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CALCIReviewProblemsExam2sols

100 a find dbdt assuming p and r are constant in terms

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Unformatted text preview: −f (2)g (2) (g (2))2 = 5(4)−3(−2) 42 = 26 16 = 13 8. 7. If you invest P dollars in a bank account at an annual interest rate of r%, then after t years you will have B dollars, where rt B =P 1+ . 100 (a) Find dB/dt, assuming P and r are constant. In terms of money, what does dB/dt represent? (b) Find dB/dr, assuming P and t are constant. In terms of money, what does dB/dr represent? Solution: r rt dB dB ln 1 + = P 1+ . The expression tells us how fast the amount of money in the bank dt 39 Page: 153100 100 10) [ex-6] dt Job: chap3-temp Sheet: (March 8, 2012 10 : is changing with respect to time for ﬁxed initial investment P and interest rate r. dB r t−1 1 dB (b) = Pt 1 + . The expression indicates how fast the amount of money changes with dr 100 100 dr respect to the interest rate r , assuming ﬁxed initial investment P and time t. (a) 3.6 THE CHAIN RULE AND INVERSE FUNCTIONS 6w51-53 153 8. Use the following graph to calculate the derivative. 3-6w59 In Problems 60–62, use Figure 3.32 to calculate the derivative. 68. Figure 3.33 shows the number of motor vehicles,7 f (t), in millions, registered in the world t years after 1965. With units, estimate and interpret (2.1, 5.3) (a) f (20) (c) f −1 (500) f (x ) (b) (d) f (20) (f −1 ) (500) 2 4 6 (millions) (2, 5) 800 ins3-6w51-53fig 8 600 ! 10 400 Figure 3.32 12 3 3-6w51 3-6w52 3-6w53 3-6w54 (a) h (2) if h(x) = (f (x)) 200 Solution: Since the point (2, 5) is on the curve, we know f (2) = 5. The point (2.1, 5.3) is on the tangent 14 3 60. h (2) if h(x) = (f (x)) line, so−1 (year) 3-6w59fig 61. k (2) if k(x) = (f (x)) ’65 ’70 ’753 − 5 ’850.3’90 ’95 2000 5. ’80 16 Slope tangent = = = 3. −1 62. g (5) if g (x) = f (x) 2.1 − 2 3.33 Figure 0.1 63. (a) (b) (c) (d) Given that Thus, f 3 ,2) = f (2). f (x) = x ( ﬁnd 3. Find f −1 (x). By the chain rule 3-6w60 69. Using Figure 3.34, where f (2) = 2.1, f (4) = 3.0, Use your answer from part (b) to ﬁnd (f −1 ) (8). 2 h (2) =f3(6)(2))27, f (2) = 4.2,5ﬁnd3 f −1 ) (8). · ( = 225. ( f = 3. · f (8) = 3 · How could you have used your answer from part (a) −1 to ﬁnd (f k)((8)? k (x) = (f (x))−1 (b) 2) if 18 20 22 3-6w55 64. (a) For f (x) = 2x5 + 3x3Sinceﬁnd f point (2, 5) is on the curve, 24 know f (2) = 5. The point (2.1, 5.3) is on the tangent Solution: + x, the (x). we (b) How can you use your answer to part (a) to deterf (x ) 24 line, so 16 mine if f (x) is invertible? 0.3 5.3 − 5 Slope tangent 8 = = = 3. (c) Find f (1). 2.1 − 2 0.1 26 (d) Find f (1). x 3-6w60fig (e) Find (f −1 ) (6). 2 4 6 8 3-6w57 65. Use the table and the fact that f (x) is invertible and differentiable everywhere to ﬁnd (f −1 ) (3). 28 Figure 3.34 30 x f (x ) f (x ) 3 1 7 3-miscw118 70. If f is increasing and f (20) = 10, which of the two options, (a) or (b), must be wrong? −1 32 Thus, f (2) = 3. By the chain rule k (2) = −(f (2))−2 · f (2) = −5−2 · 3 = −0.12. (c) g (5) if g (x) = f −1 (x) Solution: Since the point (2, 5) is on the curve...
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