Alaina's blog, 21 Nov. 2010

Chapter 3 Section 5
limits at infinity

-to find a horizontal asymptote, you take the lim x->infinity
y=ans is an asymptote

degree of top = degree of bottom => coefficient
degree of top > degree of bottom => -inifinity or infinity
degree of top < degree of bottom => 0

Ex 1: y=2x+5/(3x^2+1)
lim x->infinity 2x+5/(3x^2+1)=0
lim x-> infinity is 0
0 is a horizontal asymptote.

Chapter 3 Section 6
Curve Sketching

I can tell you how to work problems to get all of the critical information to sketch the graph but i will not show the graph.

STEPS
1. identify the domain and range
2. find the x and y intercepts, identify the vertical and horizontal asymptotes
3. do the first derivative test
4. do the second derivative test
5. plot the information and sketch

Ex: y=2(x^2-9)/(x^2=4)
1. domain: x^2-4=0; x=+/- 2
domain=(-infinity, -2)u(-2,2)u(2,infinity)
range: find horizontal asymptotes (limits approaching infinity) and set up intervals
y=2; (-infinity,2)u(2,infinity)

2. vertical asymptotes: x=+/-2
horizontal asymptotes: y=2
x intercepts: (3,0)(-3,0)
y intercepts: (0,9/2)

3.a) use quotient rule
f'(x)=20x/(x^2-4)^2=0
20x=0
x=0
b)x=+/-2
c)(-infinity,-2)u(-2,0)u(0,2)u(2,infinity)
d) f'(-3)=-ve; f'(-1)=-ve; f'(1)=+ve; f'(3)=+ve
e) decreasing; decreasing; increasing; increasing
f) x=0 is a min

4. a) second derivative(quotient rule)
f''(x)=-20(3x^2+4)/(x^2-4)^3=0
x=+/- squareroot (-4/3)
b)x=+/-2
c)(-infinity,-2)u(-2,2)u(2,infinity)
d)f''(-3)=-ve; f''(0)=+ve; f''(3)=-v3
e)decreasing; increasing; decreasing
f)concave down; concave up; concave down
x=-2,2 are points of inflection