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Calculus I                                            Test #1                          October  5, 2004

Name _________________          R. Hammack                              Score______
Directions. Answer in the space provided. Show as much work as is reasonable.
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(1)   
(a)    cos((7π)/6) =-3^(1/2)/2

(b)    cot((7π)/6) =cos((7π)/6)/sin((7π)/6) = -3^(1/2)/2/-1/2 = 3^(1/2)


(c)    Suppose  π/2≤θ≤π,  and tan(θ) = -8.  Find cos(θ).
Suppose for the moment that the problem were  tan(θ) = 8.
Since tan(θ) = OPP/ADJ = 8/1, we get a triangle that looks like this.
[Graphics:HTMLFiles/index_10.gif]
Then cos(θ) = ADJ/HYP = 1/65^(1/2).
However, in this problem θ is in the second quadrant, the answer is cos(θ) = -1/65^(1/2)


(d)   Find all solutions of the equation  cos(x) + sin(x) = 0

[Graphics:HTMLFiles/index_15.gif]
From the unit circle, you can see that the solutions are x = (3π)/4 + n π, where n is an integer.



(2) Suppose   f(x) = sin(x) + cos(x)  and   g(x) = x^(1/2) + 1.

(a)   f(π) =sin(π) + cos(π) = 0 - 1 = -1

(b)   (f∘g) (x) =f(g(x)) = sin(x^(1/2) + 1) + cos(x^(1/2) + 1)

(c)   g(f(x)) =(sin(x) + cos(x))^(1/2) + 1

(d)   f/g (x) =(sin(x) + cos(x))/(x^(1/2) + 1)

(e)   State the domain of g.   [0, ∞)

(f)   State the range of g.   [1, ∞)


(3) A line has slope 9 and passes through the point (20, 4).   What is its x-intercept?
By the point-slope formula, the equation of the line is
y - 4 = 9 (x - 20)  y = 9x - 176
Now, to find the x-intercept, set y = 0, and solve for x.
0 = 9x - 176  9x = 176  x = 176/9
(4)   Underscript[lim , x1] (x^3 - 3x + 1) =1^3 - 3 (1) + 1 = -1

(5)   Underscript[lim , x1] (x - 1)/(x^4 - 1) = Underscript[lim , x1] (x - 1)/((x^2 - 1) (x^2 + 1)) = Underscript[lim , x1] (x ... + 1)) = Underscript[lim , x1] ( 1)/((x + 1) (x^2 + 1)) = ( 1)/((1 + 1) (1^2 + 1)) = ( 1)/4


(6)    Underscript[lim , x25] (x ^(1/2) - 5)/(x - 25) = Underscript[lim , x25] (x ^(1/2) - 5)/(x - 25) (x ^(1/2) + 5)/(x ^(1/2) + 5) = Undersc ... - 25) (x ^(1/2) + 5)) = Underscript[lim , x25] 1/(x ^(1/2) + 5) = 1/(25 ^(1/2) + 5) = 1/10


(7)    Underscript[lim , x -2] (1/x + 1/( 2))/(x + 2) =Underscript[lim , x -2] (2 + x)/(2x) /(x + 2) = Underscript[lim , x -2] (2 + x ... x -2] (2 + x)/(2x) 1/(2 + x) = Underscript[lim , x -2] 1/(2x) = 1/(2 (-2)) = -1/4


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


(9)    Underscript[lim , x∞] (2x + 1)/(3x^2 + 1)^(1/2) =Underscript[lim , x∞] (2x + 1)/(3x^2 + 1)^(1/2) (1/x)/(1/x) =Underscript[lim , x∞] (2 + 1/x)/((3x^2 + 1) (1/x)^2)^(1/2) = Underscript[lim , x∞] (2 + 1/x)/(3 + 1/x^2)^(1/2) = (2 + 0)/(3 + 0)^(1/2) = 2 /3^(1/2)


(10)   Underscript[lim , x∞] (x - 8)/(x^2 - 7x - 8) =0

(11)    Underscript[lim , x∞] sin ((π x)/(2 - 3x)) =sin (Underscript[lim , x∞] (π x)/(2 - 3x)) = sin (-π/( 3)) = -3^(1/2)/2


(12)    Underscript[lim , x4] sin(x - 4)/(4 - x) =1


(13)   Underscript[lim , x0] (1 - cos(x))/x =0


(14)  Consider the function f(x) = (x^3 - 9x)/(x^3 + 3x^2).

Note f(x) = (x^3 - 9x)/(x^3 + 3x^2) = x(x^2 - 9)/x^2(x + 3) = (x(x - 3) (x + 3))/x^2(x + 3) = (x - 3)/x (provided x is neither 0 nor -3 )

(a)    At which values of x is  f  discontinuous?

0 and -3 because f is not defined there


(b) Find the vertical asymptotes (if any) of f.
The denominator is 0 for x = 0 or -3. These are the candidates for the locations of the vertical asymptotes.
Underscript[lim , x -3^+] f(x) = Underscript[lim , x -3^+] (x - 3)/x = 2, so no asymptote here.
Underscript[lim , x0^+] f(x) = Underscript[lim , x0^+] (x - 3)/x = ∞, so line x=0 is a V.A.

(c)  Find the horizontal asymptotes (if any) of f.

Underscript[lim , x∞] f(x) = Underscript[lim , x∞] (x^3 - 9x)/(x^3 + 3x^2) = 1, so line y = 1  is the horizontal asymptote.


(d) For which values of x does f(x) = -2 ?
We need to solve
f(x) = -2  (x - 3)/x = -2  x - 3 = -2x  3x = 3  x = 1
Answer: x = 1