Calculus I,    B-Track  
Test #1
September  30, 2002
R.  Hammack
Score ______

(1) (10 points)

(a)  [Graphics:Images/T1BF02sol_gr_1.gif] [Graphics:Images/T1BF02sol_gr_2.gif]

(b) Find all the x values that are not in the domain of the function [Graphics:Images/T1BF02sol_gr_3.gif].
The denominator will be zero for those values of x for which sin(x) =1/2.
Therefore the values x = [Graphics:Images/T1BF02sol_gr_4.gif] and  x = [Graphics:Images/T1BF02sol_gr_5.gif] (where n is an integer) are not in the domain of f.

(2) (10 points)

(a) Sketch the graph of the equation    [Graphics:Images/T1BF02sol_gr_6.gif].
To find the x-intercept, set y = 0 and solve for x. You get x = -2.
To find the y-intercept, set x = 0 and solve for y. You get y  = 4.


(b) Find the equation of the line that is parallel to the graph of   [Graphics:Images/T1BF02sol_gr_8.gif] (from Part a, above) and which passes through the point [Graphics:Images/T1BF02sol_gr_9.gif]. Put your final answer in slope-intercept form, and simplify as much as possible.

Looking at the graph above, you see that the line has slope 2. We seek a line that is parallel to this one, so its slope will be 2 also. Moreover, it is to pass through (-2,-3), so the point-slope formula gives

(3) (10 points)  Consider the functions [Graphics:Images/T1BF02sol_gr_14.gif] and   [Graphics:Images/T1BF02sol_gr_15.gif]

(a) [Graphics:Images/T1BF02sol_gr_16.gif] [Graphics:Images/T1BF02sol_gr_17.gif][Graphics:Images/T1BF02sol_gr_18.gif]

(b) [Graphics:Images/T1BF02sol_gr_19.gif] [Graphics:Images/T1BF02sol_gr_20.gif]

(4)  (20 points)  Evaluate the following limits.

(a)    [Graphics:Images/T1BF02sol_gr_21.gif] [Graphics:Images/T1BF02sol_gr_22.gif]

(b)  [Graphics:Images/T1BF02sol_gr_23.gif] [Graphics:Images/T1BF02sol_gr_24.gif][Graphics:Images/T1BF02sol_gr_25.gif]0

(c)    [Graphics:Images/T1BF02sol_gr_26.gif][Graphics:Images/T1BF02sol_gr_27.gif](0)(1) = 0

(d)    [Graphics:Images/T1BF02sol_gr_28.gif] [Graphics:Images/T1BF02sol_gr_29.gif][Graphics:Images/T1BF02sol_gr_30.gif][Graphics:Images/T1BF02sol_gr_31.gif][Graphics:Images/T1BF02sol_gr_32.gif]

(5) (35 points) This problem concerns the function  [Graphics:Images/T1BF02sol_gr_33.gif].

(a) State the domain of f.
Observe [Graphics:Images/T1BF02sol_gr_34.gif]=[Graphics:Images/T1BF02sol_gr_35.gif][Graphics:Images/T1BF02sol_gr_36.gif][Graphics:Images/T1BF02sol_gr_37.gif](provided x is neither 0 nor 2)

Note that x = 0 or x = 2 will give a zero in the denominator, so these values are not in the domain. However, any other value of x will work. Therefore the domain of f is all real numbers except 0 and 2.

(b)    [Graphics:Images/T1BF02sol_gr_38.gif] [Graphics:Images/T1BF02sol_gr_39.gif](2+2)/2 = 2

(c)    [Graphics:Images/T1BF02sol_gr_40.gif][Graphics:Images/T1BF02sol_gr_41.gif] (Note highest power of x occcurs on both top and bottom, each with coefficient 1.)

(d)    [Graphics:Images/T1BF02sol_gr_42.gif][Graphics:Images/T1BF02sol_gr_43.gif](Note, top gets closer to 2, and bottom gets closer to 0, but positive.)

(e)    [Graphics:Images/T1BF02sol_gr_44.gif][Graphics:Images/T1BF02sol_gr_45.gif]

(f) List the vertical asymptote(s) of f  (if any).  (Feel free to use any relevant information from parts a-d above)
By part, (d), the line x=0 is a vertical asynptote,

(g) List the horizontal asymptote(s) of f  (if any).   (Feel free to use any relevant information from parts a-d above)
By part, (c), the line y=1 is a horizontal asynptote,

(6) (10 points)  This problem concerns the function [Graphics:Images/T1BF02sol_gr_46.gif]
(a) [Graphics:Images/T1BF02sol_gr_47.gif] [Graphics:Images/T1BF02sol_gr_48.gif]
(b) [Graphics:Images/T1BF02sol_gr_49.gif] [Graphics:Images/T1BF02sol_gr_50.gif]
(c) [Graphics:Images/T1BF02sol_gr_51.gif] [Graphics:Images/T1BF02sol_gr_52.gif]
(d) [Graphics:Images/T1BF02sol_gr_53.gif] D.N.E. because right- and left-hand limits do not agree.
(e) Is f continuous at x = 0? (Explain, using above information.)
NO, because [Graphics:Images/T1BF02sol_gr_54.gif] does not exist, so it's impossible that [Graphics:Images/T1BF02sol_gr_55.gif]

(f) [Graphics:Images/T1BF02sol_gr_56.gif][Graphics:Images/T1BF02sol_gr_57.gif]
(g) [Graphics:Images/T1BF02sol_gr_58.gif][Graphics:Images/T1BF02sol_gr_59.gif]
(h) [Graphics:Images/T1BF02sol_gr_60.gif] [Graphics:Images/T1BF02sol_gr_61.gif]
(i) [Graphics:Images/T1BF02sol_gr_62.gif] 0, by parts g and h above
(j) Is f continuous at x = 1? (Explain, using above information.)
YES, because the above shows  [Graphics:Images/T1BF02sol_gr_63.gif]

(7) (5 points) Sketch the graph of a function f  satisfying the following properties:  [Graphics:Images/T1BF02sol_gr_64.gif],   [Graphics:Images/T1BF02sol_gr_65.gif],   [Graphics:Images/T1BF02sol_gr_66.gif],  [Graphics:Images/T1BF02sol_gr_67.gif],   and   [Graphics:Images/T1BF02sol_gr_68.gif],  There are many correct answers.