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.
Russell Campbell
Sun Nov 09 1997