Students and instructors have urged that answers be available for these problems. "The management" is happy to cooperate. Answers supplied by students (along with comments and corrections, if necessary) will be posted. Students who supply material for this page will be credited using their initials. So please send answers, comments, and corrections. Graphs should be described as well as possible. This isn't supposed to be a complete description of solutions, but just a place where someone who has attempted the problems can check their final conclusions.
A set of solutions were received today (Monday, October 6, 1997) via
e-mail anonymously (!) so there's no way to thank those
responsible. A few sections are still missing, and there may be
better descriptions and solutions possible than are given here. Note
that ^ is used to denote exponentiation, so, for example, 3^2 is
9. Also, sqrt{x} is used to denote square root, so, for example,
sqrt{9} is 3.
Additional material is dated and credited as shown.
Problem #1:
a) xsqrt{x} (CREDIT: MWT, 10/7/97)
b) [(72)^(1/2)]/11
c) NO CORRECT ANSWER YET GIVEN! Please see the comment below.
9/27/97: Comment on part c provided by the management: as originally proposed on 9/23, this problem asked students to compute tan(arctan(7.75 Pi)) exactly. As revised on 9/26, it asks for arctan(tan(7.75 Pi)) exactly. The former question is easy (what is the answer, though?), but the latter question is a little bit tricky. Let me give a hint by asking a question about another function and its well-known inverse:
Suppose S(x)=x^{2}. Then the "inverse function" everyone thinks of is T(x) = the square root of x. Compute S(T(7)) and T(S(7)) and T(S(-7)). The logic of these computations are relevant to 1c. |
Problem #2:
a) (x-2)^2+y^2=9
b) m=-2, 2x+y=6
c) substitute:
(x-2)^2+(6-2x)^2=9
x^2-4x+4+36-24x+4x^2=9
5x^2-28x+31=0
x=1.5, 4.1
(1.5,3) and
(4.1,-2.2)
Alternatively and more accurately (this is the "exact" answer):
the intersection points are ([14-(41^1/2)]/5, [2+2(41^1/2)]/5), and
([14+(41^1/2)]/5, [2-2(41^1/2)]/5) (CREDIT: MWT, 10/7/97).
Problem #3: use substitution:
7= A(ln10^10)+B
-9=-A(ln10^100)-B
add to get
-2=A(ln10^10)-A(ln10^100)
-2=A[ln10^10-ln10^100]
-2=Aln[(10^10)/(10^100)]
A=.00965
B=6.7778
These last two approximations are contributed by the management, since
the answers supplied for the approximations were wrong. (10/8/97)
The "exact" solutions:
A = 1/(45 ln 10) (CREDIT: MWT, 10/7/97)
B=61/9 (CREDIT:IMD, 10/8/97)
Problem #4:
No answer supplied.
Problem #5:
a) No answer supplied.
b) Z(2)=4 & Z(4)=7
c) N(8)=5 & N(2)=1
d) No answer supplied.
Problem #6:
Domain: 103 <= p <= 130. (CREDIT: MWT, 10/7/97)
No formula supplied for n as a function of p.
10/7/97: Although the following phrase does represent what the formula
gives: "this is an inverse function" it isn't what the management
intended (GOOD! This will help the management learn to write better
questions, or maybe learn to write questions better). Perhaps this
part of the question should be rewritten as: "Describe what the
formula you have obtained represents in terms of detergent and
production costs. Be as specific as you can."
No answer supplied.
Problem #7:
a) 0
b) =lim(1/(x+2))=1/4
c) 0
d) =lim sqrt{x} + 2 as x approaches 4=4
Problem #8:
a) Does Not Exist
b) -infinity
c) Pi/2
d) Pi/2
e)(Pi/2)^2
Problem #9:
m=f'=6x, at x=1, m=6
TanLine: 6x-y=-1
Problem #10:
delta x will be denoted as dx.
f(x)=3x^-1
f'(x) = (lim as dx -> 0) (3/(x+dx) - 3/x) / dx = (lim as dx -> 0)
[3x/(x(x+dx)) -3(x+dx)/(x(x+dx))] /dx = (lim as dx -> 0) [(3x-3x
-3dx)/(x(x+dx))]/dx = (lim as dx -> 0) -3/(x(x+dx)) = -3/x^2.
Problem #11:
a) y'=3x^2-3
b) e^x(sin x + cos x)
10/7/97: the management comments that answers to such questions should
be written and not "simplified". Thus, the management (!) would leave
the answer to this question as (e^x) sin x + (e^x) cos x.
c) NO CORRECT ANSWER SUPPLIED!
d) NO CORRECT ANSWER SUPPLIED!
Problem #12:
a) ? = arctan (10/15) = arctan (2/3) = .588 radians
10/6/97: Management comment: the first and second answers are
fine. The third is actually just an approximation, so that equality is
not correct!
b) alpha=arctan (10/A)
Problem #13:
a)A=2
b) f is not continuous at x=2 because at x=2 the lim f(x) as x
approaches 2 does not exist
lim does not exist because (lim x->2+) f(x)=5 while (lim x->2-) f(x)=4
(CREDIT: MWT, 10/7/97)
Problem #14:
a) yes, no. lim g(x) as x->A exists, lim g as x->A- is different than
lim g as x->A+
10/6/97: Management comment: why ISN'T this function differentiable at
A?
b) yes, yes. lim g(x) as x->B exists, lim g as x->B- = lim g as x->B+
10/6/97: Management comment: why IS this function differentiable at B?
c) no, no. lim g(x) as x->C does not exist, lim g as x->C- is
different than lim g as x->C+
Problem #15:
graph: a parabola with minimum value at (2,1) and vertical asymptotes at
x=0 and x=3, and a portion on the interval (3,5) of the curve with
maximum value at (4,2) with a vertical asymtote at x=3, which
approaches the y value 1 on the right hand side of the point (4,2).
10/6/97: Management comment: there's an "empty circle" on the graph at
(5,1). Of course, a "real" parabola doesn't have asymptotes, but the
darn thing looks parabolic (parabolish?).
Problem #16:
i) [2.5, 3.5], it begins to take longer amounts of time for the
enzyme concentration to increase, the slope is not increasing very
rapidly.
ii) [1.5, 2.5], the concentration is increasing rapidly over a
smaller amount of time, the slope is increasing rapidly.
Problem #17:
a) F(1) = f(1)g(1) = 2*5 = 10
b) F'(x) = f(x)g'(x) + g(x)f'(x)
F'(1) = f(1)g'(1) + g(1)f'(1) = 2*7+5*-3 = 14-15 = -1
c) G(1)= f(1)/(g(1)+1) = 2/(5+1) = 1/3
d) G'= [(g+x^2)(f')-(f)(g'+2x)]/[(g+x^2)^2]
G'(1) = [(5+1)(-3)-2(7+2)]/[(5+1)^2] = -36/36 = -1
10/6/97: Management comment: these numbers have not been checked!