(x-101)\right]+25$$ and is graphed in the accompanying figure. a. On what day is the temperature increasing the fastest? b. About how many degrees per day is the temperature increasing when it is increasing at its fastest?

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