180-2 f(x1) \geq f(x2) and g(x1) \geq g(x2) implies f(x1)+g(x1) \geq f(x2)+g(x2) If either inequality is strict, the resultant inequality will be strict. 8 Geometrically, the graph of f lies above the graph of g, and both are increasing to the right. Hence a horizontal line which intersects both graphs will intersect f to the left of g. y = f(f^{-1}(y)) \gt g(f^{-1}(y)); y = g(g^{-1}(y)) \lt f(g^{-1}(y)) hence f(f^{-1}(y)) \lt f(g^{-1}(y)) and since f is increasing, the left hand argument is less than the right hand argument. 9 f(x)=x maps rational numbers to rational numbers; f(x)=(1-x) maps irrational numbers to irrational numbers; hence it suffices to consider f seperately on those two subdomains. Both x and 1-x are stricly monotone, hence injective. 1-(1-x)=x, and the identity is known to be its own inverse. If |x-.5| LT epsilon both |x-.5| and |(1-x)-.5| are LT epsilon, so since f(.5)=.5, f(x) is continuous at .5. At a point c not equal to .5, let epsilon equal |c-(1-c)|/4, and since f(x) will be arbitrarily close to both c and (1-c) in any delta neighborhood of c the values will differ by (2c-1) and will not remain in any epsilon neighborhood as defined above. 194-1c (sqrt(c)-sqrt(x))/(c-x) = (c-x)/((c-x)(sqrt(c)+sqrt(x)) = 1/(2sqrt(c) in the limit as x goes to c 6 f'(x)=0 for x LT 0. For x GT 0, f'(x) = nx^(n-1) which is 0 at 0 only for n GT 1; f(x) is continuous at 0 for n GT 0. lim (x goes to 0+) (x^n-0)/(x-0) = x^(n-1) which is 1 for n=1 which does not equal the left hand limit of 0; it is not defined for n LT 1; hence the derivative is continuous where it exists, which is n GT 1. [Sorry, this problem was for n in N (we had not yet defined real eaponents?); still problems with n=0 since x^0 = 1 for x=0, n=1 gives a continuous function but the derivative is not continuous (it does not exist at 0), n GT 1 has everything hunky-dory. 11a L'(2x+3) = (1/(2x+3))(2) (actually valid for x GT -1.5) 205-4 Since the leading term is nx^2, there is a minimum. taking the derivative which equals \sum 2(ai-x) , we see that the minimum occurs at \sum ai/n 9 2 GT sin(1/x), hence x^4(2-sin(1/x)) GT 0 for x notequal 0 (and equal 0 at 0). The derivative is 8x^3 + 4x^3sin(1/x) + x^4cos(1/x)(-1/x^2) Since cos(1/x) = 1 and -1 in every nbhd of 0, and the third summand has the lowest power of x (x^2), it determines the signum near 0 when cos(1/x) = +/- 1. since this takes + and - values we are done. 18 (f(x)-f(y))/(x-y) = (f(x)-f(c)+f(c)-f(y))/(x-y) = ((f(x)-f(c))/(x-c))((x-c)/(x-y)) + ((f(c)-f(y))/(c-y))((c-y)/(x-y)) [One can easily treat the cases with x=c or c=y seperately.] One can also multiply f'(c) by [(x-c)/(x-y) + (c-y)/(x-y)) and use the triangle inequality to provide a full answer.

Sun Nov 09 1997