Unformatted text preview: Solutions to Practice Problems. True or False:
1F2F3T4F5F6T7T8F9F10T11F12T13T14F15T
16F 17T 18 F 19 F 20F 21T 22F 23F 24F 25F 26 T 27F 28F 29T 30T 31F
Multiplechoice:
10 2D 30 4A 5B 60 7C 8 B 9B 10D 11A 12D 13C 14B 15D 160 170 18B19A 200
21A 22C 23C 24A 25A 260 27A 28B 29B 300 31B 32D 33A 34D 35B 36A 37E 380 39B 40B
41B 420 43B 44A 45B 46B 47D 480 49D 50B 51B 52C 11. A
From the demand equation 50 — ,'p(:r2 + 1) = 0, we can get 19 = $331
So the revenue function is R(x) = p  as = (IE—31):: = xiii. 12D.
The marginal revenue function is the ﬁrst derivative of the revenue function, that is.
R’(:r) = W = % (by using quotient rule). 13. C
Method 1: The slope function is equal to the ﬁrst derivative function 3—: = 2:1: — 3.
So, at point (1, —2), the slope is m = 2(1) — 3 = —1.
Therefore the equation of the tangent line to the graph of y = 332 — 3:13 at the point (1, —2)
is y — (—2) = (—1)(:r — 1}, that is y = —:r; — 1. Method 2: Similarly to Method 1, we can ﬁnd the slope is m = —1. Now, we suppose
the equation is y = —:L‘ + I).
Since when m = 1, y = —2, we get —2 = —(1)+ h.
Then 5 = —1. So the equation of the tangent line to the graph of y = x2 — 3:12 at the point
(1, —2) is y = —:r: — 1. 14B
The slope function is equal to the ﬁrst derivative function 3—: = 11156 + 3%) = lnx + 1 (by
using product rule and ﬁﬂn x] = i).
When at = 1, the slope is m=ln1+ 1 = 1.
Suppose the equation is y = I + 1).
Since when at = 1, y = 0, we get 0 = (1)+ b.
Then b = —1. So the equation of the tangent line to the graph of y = aclnzr at the point (1,0)isy=9:—1. 15. D NOTICE THAT: iﬂn my = (ﬁﬁymn. ...
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 Fall '08
 Bayer,Margaret
 Calculus

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