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386 Answers to Exercises and Solutions to Problems

40. The local maximum and minimum values are those values of k for
    which y = k is tangent to y = x3 + 3px2 + 3qx + r. This occurs only if the
    equation f(x) = 0 has a double zero, where

f(x) = x3 + 3px2 + 3qx + (r - k)

= (x + p)” + 3(q - p2)(x + P) + (2~~ - 3pq + r - k).

Now f’(x) = 3[(x + P)~ + (q - p2)]. S' mce j(x) has three real zeros, f’(z)

has real zeros and p2 - q > 0. We have, by division, that

3f(z) = (x + p)f’(x) + 3[2(q - p2>(x + P) + (2~~ - 3pq + r - k)].

Since a double zero of f(x) is a zero of f’(x), a zero u of f(z) is a double

zero a

2(q-p2)(x+p)+(2p3-3pq+r-k)=O

and (x + P)~ + (q - p2) = 0

+ 4(p2 - q)3 = 4(q - P~)~(x + p)” = (2p3 - 3pq + r - k)2

• T2(P2 - q) 3/2 = 2p3 - 3pq + r - k
  e k = 2p3 - 3pq + r f 2(p2 - q)3/2.

41. (a)

2(ub - cd)

a-b+c-d =

[u” + b2 - c2 - d2] - [(u - b)2 - (c - d)2]

a-b+c-d

= (u” + b2 - c2 - d2)/(u - b + c - d) - [(u - b) - (c - d)]

= (u2 - b2 - c2 + d2)/(u + b + c + d) - [(u + d) - (b + c)]

[u2 - b2

=

• c2 + d2] - [(u + d)2 - (b + c)“] 2(bc - ad)
  u+b+c+d = u+b+c+d’
  (b) (a, b,c,d) = (5,3,2,6) works.

42. Suppose, if possible, that the degree of the polynomial f(z) is a number
    n exceeding 1. From Lagrange’s Formula,

f(x) = 2 f(k)z(z - 1)... (x=k)... (x - n)

k=O

k(k - l)...(k -n)

it follows that all the coefficients must be rational. Hence, the polynomial
df(x) obtained by multiplying f(x) by the least common multiple d of the
denominators of its coefficients is a polynomial with the property of the
problem which has integer coefficients. Suppose

df(x) = u,,x” + a,-lx”-’ +. 1. + ulx + uo.