I do not curve exams and I am not curving this class. The
scale out lined on the syllabus tells us that the grade scale when scores
are given out of 60 is as follows.
A+ , A, A-
B+, B, B-
D+, D, D-
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.
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,
Inter changing x and y , we
Remark 2. Most people did not take into account that x is negative,
and so we must take the negative square root.
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
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:
Hence, when the half-life is 4 years, we have to wait
the mass is 20% of what it was initially.
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,
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.