Unformatted text preview: Math 191 FINAL EXAM SOLUTIONS Fall 2000 1. (a) Using l’Hopital’s rule, the limit is 1.
(b) The only critical point is m = 6—1. (f’(a:) = (1 + ln ﬂew”) (c) The absolute minimum occurs when a: = 8‘1 where the value is 6—1/6. There i
no absolute maximum since lim 2 oo. (II—+00 (d) The linearization of f at x m 1 is the function 1(x) m 1 + (a: — 1) = 11:. (e) The approximation from Newton’s method is 331 I 101 = 1 — _—01.01' 2. Separate variables and use partial fractions: / Zdt 4 =fsinzrda:
y a. to obtain 1 y _F 2
11mm = —cos:c+ 0
Substitute the initial condition to compute C = 1 + iln %
3. (a) (b) This problem is more diﬂicult than intended. The approximation is an under
estimate of ﬂan—1(1) z 1.57. The function 1—4235 is concave downwards on 13h!
interval (—1 / {3—, 1J3) and concave upwards for t > iﬁ). The parts that an
above the trapezoids are larger than those that are below. ' If one evaluates th: difference between f(i1/4) and f (:lz3 / 4) and the points on the secants, then a'
:l:1/4 is about four times as large as the magnitude of the difference at 2E3/4. .A in”?
g 4. (a) (b) Two equations are 2
deﬁning the function Ml) implicitly and d(h2 + (11+ 332?) _ dh 32 ...
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 Spring '07
 BERMAN
 Math

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