Sample Exam 3Solutions PDF

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excellent sample/practice exams to thoroughly prepare for the midterm and final exam...

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Judge Exam 3 solutions 1) There appears to be a critical point (flat) at x=-4 and x=0 and an inflection point at x =-2, so (C) 2) Global maxs or mins can come from critical points or endpoints. Thus (A) (again on these graphs, you can eyeball the approximate coordinates and assume the graph is relatively smooth). 3) By the 1st derivative test, a local min goes from decreasing to increasing, thus the derivative goes from – to +. This occurs at t=6. Likewise a max occurs when the derivative switches from + to -, which is somewhere between t= 12 and t=14. (A) " ( q) = R ! C = 100 q ! (1000 + 2 q 2 ) = !2 q 2 + 100 q ! 1000 " ' ( q) = !4q + 100 = 0 4) (B) q = 25 units " (25) = ! 2(25) 2 + 100(25) ! 1000 = $250 **Watch your units, note how he has A with $25 instead of 25 units** ln 2 & ln C L # 5) Greatest rate occurs at $ ! 20 , so , ! and since we want years, use t = .0346 % k 2" around 1970+20 or 1990 (C)

6) As written, this problem is impossible to answer as there is a critical scaling error. The units from what appears for be 0 to 10 takes two notches, but then takes two notches for every 5 units afterwards until 20, when it takes two per 10 again. Since we don’t have 0 or an axis labeled, we can’t really know if our 1st number is 0 or 5 (if it is 5, any answer would be possible). Assuming this wasn’t there, I would set up a table, eyeballing the heights and then averaging the left and right Riemmann sums. Assuming the graph starts at 0, we get: x f(x)

0 16

5 18

10 20

12.5 17.5

15 15

17.5 10

20 7

25 5

30 7.5

80

90 90

50 100

43.75 43.75

37.5 37.5

25 25

35 17.5

25 25

37.5

Left= Right=

386.25 376.25

Best=

381.25

Thus I have no clue what the answer was meant to be or if this error was corrected during the exam. If this were from 0 to 40 instead, the correct answer would be around 525.

7) Demand is given by q = mp+b, so if we find two points: 3100 ! 3000 = !1000 items / $ 2.9 ! 3 q = !1000 p + b

m= ( p, q)

then we can find the demand by 3000 = !1000(3) + b (2.90,3100) 6000 = b q = !1000 p + 6000

(3,3000)

R = pq = p( !1000 p + 6000) = !1000 p2 + 6000 p R' = !2000 p + 6000 = 0 p = $3

8) Since this is a graph of his velocity, to find total distance, we need to look at the integral. Since the integral from 12pm to 1:30 is greater in absolute value than the integral from 1:30 to 3, he traveled further away from home then he did back towards it. Since after 3 he starts traveling away from home again, he never gets back home. (E) 9) Upper: (18+22+22)*5=310 Lower: (10+18+16)*5 (D) *Remember that on upper estimates, you need to look at each interval and determine which number is higher, even if it means using some numbers twice or skipping some numbers. 10) If you remember that integrals for functions above the x-axis are positive, and those below are negative, then this is a pretty easy problem. b

A)

c

! g ( x) dx

is larger than ! g ( x) dx because

a

a

c

! g ( x )dx < 0 b

11) Using a left and right Riemann sum, we get Left: (15+18+21)*3=162 Right: (18+21+12)*3=153 Best: (153+162)/2=157.5 (C) 10

12) f (10) " f (0) =

! (2000 " 6t )dt = $18,000 2

0

A) f ' ( x) = 3 x 2 ! 6 x ! 9 B ) f ' ' ( x) = 6x ! 6 13)

C)3 x 2 ! 6 x !9 = 0 3( x 2 ! 2 x ! 3) = 0 3(x ! 3)(x + 1) = 0 x = 3, x = !1

**These two critical points are in our interval

D) Using the 2nd derivative test, we see f ‘’(3)=12>0, so we have a local min and f ‘’(-1) = -12, so it is a local max. f ' ' ( x) = 0 E) 6 x ! 6 = 0 x =1 x

f(x) -5

-140

-1 3 4

20 -12 -5

Global Min Global Max...