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Math 123 Calculus I
Summer 2006 HW #2
Due Thursday  work with at most one partner
1. This problem concems the polynomial
f(x) = x“ + 2% —17x2 fl8x+72
a. Use synthetic division to evaluate f (4). ,F( f); ”1 b. Use synthetic division to give a polynomial which is equivalent to x4+2x3—17x2—18x+72 C 1C3v‘K‘L‘H‘X? (#24.
x+3 except at x = —3. 2. There is a general formula for solving ax2 +bx+c =0 (in terms of radicals) for given
values of a, b, and c. This formula is called the ADUWM " formula. @r False: There is a general formula for solving ax3 +be + cx+ d =‘ 0 (in terms of
ra cals) for given values of a, b, c, and d. ' / Or False: There is a general formula for solving ax‘ +ch3 +ch +dx+ e = 0 (in
Y ' terms of radicals) for given values of a, b, c, d, and e. 5. True [email protected] There is a general formula for solving ax5 +bx‘ + cx3 +de +ex+ f :0
M (in terms of radicals) for given values of a, b, c, d, e, and f. 6. This problem concerns the polynomial
f(x) =x7 +3x5 +x—4
Note that f(0)=—4<0 and f(1)=l>0. a. Isfcontinuous on [0,1]? Va; (and W (I “L éw/ 09)
/ W 7/ b. What theorem (give a name) guarantees at least one value x between 0 and 1 so that g f (x) = 0? (Such an x is called a “root” or “zero” off.) a i 5 (741:, V4 /v¢ 4) g c. Use bisection to ﬁnd a root off to within 1/ 128. Hint: Use a sequence of 11W
1 as we did in a similar example in class. g 727* (3 1L [Izl/ng/ 0/5413 = [.7453/15, . 753/247
\L l / (my ale/gum»..— Juio ﬁe Sb/ud‘ﬂk a5 .95/6H?z/) ‘ OCH): 765+ 27:31171'127: 4—72. 8(4) ., ? f . I Z “I? ~13, 7'1 @ [OE/7‘): 73.4 37:3; ’R‘F
((2/0): "f co
(Flt): / >5 ...
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 Summer '06
 JOHNSON
 Calculus

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