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Calculus I Test
#1 October 6,
2003
Name____________________ R. Hammack Score
______
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(1) (12 points)
(a)
(b) 1
(c) Find all
solutions of the equation .
This is equivalent to .
Looking at the unit circle, we see that and make
this equation true. But a multiple of
could be added to these numbers, and the equation would still hold. Thus the
solutions are all numbers of form
or .
(2) (8 points)
(a) Sketch the graph of the equation .
(b) Find the equation of the line that
is perpendicular to the graph of
(from Part a, above) and which has an x-intercept
of 1. Put your final answer in slope-intercept form, and simplify
as much as possible.
The above line has slope 1/2, so the slope of the line we seek is the negative
reciprocal of this, namely -2.
Since the line we seek has x-intercept
1, it passes through the point (1, 0).
Using the point-slope formula,
(3) (40 points) The problems
on this page concern the functions
and
(a)
(b)
(c) State the domain of
Note: the denominator factors as 3x(x - 3).
Domain:
All real numbers except 0 and 3.
(d) List the values of x
at which f is not
continuous. 0 and 3
(e)
(f)
(g)
(because
top approaches 3, bottom approaches 0, and is positive)
(h)
(i) List the vertical asymptote(s) of
f (if any). (Feel
free to use any relevant information from parts a-d above)
By part (g) above, the line x = 3 is
a vertical asymptote.
(j) List the horizontal asymptote(s)
of f (if any). (Feel
free to use any relevant information from parts a-d above)
By part (h) above, the line y = 1/3
is a vertical asymptote.
(4) (16 points) Evaluate
the following limits.
(a)
(b)
(c) 1
(d)
(5) (20 points) This problem
concerns the function
(a)
0
(b)
(c)
(d) Sketch a graph of the function f.
(e) Is the function f
continuous at x = 1? Yes,
because by
the above work.
(6) (4 points)
(Note that x is negative in this problem,
since it's approaching negative infinity. Therefore we must multiply the 1/x
terms by -1 to convert them to positive quantities that can be brought into
the radical.)