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May 23rd

May 23rd

# MATH 1050 MIDTERM 2 SOLUTIONS AND COMMENTARY

Overview

The exam did not go as well as I'd hoped. Scores were a bit lower
than I would have liked, but overall they were not too bad. 7 people
took the exam, and the average was 36.

The following table shows the score distribution .

 Score Number of Students 42 1 40 2 38 1 34 1 30 1 26 1

I do not curve exams and I am not curving this class. The preset
scale out lined on the syllabus tells us that the grade scale when scores
are given out of 60 is as follows.

 Score Grade 54-60 A+ , A, A- 45-53 B+, B, B- 39-44 C+, C 36-38 C- 27-35 D+, D, D- 0-26 E

Every one still has a good chance to pass the class, since the final is
worth 30% of the grade, most of you are passing the course, and those
who aren't are still pretty close to passing. Saying that, I look for
substantiative improvement on the remaining exams and homework.

Problem 1

For x ≤ 0, let f(x) = x2 - 2: Find f -1.
Solution 1. Let y = f(x). Then y = x2 - 2 and so y + 2 = x2 Since
x ≤ 0, we take the negative square root of the left hand side: that is,
Hence, Inter changing x and y , we
get

Remark 2. Most people did not take into account that x is negative,
and so we must take the negative square root.

Problem 2

A certain substance decays exponentially. If A0 is the initial mass
of the substance, h is the half-life in years, t is the time in years, and
A(t) is the mass at time t, then

(1)

If h = 4, how long do we have to wait until only 20% of the original
mass is left?

Solution 3. We want to know what t is when
and h = 4. We plug this into the equation and solve for t:

(2)

Notice that   Similarly,
and so

(3)

Hence, when the half-life is 4 years, we have to wait years before
the mass is 20% of what it was initially.

Remark 4. The last line in Equation 2 was what I was really looking
for. The rest of my solution is essentially simplification .

Problem 3

Sketch the graph of labeling your axes, the
graph of the function, and the x-intercept. In the course of your expo-
sition, you will explicitly find the x-intercept.

Solution 5. The graph of y = 3x looks like a standard exponential
graph. The graph of is a reflection of this exponential graph
through the line y = x. The graph of y = f(x) shifts the graph of
up by 2 units. Hence, the y-axis is a vertical asymptote for
f. The x- intercept occurs when f(x) = 0. That is,

(4)

We sketch the graph of f in figure 1.

Figure 1. Illust ration for Problem 3

Remark 6. Finding the x-intercept seemed to cause problems, espe-
cially since most people sketched the graph of an exponential function
rather than that of a logarithmic function .

Problem 4

A bacteria population in a petri dish grows exponentially. If t is the
time variable, A0 is the initial population, and A(t) is the population
at time t, then

(5)

for some positive constant, k. If A(2) = 1 and A(5) = 3, find k.

Solution 7. We are given that

(6)

and

(7)

From the first equation, Plugging this into the second
equation gives

(8)

Remark 8. No one seemed to know how to do this problem. However,
I do not think that it was an unreasonable problem. As can be seen,
the answer is quite short and the techniques should be familiar.

Problem 5

Show that is irrational. (Hint: use the rational roots theorem on
a polynomial which has as a root.)

Solution 9. Let P(x) = x2 - 5. Then P is a polynomial with rational
coefficients which has as a root. We can apply the rational roots
theorem to P. From the theorem, the only possible rational roots of P
are 1, -1, 5, and -5. But P(±1) = -4 ≠ 0 and P(±5) = 20 ≠ 0. Thus,
P has no rational roots and hence must be irrational as desired.

Remark 10. The polynomial is not helpful here, because we
do not know a priori whether is rational or irrational, and so we
cannot apply the rational roots theorem to this polynomial.

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