|Calculus I, B-Track||
September 30, 2002
(1) (10 points)
(b) Find all the x values that are not in the domain of the function .
The denominator will be zero for those values of x for which sin(x) =1/2.
Therefore the values x = and x = (where n is an integer) are not in the domain of f.
(2) (10 points)
(a) Sketch the graph of the equation .
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.
(3) (10 points) Consider the functions and
(4) (20 points) Evaluate the following limits.
(c) (0)(1) = 0
(5) (35 points) This problem concerns the function .
(a) State the domain of f.
Observe =(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) (2+2)/2 = 2
(c) (Note highest power of x occcurs on both top and bottom, each with coefficient 1.)
(d) (Note, top gets closer to 2, and bottom gets closer to 0, but positive.)
(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
(d) 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 does not exist, so it's impossible that
(i) 0, by parts g and h above
(j) Is f continuous at x = 1? (Explain, using above information.)
YES, because the above shows
(7) (5 points) Sketch the graph of a function f satisfying the following properties: , , , , and , There are many correct answers.