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390 Answers to Exercises and Solutions to Problems

(3) f’(x) = (x - k)g(x) where

g(x) = 4x2 + 4kx + 2k(2k + 1) = (2x + k)2 + k(3k + 2).

(4) f”(x) = 2(6x2 + k).

(5) f(x) and f’(x) h ave a zero in common _ k = -l/2, -l/3,^0 or

else g(x) and x3 + Lx - 2k2(2k + 1) have a zero in common. In the

last case we have that

(x3 + kx) + (4kx2 + 4k2x) = 0 + x2 + 4kx + (4k2 + k) = 0.

Since also x2 + kc + (k2 + k/2) = 0, we obtain 3kx + (3k2 + k/2) = 0

or x = -(k + l/6). Plugging this into the equation f(z) = 0 yields

36k2+24k+1=Oor6k=(-2ffi).

(6) f’(x) has three real zeros _ -213 5 k 5 0. Let the zeros of y(x)

be k, r, s with r 5 s.

(7) k is a double zero of j’(x) G+ 0 = g(k) = 2k(6k + 1) u k =

0, -l/6.

We consider various ranges of values for k:

(a) k < -213: f x has a sin le minimum value at x = k and y(x) <

o*-&<x,&6.

(b) k = -213: r = s = l/3; f”(x) < 0 9 -l/3 < t < l/3.

(c) -213 < k < (-2 - d)/6: Since r + s = -k > 0, rs = k(k + l/2), r

and s are both positive.

(d) k = (-2 - d)/6: r = l/6, s = (1 + fi)/6, j(s) = 0.

(e) (-2 - &)/6 < k < -l/2: r and s are both positive.

(f) k = -l/2: r = 0, s = l/2.

(g) -l/2 < k < -l/3.

(h) k = -l/3.

(i) -l/3 < k < -l/6.

(j) k = -l/6.

(k) -l/6 < k < (-2 + &)/S.

(1) k = (-2 + &)/6.

(m) (-2 + &)/6 < k < 0.

(n) k = 0.

(0) k > 0.