ch02_03 - 170 CHAPTER 2 . Differentiation 2-26...

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170 CHAPTER 2 . . Differentiation 2-26 intersections.) Sketch a graph where F ± (1) < 1 and there are no solutions to the equation F ( x ) = x between 0 and 1 (although x = 1i sa solution). Solutions have a connection with the probability of the extinction of animals or family names. Sup- pose you and your descendants have children according to the following probabilities: f 0 = 0 . 2is the probability of having no children, f 1 = 0 . 3is the probability of having exactly one child, and f 2 = 0 . 5is the probability of having two children. Define F ( x ) = 0 . 2 + 0 . 3 x + 0 . 5 x 2 and show that F ± (1) > 1. Find the solution of F ( x ) = x between x = 0 and x = 1; this number is the probability that your “line” will go extinct some time into the future. Find nonzero values of f 0 , f 1 and f 2 such that the corresponding F ( x ) satisfies F ± (1) < 1 and hence the probability of your line going extinct is 1. 4. The symmetric difference quotient of a function f centered at x = a has the form f ( a + h ) f ( a h ) 2 h .If f ( x ) = x 2 + 1 and a = 1, illustrate the symmetric difference quotient as a slope of a secant line for h = 1 and h = 0 . 5. Based on your pic- ture, conjecture the limit of the symmetric difference quotient as h approaches 0. Then compute the limit and compare to the derivative f ± (1) found in example 1.1. For h = 1 , h = 0 . 5 and h = 0 . 1, compare the actual values of the symmetric difference quotient and the usual difference quotient f ( a + h ) f ( a ) h . In general, which difference quotient provides a better esti- mate of the derivative? Next, compare the values of the dif- ference quotients with h = 0 . 5 and h =− 0 . 5tothe deriva- tive f ± (1). Explain graphically why one is smaller and one is larger. Compare the average of these two difference quotients to the symmetric difference quotient with h = 0 . 5. Use this result to explain why the symmetric difference quotient might provide a better estimate of the derivative. Next, compute sev- eral symmetric difference quotients of f ( x ) = ± 4i f x < 2 2 x if x 2 centered at a = 2. Recall that in example 2.7 we showed that the derivative f ± (2) does not exist. Given this, discuss one major problem with using the symmetric difference quotient to approximate derivatives. Finally, show that if f ± ( a )exists, then lim h 0 f ( a + h ) f ( a h ) 2 h = f ± ( a ). 2.3 COMPUTATION OF DERIVATIVES: THE POWER RULE Youhavenow computed numerous derivatives using the limit definition. In fact, you may have computed enough that you have started taking some shortcuts. In exploratory exercise 1 in section 2.2, we asked you to compile a list of derivatives of basic functions and to generalize. We continue that process in this section. The Power Rule We first revisit the limit definition of derivative to compute two very simple derivatives. Forany constant c , d dx c = 0 . (3.1) y x a c y ± c FIGURE 2.20 A horizontal line Notice that (3.1) says that for any constant c , the horizontal line y = c has a tangent line with zero slope. That is, the tangent line to a horizontal line is the same horizontal line (see Figure 2.20).
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ch02_03 - 170 CHAPTER 2 . Differentiation 2-26...

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