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Calculus I                                                              Test #1                                    March  8, 2004

Name____________________                       R.  Hammack                                Score ______
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(1)
(a)  cos(-(2π)/3) = -1/2

(b)  sec(-(2π)/3) =1/cos(-(2π)/3) = 1/(-1/2) = -2


(c) Find the domain of the function    f(x) = 1/(2^(1/2) sin(x) - 1).

We can't have 2^(1/2) sin(x) - 1 = 0, or, in other words, we can't have sin(x) =1/2^(1/2),
for such values make the denominator of f 0
Now, the values of x between 0 and 2π for which sin(x) =1/2^(1/2) are x = π/4and x = (3π)/4.
Of course, we could add any integer multiple of 2π to them and still  sin(x) =1/2^(1/2).
Thus the domain of f is the set {x | x≠π/4 + k 2π,   x≠ (3π)/4 + k 2π,   where k is an integer }


(2)
(a) Sketch the graph of the equation    y = -1/3x - 2 .
[Graphics:HTMLFiles/test1sol_16.gif]

(b) Find the equation of the line that is parallel to the graph of   y = -1/3x - 2 (from Part a, above) and passes through the point (3, 4).  Put your final answer in slope-intercept form, and simplify as much as possible.

By point-slope formula,
y - 4 = -1/3 (x - 3)
y - 4 = -1/3x + 3
y = -1/3x + 5

(3)   Suppose  f(x) = x/(x + 1) and   g(x) = 1/x .

(a)   f∘g(x) = FormBox[RowBox[{f(g(x)), =, RowBox[{(1/x)/(1/x   + 1), =, RowBox[{(1/x)/(1/x &n ... x), =, RowBox[{(1/x)/((1 + x)/x), =, RowBox[{1/(1 + x), Cell[], Cell[]}]}]}]}]}], TraditionalForm]

(b)   g∘f(x) = g(f(x)) = 1/x/(x + 1) = (x + 1)/x

(4) Sketch the graph of a function f(x) satisfying the following properties.
f(0) = 2,   f(2) = 1,    Underscript[lim , x2^-] f(x) = 3,   Underscript[lim , x2^+] f(x) = 1,   Underscript[lim , x -2] f(x) = ∞,    Underscript[lim , x∞] f(x) = ∞.  and   Underscript[lim , x -∞] f(x) = 0.


[Graphics:HTMLFiles/test1sol_37.gif]
Note: It's hard to draw this graph with the program I am using.
There should be a solid dot at the point (2, 1) and a hollow dot at (2, 3).
The vertical line between these points should not be there.



(5)  The graph of a function f(x) is sketched. Use this information to find the following limits.
[Graphics:HTMLFiles/test1sol_39.gif]
(a)     Underscript[lim , x2] 4f(x) = Underscript[4 lim , x2] f(x) = 4 (-2) = -8

(b)     Underscript[lim , x -1] (f(x) + 2)/(f(x) + 3) = (1 + 2)/(1 + 3) =3/4

(6)   Evaluate the following limits.

(a)    Underscript[lim , x -1] (2x^3 - 4x - 5) = 2 (-1)^3 - 4 (-1) - 5 = -3

(b)  Underscript[lim , x2] (2x^2 - 8)/(x - 2) = Underscript[lim , x2] (2 (x^2 - 4))/(x - 2) = Underscript[lim , x2] (2 (x - 2) (x + 2))/(x - 2) = Underscript[lim , x2] 2 (x + 2) = 2 (2 + 2) = 8

(c)    Underscript[lim , x2] (2x^2 - 8)/(x - 3) =(2 (2^2) - 8)/(2 - 3) = 0

(d)   Underscript[lim , w25] (w - 25)/(w^(1/2) - 5) = Underscript[lim , w25] ((w^(1/2) - 5) (w^(1/2) + 5))/(w^(1/2) - 5) = Underscript[lim , w25] (w^(1/2) + 5) = 25^(1/2) + 5 = 10

(e) Underscript[lim , x2] cos(π/(4 + x)) = cos(π/(4 + 2)) = cos(π/6) = 3^(1/2)/3

(f) Underscript[lim , x0] 1/(3x cot(x)) = 1/3 Underscript[lim , x0] 1/(x cos(x)/sin(x)) = 1/3 Underscript[lim , x0] 1/( ... ;0] sin(x)/(cos(x) x) = 1/3 Underscript[lim , x0] sin(x)/( x) 1/cos(x) = 1/3 (1) 1/1 = 1/3


(7)  This problem concerns the function f(x) = {   2sin(x) ------- x ... sp;x≠0 1 if  x = 0

(a) f(0) = 1

(b) Underscript[lim , x0] f(x) = Underscript[lim , x0] (2sin(x))/x = 2Underscript[lim , x0] sin(x)/x = (2) (1) = 2

(c)  At which x values (if any) is the function f(x)  discontinuous? Explain  your answer.

The function is not continuous at x = 0, because, as the previous two answers demonstrate, Underscript[lim , x0] f(x) ≠f(0).
Thus the definition of continuity is not satisfied.

(d)   Find the horizontal asymptotes (if any) of  f(x) .  Be sure to explain your work.

Note that Underscript[lim , x∞] f(x) = Underscript[lim , x∞] (2sin(x))/x = 0 (Because, while the numerator is bounded between -1 and 1, the denominator grows without bound.)
Likewise the limit as x approaches negative infinity is 0.

THUS the line y = 0 is a horizontal asymptote.