`f(x) = 1 + 1/x - 1/(x^2)` (a) Find the vertical and horizontal asymptotes. (b) Find the intervals of increase or decrease. (c) Find the local maximum and minimum values. (d) Find the intervals of concavity and the inflection points. (e) Use the information from parts (a)-(d) to sketch the graph of `f`.
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=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
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 -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.
f'(a) >0 and f'(b) <0 in the interval a<c<b implies concave down and local maximum point occurs at x=c.
f'(a) <0 and f'(b) >0 in the interval a<c<b implies concave up and local minimum point occurs at x=c.
Note 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` .
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`
Possible inflection point occurs at x=a when f"(a)=0.
`2x-6 = x^4*0`
2x-6 = 0
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)= `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` )