Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 45

a) For a rational function: R(x) = (N(x))/(D(x)) , a vertical asymptote (VA) occurs at x=c when D(c)=0. It is the value of x that makes the function undefined.
Express f(x) = 1 + 1/x -1/x^2 as one fraction:
Least common denominator (LCD) = x^2.
Multiply each term by the missing factor:
f(x) = 1*(x^2/x^2) +(1/x)(x/x)-1/x^2
f(x) = x^2/x^2 +x/x^2 -1/x^2
f(x) = (x^2+x-1)/x^2

In the function f(x), we have D(x) =x^2.
Solve for x when D(x)=0
x^2=0
sqrt(x^2) =sqrt(0)
x=0 as the vertical asymptote

To solve for horizontal asymptote we compare the highest degrees of the leading coefficient from numerator N(x) and denominator D(x).
an = first integer of the leading coefficient in N(x): anxn +.....+ao
bn = first integer of the leading coefficient in D(x): bnxn +.....+bo
Conditions:
n < m horizontal asymptote: y=0
n= m horizontal asymptote: y =a_n/b_n
n>m horizontal asymptote: NONE

In f(x) =(x^2+x-1)/x^2 , we have n=2 and m=2 that shows n=m
then y=1/1 or y =1 as the horizontal asymptote.

b) To find the intervals of increase or decrease, recall:
--> f'(x) = positive value implies increasing f(x) of an interval I.
--> f'(x) = negative value implies decreasing f(x) of an interval I.

Applying quotient rule derivative:
f'(x) = ((2x+1)(x^2)-(x^2+x-1)(2x))/(x^2)^2
f'(x) = (2x^3+x^2 -2x^3-2x^2+2x)/(x^4)
f'(x) = (-x^2 +2x)/(x^4)
f'(x) = (-x+2)/x^3

Solve for critical point(s) by letting f'(x)=0:
(-x+2)/x^3=0
(-x+2)=x^3*0
-x+2=0
-x+2+x =0+x
2=x.
Sign table:
x -1 0 1 2 3
f'(x) -3 und. 1 0 -1/(27)
It shows the function has increasing interval: (-oo , 0) and (2, +oo) while decreasing interval: (0,2).

c) The local extrema occurs at critcal value: x=c such that f'(c)=0.
f'(2)= 0 then local extrema occurs at x=2.
Conditions:
f'(a) >0 and f'(b) <0 in the interval a f'(a) <0 and f'(b) >0 in the interval aNote that f'(1)=1 and f'(3)=-1/(27) then it implies a local maximum occurs at x=2.
Plug-in x=2 in f(x)=(x^2+x-1)/x^2 .
f(2)= (2^2+2-1)/2^2
local maximum value: f(2) = 5/4 or 1.25
local minimum value: none
d) Use second derivative test to find concavity and inflection point.
Apply quotient rule derivative on f'(x)=(-x+2)/x^3
f"(x)=((-1)(x^3)-(-x+2)(3x^2))/(x^3)^2
f"(x)=(-x^3 +3x^3-6x^2)/(x^6)
f"(x)=(2x^3-6x^2)/(x^6)
f"(x)=(2x-6)/(x^4)
Possible inflection point occurs at x=a when f"(a)=0.
Let f"(x)=0.
(2x-6)/x^4=0
2x-6 = x^4*0
2x-6 = 0
2x =6
x =3.
Real inflection point occurs when there is a change of concavity before and after x=3.
x 2 3 4
f"(x) -1/8 0 1/(128)
Concavity: Up Down.
Plug-in x=3 in f(x)=(x^2+x-1)/x^2.
f(3) =(3^2+3-1)/3^2
f(3)= 11/9 or 1.22.
Inflection point (IP) : (3,(11)/9 )
Intervals of concavity :
x -1 0 1 2 3 4
f"(x) -8 und. -4 -1/8 0 1/(128)
Concave down: (-oo , 0) and (0, 3)
Concave up: (3, +oo )
e) Graph.