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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)    sin((5π)/4) =-1/2^(1/2)


(b)    csc((5π)/4) =1/sin((5π)/4) = 1/-1/2^(1/2) = -2^(1/2)


(c)    Suppose  0≤θ≤π/2,  and sec(θ) = 8.  Find tan(θ).

From sec(θ)=8, we get 1/cos(θ) = 8, or cos(θ) = 1/8.
Also, sin^2(θ) + cos^2(θ) = 1, meaning sin(θ) =  (1 - cos^2(θ))^(1/2) =  (1 - (1/8)^2)^(1/2) =  (1 - 1/64)^(1/2) =  63/64^(1/2) =  63^(1/2)/8
Then tan(θ)=sin(θ)/cos(θ) =  63^(1/2)/8/1/8 =  63^(1/2).  But, as θ is in the first quadrant, tan(θ) is the POSITIVE square root of 63.


(d)   Find all solutions of the equation  cos(x) = 1/2

[Graphics:HTMLFiles/T1F04A_16.gif]

Looking at the unit circle, you can see that the solutions are all values of x of the form
π/3 + 2n π  or   (5π)/3 + 2n π  , where n is an integer.




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

(a)   f(π) =sin(π)+1 = 1


(b)   (f∘g) (x) =FormBox[RowBox[{f(g(x)), =, RowBox[{sin(x^(1/2) + x), +, RowBox[{1, Cell[]}]}]}], TraditionalForm]


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


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


(e)   State the domain of f.   All real numbers.


(f)   State the range of f.    [0, 1]


(3) Find the equation of the line having slope 2/3 and passing through (4, 5). Put your answer in slope-intercept form.

y - 5 = 2/3 (x - 4)  y - 5 = 2/3x - 8/3  y = 2/3x - 8/3 + 5  y = 2/3x - 8/3 + 15/3  y = 2/3x + 7/3

(4)   Underscript[lim , x1] (x^3 - 3x + 1) =1^3 - 3 (1) + 1 = -1



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



(6)    Underscript[lim , x16] (x ^(1/2) - 4)/(x - 16) =Underscript[lim , x16] (x ^(1/2) - 4)/(x - 16) (x ^(1/2) + 4)/(x ^(1/2) + 4) = Undersc ...  - 16) (x ^(1/2) + 4)) = Underscript[lim , x16] 1/(x ^(1/2) + 4) = 1/(16 ^(1/2) + 4) = 1/8



(7)    Underscript[lim , x3] (1/3 - 1/( x))/(x - 3) =Underscript[lim , x3] (x - 3)/(3x) /(x - 3) = Underscript[lim , x3] (x - 3)/(3 ... lim , x3] (x - 3)/(3x) 1/(x - 3) = Underscript[lim , x3] 1/(3x) = 1/(3 (3)) = 1/9



(8)    Underscript[lim , x3] (x^2 - 9)/(x - 3) = Underscript[lim , x3] ((x + 3) (x - 3))/(x - 3) = Underscript[lim , x3] (x + 3) = 3 + 3 = 6



(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∞] (8 - x^3)/(x - 10) =-∞

(11)    Underscript[lim , x0^+] sin(x)/(5x^(1/2)) =Underscript[lim , x0^+] sin(x)/(5x^(1/2)) x^(1/2)/x^(1/2) = Underscript[lim , x0^+] sin(x)/(5x) x^(1/2) = 1/50^(1/2) = 0


(12)    Underscript[lim , x0] cos(x + π) =   cos(Underscript[lim , x0] (x + π)) = cos(π) = -1


(13)   Underscript[lim , x0] (x sin(x) - x^3)/(4x^2) = Underscript[lim , x0] ((x sin(x))/(4x^2) - x^3/(4x^2)) = Underscript[lim , x0] (sin(x)/(4x) - x/4) = 1/4 - 0 = 1/4


(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