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Calculus I                                                                           Test #1                                              October  6, 2003

Name____________________                                   R.  Hammack                                              Score ______
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(1) (12 points)

(a)  sin(-(3π)/4) =-1/2^(1/2)

(b) tan(-(3π)/4) =1

(c) Find all solutions of  the equation    2cos(x) = 3^(1/2).
This is equivalent to  cos(x) = 3^(1/2)/2. Looking at the unit circle, we see that x = π/6and  x = -π/6make this equation true. But a multiple of 2π could be added to these numbers, and the equation would still hold. Thus the solutions are all numbers of form x = π/6 + n 2π or  x = -π/6 + n 2π.

(2) (8 points)

(a) Sketch the graph of the equation    2x - 4y = -8.


[Graphics:HTMLFiles/T1F03Csol_13.gif]

(b) Find the equation of the line that is perpendicular to the graph of   2x - 4y = -8 (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,
y - 0 = -2 (x - 1)

y = -2x + 2

(3) (40 points)  The problems on this page concern the functions f(x) = x^2/(3x^2 - 9x) and   g(x) = cos(x) .

(a) f∘g(x) = FormBox[RowBox[{f(g(x)), =, RowBox[{cos^2(x)/(3cos^2(x) - 9cos(x)), Cell[], Cell[], Cell[]}]}], TraditionalForm]

(b) g∘f(x) = g(f(x)) = cos(x^2/(3x^2 - 9x))

(c) State the domain of f(x) = x^2/(3x^2 - 9x) 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)     Underscript[lim , x1] x^2/(3x^2 - 9x) =1^2/(3 (1^2 ) - 9 (1)) = -1/6

(f)     Underscript[lim , x0] x^2/(3x^2 - 9x) =  Underscript[lim , x0] x^2/x(3x - 9) =Underscript[lim , x0] x/(3x - 9) = 0/(3 (0) - 9) = 0

(g)   Underscript[lim , x3^+] x^2/(3x^2 - 9x) = Underscript[lim , x3^+] x^2/x(3x - 9) =Underscript[lim , x3^+] x/(3x - 9) = ∞  (because top approaches 3, bottom approaches 0, and is positive)

(h)   Underscript[lim , x∞] x^2/(3x^2 - 9x) =1/3

(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)    Underscript[lim , x3] (x^2 - 3x + 4) =3^2 - 3 (3) + 3 = 4

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

(c)    Underscript[lim , x0] sin(x)/x =1

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

(5) (20 points)  This problem concerns the function f(x) = {   2 1 - x if  x<1 x - 1 if  1≤x
(a) f(1) = 0

(b) Underscript[lim , x1^-] f(x) = Underscript[lim , x1^-] 1 - x^2 = 0

(c) Underscript[lim , x1^+] f(x) =Underscript[lim , x1^-] x - 1 = 0

(d) Sketch a graph of the function f.

[Graphics:HTMLFiles/T1F03Csol_52.gif]
(e)  Is the function f continuous at x = 1?  Yes, because Underscript[lim , x1] f(x) = 0 = f(1)by the above work.



(6) (4 points)
Underscript[lim , x -∞] x/(4x^2 - 6 )^(1/2) =Underscript[lim , x -∞] x/(4x^2 - 6 )^(1/2) -1/x/-1/x = Underscript[lim , x -∞] -1/((4x^2 - 6) (-1/x) ^2)^(1/2) = Underscript[lim , x -∞] -1/(4 - 6/x^2 )^(1/2) =-1/(4 - 0 )^(1/2) = -1/2
(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.)