1.  Cot(x) has vertical asymptotes at x=kπ, where kεZ.

2.  A horizontal asymptote is the behavior of a function when it heads to ±∞. Tangent has a domain from -π/2 to π/2, so arctan is limited to that range. Here, we want to know the range of 2 arctan x, so the answer is -π and π

3.  Vertical asymptotes are when there is an invalid point in the domain, such as a division-by-zero point. Since 9x4+1 is never zero, there are no vertical asymptotes. Check for horizontal asymptotes by plugging in x=±∞ and see what happens. The constants cease to be important, so we have (in essence) 9x² / 3x² = 3, so the horizontal asymptote in both directio0ns is y=3

4. 

a.  The domain is all R except for x=2.

b.  To find the x-intercepts, set y=0 and solve. x²-2x+4=0 → [2±√(4-16)]/2 gives a complex value, so there are no x-intercepts

c.  To find the y-intercept, set x=0 and get 4=-2 = -2

d.  The critical points are where the f' = 0, so find f'. gf'-fg'/g² = [(x-2)(2x-2)-(1)(x²-2x+4)]/(x-2)² = 2x²-6x+4 - (x²-2x+4) / (x-2)² = 2x²-6x+4-x²+2x-4 / (x-2)² = x²-4x / (x-2)² = x(x-4)/(x-2)². When is that equal to 0? When x(x-4)=0 as long as x != 2. Well, that gives x=0 and x=4, so the points are (0, -2) and (4, 6)

e.  So, now we want to know when x(x-4)/(x-2)² is positive or negative. The bottom is always positive, except at x=2. The top two terms are the same sign (and therefore the product is positive) when x<0 or x>4, so it is increasing on the domain (-∞,0)U(4,∞) and decreasing on the domain (0, 4)

f.  Now, we need the second derivative. gf'-fg'/g² = (x²-4x)/(x-2)² gives (x-2)²(2x-4) – (2(x-2))(x²-4x) / (x-2)4. Factor out (x-2)/(x-2) to get (x-2)(2x-4)-[2(x²-4x)]/(x-2)³ = (2x²-8x+8 – 2x²-8x) / (x-2)³ = 8 / (x-2)³. Evaluate that at x=0 to get 8/-8=-1, so a local maxima. Evaluate at x=4 to get 8/8=1, so local minima.

g.  To determine concavity, we observe when f''>0 or f''<0 or does not exist, so when is 8/(x-2)³ positive? When (x-2)³>0, and that happens when x>2. It is negative when x<2. Therefore, the function is concave down when x<2 and concave up when x>2

h.  Obviously, there is a vertical asymptote at x=2. There is no horizontal asymptote since ∞/∞ is not a number, but there is an oblique asymptote. Divide (x²-2x+4)/(x-2) and you get x + 4/(x-2), so there is an oblique asymptote at y=x

7.  L'Hopital's rule works here. Derivative of top and bottom reduces this to (2-cos x)/cos x. As x goes to 0, this goes to (2-1)/1 = 1

8.  This is one of the basic definitions, but let's run it through.
Limx→∞ (1-1/x)x = Limx→∞ e(x ln(1-1/x))

Since ex is continuous, we can pull it out of the limit. It looks a bit strange, but...

eLim x→∞ ln(1-1/x)/(1/x)

Here we need L'Hopital again.

a.  d/dx ln(1-1/x) = d/dx (1-1/x) / (1-1/x) = 1/x²(1-1/x).

b.  d/dx 1/x = -1/x².

c.  [1/x²(1-1/x)]/[-1/x²] = -x² / x²(1-1/x) = -1 / (1-1/x)

eLim x→∞ -1 / (1-1/x)

e-1/(1-0)

e-1/1

e-1 or 1/e

9.

Differentiate top and bottom of the fraction (L'Hopital's Rule, yet again) to get pepx / 3 = 5. Multiply both sides by 3 to get pepx=15. When x goes to 0, epx heads to e0=1, so the left side when x=0 becomes just 1*p, so p=15

Atul N Roy Page 2