151 - 1c The numerator is continuous because sin(x) is cont., the |.| of a cont. function is cont., the sum of cont fntns is cont, the sqrt of a positive cont function is cont. The ratio is continuous except where x=0, because ratios of continuous functions are continuous except where the denominator is 0. 8 If x is irrational, there is a sequence of real numbers (x_n) which approaches it. Since f(x_n)=g(x_n) for all n, they have the same limit (which exists since f and g are continuous) and that limit is f(x)=g(x) because f and g are continuous. 15 If f(x) GT g(x), .5(f+g) + .5|f-g| = f, if f LT g, .5(f+g) + .5|f-g| = g, hence h is as given. Since the sum, difference, and absolute value of continuous functions is continuous, h is continuous. 159 - 1 Because f is cont, it obtains its minimum on a closed bounded interval (Thm 5.3.4). That minimum is GT 0, because all values of f are on that interval. Let alpha = half the minimum. 8 By continuity, f(w) is LT or equal to 0. (If w in W, f(w) LT 0, if w not in W, there is a sequence in W which approaches it, and f(w) LT or equal to 0 by continuity.) If f(w) LT 0, let f(w) = y LT 0. The IVT implies there exists an x between w and b with f(x) = y/2. This contradicts that w is the sup of W. 12 A open [respectively closed] interval with both endpoints positive will be mapped to an open [respectively closed] interval since the endpoints which are not [respectively are] included are mapped to the enpoints which are not [respectively are] included. The same argument holds if both endpoints are negative. (Or if one enpoint is zero.) If the enpoints span 0, 0 will be an included endpoint in the image, and the inclusion of the endpoint which is greatest in magnitude will determine the inclusion of the image endpoint. 171 - 1 (Look at |1/x - 1/c| = |(c-x)/xc|. If delta = min(.5, epsilon/2), xc GT .5, hence |(c-x)/xc| LT epsilon if |x-c| LT delta. 10 Let M be an upper bound for A and m a lower bound. Let epsilon equal 1. Let delta be the associated delta. Then there are (M-m)/delta intervals on which f varies byy 1 or less. One can choose any point in each interval where f is defined, and evalute f there, then take the supremum and infinum of those evaluations (which will be among the evaluations, since there is a finite number). 1 less than the imfinum and 1 more than the supremum provide bounds on f. [N.B., I was worried about gaps in A where f is not defined which would preclude asserting f varies by less than (M-m)/delta over A.] 12 If f is continuous on [0, infty), it is uniformly continuous on [0,2a]. For any epsilon, choose delta = min (.5a, delta(for [0,2a]), delta(for [a, infty) ) ). That delta demonstrates uniform continuity, since it works on [0, infty).

Sun Nov 02 1997