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Math 414 Professor Lieberman April 22, 2003 EXAM #3 SOLUTIONS 1. By freshman calculus, 0 fr,s (x) = rxr−1 sin(x−s ) − sxr−s−1 cos(x−s ) for x 6= 0. In addition, hr sin(h−s ) = lim hr−1 sin(h−s ), h→0 h→0 h 0 fr,s (0) = lim 0 (0) exists for r > 1. so fr,s 0 is continuous at zero if r > 1 and r > s + 1, for example, r = 3 and s = 1. (a) Therefore fr,s 0 0 (0) exists if r > 1 and r ≤ s + 1, for example, r = 3 (b) fr,s is discontinuous at 0 but fr,s and s = 2. 2. Abbreviate f (x) = x/(1 + x2 ), g(x) = arctan x and h(x) = x. Then f 0 (x) = 1 2x2 1 − , g 0 (x) = , h0 (x) = 1, 2 2 2 1+x (1 + x ) 1 + x2 so f 0 (x) < g 0 (x) < h0 (x) if x > 0, so f − g and g − h are strictly decreasing. Because f (0) = g(0) = h(0), it follows that f (x) < g(x) < h(x) for x > 0. 3. The easiest such function is ( 0 f (x) = 1 if x 6= 12 , if x = 12 . It isn’t monotone because f takes on its maximum at an interior point and it is clearly discontinuous at x = 1/2. To see that it’s integrable, note that the lower sum for any partition is 0 and we can make the upper sum less than a given ε > 0 by taking the partition P = {x0 , x1 , x2 , x3 } with x0 = 1, x1 = max{1/4, (1 − ε)/2}, x2 = min{3/4, (1 + ε)/2}. Another good example is Dirichlet’s function: ( 1 if x = pq in lowest terms, f (x) = q 0 otherwise. This was shown to be integrable in a homework problem. 4. From the definition, f (x) − f (t) . t→x x−t f 0 (x) = lim By hypothesis, f (x) − f (t) ≤ |x − t| x−t and limt→x −|x − t| = limt→x |x − t| = 0, so the squeeze theorem implies that f 0 (x) = 0 for all x. Therefore f is constant. −|x − t| ≤ 1 2 5. Actually, there is no such function, but here is a solution that would get full credit: If f 0 (x) = 1/f (x), then f 0 (x)f (x) − 1 = 0, and f 0 (x)f (x) − 1 is the derivative of f (x)2 /2 − x, so f (x)2 /2 − x must be constant. Therefore √ f (x) = 2x + C for some constant C. 6. Method 1: From the chain rule, sin(ln x) d (− cos(ln x)) = . dx x Because these functions are both continuous on [1, 2], the fundamental theorem of calculus says that Z 2 2 sin(ln x) dx = − cos(ln x) = 1 − cos(ln 2). x 1 1 Method 2: Set f (x) = sin x and g(x) = ln x. Then g is differentiable and g 0 is Riemann integrable on [1, 2] and g([1, 2]) = [0, ln 2]. Also f is continuous on [0, ln 2], so the Change of Variable Theorem says that Z 2 Z ln 2 ln 2 sin(ln x) dx = sin x dx = − cos x = 1 − cos(ln 2). x 0 1 0