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Chapter 2: Problem 73

Using the Alternative Form of the Derivative In Exercises \(65-74,\) use the alternative form of the derivative to find the derivative at \(x=c\) (if it exists). $$ h(x)=|x+7|, \quad c=-7 $$

Short Answer

Expert verified
The derivative of the function \(h(x)=|x+7|\) at \(x=-7\) does not exist.

Step by step solution

01

Understand the absolute value function

The absolute value function can be rewritten as a piecewise function in terms of standard functions to facilitate differentiation. For \(x \geq -7\), \(h(x) = x+7\). For \(x < -7\), \(h(x) = -(x+7)\).
02

Derive the two pieces separately

The derivative of \(h(x)\) when \(x<-7\) is obtained as follows: \(h'(x)=-1\). On the other hand, the derivative of \(h(x)\) when \(x \geq -7\) is obtained as: \(h'(x)=1.\)
03

Evaluate limit for \(x = -7\)

Verify if the limit as \(x\) approaches \(c\) from the left (denoted as \(-\)) and limit as x approaches c from the right (denoted as \(+\)) are equal. For \(x=-7\), we have \(lim_{{x \to -7^{-}}} h'(x) = -1\) and \(lim_{{x \to -7^{+}}} h'(x) = 1\). Since these two one-sided limits are not equal, the derivative at \(x = -7\) does not exist.

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Key Concepts

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

Understanding the Absolute Value Function
The absolute value function, denoted by \(|x|\), represents the distance of a number \(x\) from zero on the number line. Its value is always non-negative. In the given exercise, we're dealing with the function \(h(x) = |x + 7|\). Here, the expression inside the absolute value shifts the entire graph of \(|x|\) by 7 units to the left. Thus, when working with absolute value functions, it's useful to express them as piecewise functions. This breaks the function into different cases that are easier to handle when differentiating.
So, for \(h(x) = |x + 7|\):
  • For \(x \geq -7\), \(h(x) = x + 7\)
  • For \(x < -7\), \(h(x) = -(x + 7)\)
This piecewise form sets the stage for finding the derivative using straightforward techniques.
The Role of Piecewise Functions
When dealing with absolute value functions like \(|x + 7|\), it's necessary to use a piecewise function to transition between different behavior on either side of the critical point \(x = -7\).
Piecewise functions are defined by multiple sub-functions, each applying to a specific interval of the main function's domain.
  • For ##x < -7##, we have ##h(x) = -(x + 7)##, which simplifies to ##h(x) = -x - 7##.
  • For ##x \geq -7##, it's ##h(x) = x + 7##.
This transformation is essential because it allows us to apply standard derivative rules on each segment separately. The absolute value function changes its behavior at the point where the inside of the absolute value equals zero, here at \(-7\). By rewriting it piecewise, we manage this change effectively when calculating derivatives.
Calculating One-Sided Limits
One-sided limits are pivotal in determining the behavior of a function as it approaches a particular point from either direction. These are crucial when investigating the existence of a derivative at a specific point.
In our context, we consider the point \(x = -7\), and calculate:
  • The limit from the left, \(\lim_{{x \to -7^{-}}} h'(x) = -1\).
  • The limit from the right, \(\lim_{{x \to -7^{+}}} h'(x) = 1\).
To find these limits, we look at the derivatives from each piece of the piecewise function as we approach \(-7\). These one-sided limits are not equal \((-1 eq 1)\), crucial to our next concept鈥攖he non-existence of the derivative at this point.
Understanding the Non-existence of Derivative
The derivative of a function represents the rate at which it changes. For a derivative to exist at a point \(c\), the function must have the same rate of change from both sides at \(c\).
When evaluating the derivative of a piecewise function like \(h(x) = |x+7|\) at \(x = -7\), we saw:
  • \(\lim_{{x \to -7^{-}}} h'(x) = -1\)
  • \(\lim_{{x \to -7^{+}}} h'(x) = 1\)
The mismatch between these one-sided limits tells us that the slope of the tangent line on the left side isn't equal to the slope on the right side. Therefore, the derivative does not exist at \(x = -7\).
This lack of a consistent slope across the critical point provides a deeper insight into functions with sharp turns, a common issue with absolute value functions at their vertex.

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Most popular questions from this chapter

Using Relationships In Exercises \(103-106,\) use the given information to find \(f^{\prime}(2) .\) $$ \begin{array}{l}{g(2)=3 \quad \text { and } \quad g^{\prime}(2)=-2} \\\ {h(2)=-1 \quad \text { and } \quad h^{\prime}(2)=4}\end{array} $$ $$ f(x)=\frac{g(x)}{h(x)} $$

Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\cos x ; \frac{d x}{d t}=4 \text { centimeters per second }} \\ {\begin{array}{llll}{\text { (a) } x=\frac{\pi}{6}} & {\text { (b) } x=\frac{\pi}{4}} & {\text { (c) } x=\frac{\pi}{3}}\end{array}}\end{array} $$

Modeling Data The table shows the health care expenditures \(h\) (in billions of dollars) in the United States and the population \(p\) (in millions) of the United States for the years 2004 through 2009 . The year is represented by \(t,\) with \(t=4\) corresponding to 2004 . (Source: U.S. Centers for Medicare \& Medicaid Services and U.S. Census Bureau) $$ \begin{array}{|c|c|c|c|c|c|}\hline \text { Year, } & {4} & {5} & {6} & {7} & {8} & {9} \\ \hline h & {1773} & {1890} & {2017} & {2135} & {2234} & {2330} \\\ \hline p & {293} & {296} & {299} & {302} & {305} & {307} \\\ \hline\end{array} $$ (a) Use a graphing utility to find linear models for the health care expenditures \(h(t)\) and the population \(p(t) .\) (b) Use a graphing utility to graph each model found in part (a). (c) Find \(A=h(t) / p(t),\) then graph \(A\) using a graphing utility. What does this function represent? (d) Find and interpret \(A^{\prime}(t)\) in the context of these data.

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet. (a) Water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when the depth \(h\) is 1 foot? (b) The water is rising at a rate of \(\frac{3}{8}\) inch per minute when \(h=2 .\) Determine the rate at which water is being pumped into the trough.

True or False? In Exercises \(129-134\) , determine whether the statement is true or false. If is false, explain why or give an example that shows it is false. If the velocity of an object is constant, then its acceleration is zero.
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