Answers to Exercises; Chapter 2 2653.5. (4 (~1~2.. -Pn)‘= ~;~~PlP2-P~-Pn.
(e) @q)(t) = Wq(t)r. D’ff^1 erentiating yields (poq)‘(t) = Cru,q(t)r-lq’(t)
from which the result follows.
3.15. (a) Prove the result by induction on the degree of p. If degp = 0,
the result holds for each c with r = 0. Suppose the result holds for all
polynomials of degree not exceeding n - 1 1 0. Let degp = n. Then, if
p(c) # 0, we may take r = 0. If p(c) = 0, then p(t) = (t - c)u(t) and we
may apply the induction hypothesis to achieve u(t) = (t - c)‘-‘q(t) where
q(c) # 0 for some positive integer r.
(b) Let p(t) = (t - c)‘q(t) w h ere r 1 1, q(c) # 0. Then p’(t) =
(t - c)‘-l[rq(t) + (t - c)q’(t)]. Th e q uantity in square brackets does not
vanish when t = c. Hence p’(c) = 0 _ r - 1 1 1 _ r 2 2. The result
follows.
(d) p(‘+)
-+‘- CY.
4.3. (e) The quadratic must assume either a maximum or minimum value
between the two zeros. At this extremum, the derivative has a zero.
4.4. (f)a>0J,/ /
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//4.5. Set s = x + b. a. The change of variables represents a horizontal shift
of the origin of coordinates.
4.6. The graph of any cubic polynomial is a horizontal translate of the
graph of a polynomial of the form uz3 + cx + d, which in turn is a vertical
translate of the graph ax3 + cx. In the case of ax3 + cx, the inflection point
is at the origin and (x, y) satisfies the equation y = ax3 + cx if and only if
(-x, -y) does. The result follows.
4.7. (a) Let the polynomial be ax3+bx2+cx+d = x3(a+b/x+c/x2+d/x3).
When 1x1 is large enough, the quantity in parenthesis has the same sign as
a, and the sign of the polynomial is the same as the sign of ux3. Hence the
polynomial takes both positive and negative values, and so must vanish at
some point because of the continuity.
/ )Y
y=x3+ax a<0/ / \ \
/ / \ \
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X7y =X3 +axJ \ X
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