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By B. Dacorogna

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Division by x - 2 yields the factorization (x - 2)(*3 + x2 8x - 12). It turns out t h a t - 2 is another root; division of x3 + x2 - 8x - 12 by x + 2 gives (x-2)(x + 2)(x2 -x-6) = (x- 2)(x + 2)(x -3)(x + 2). So, the roots are 2, -2, and 3. 79 x3 - 2x2 + x - 2. x3 - 2x2 + x - 2 = x\x -2) + x-2 = (x- 2)(x2 + 1). Thus, the only real root is 2. 80 x3 + 9x2 + 26x + 24. Testing the divisors of 24, reveals the root -2. Dividing by x + 2 yields the factorization (x + 2)(x2 + Ix + 12) = (x + 2)(x + 3)(;t + 4).

Is f continuous over [0,1]? Yes. When continuity over an interval is considered, at the endpoints we are concerned with only the one-sided limit. So, although/is discontinuous at x = l, the left-hand limit at 1 is 2 and /(I) — 2. 26 continuous over [1,2]? No. The right-hand limit at x=l is lim (* - 1) = 0, whereas /(I) = 2. 28 Let if if if Determine c and d so that/is continuous everywhere (as indicated in Fig. 7-8). Fig. 7-8 Since i-*0~ lim 3x2 -1 = — 1, the value of ex + d at x = 0 must be — 1, that is.

57 f(x) = mx + b for all x, where m^O. Assume f(u)=f(v). Then, mu + b = mv + b, mu = mv, u = v. Thus, /is one-one. 58 /(*) = Vx for all nonnegative x. Assume f ( u ) = f ( v ) . Then, Vu = Vv. Square both sides; u = v. Thus,/is one-one. 59 f(x) = x2 for all x. ). Hence, /is not one-one. 60 f(x) = - for all nonzero x. I Assume f(u)=f(v). 61 Hence, u = v. Thus,/is one-one. f(x) = \x\ for all x. /(—I) = 1 = /(I). Hence, /is not one-one. 62 f(x) = [x] for all x. /(O) = 0 = /(|). Hence, /is not one-one.

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