Question

How do you sketch the graph by determining all relative max and min, inflection points, finding intervals of increasing, decreasing and any asymptotes given `f(x)=(x-4)^2/(x^2-4)`?

Answer 1

Relative maximum `f(1)=-3` and relative minimum f(4)=0. Asymptotes : `uarr x=+-2 darr. POI : (5.034, 0.0500). f darr` : x in (1, 2) and (2, 4)`and`f uarr :, x in (-oo, -2), (-2, 1) and (4, oo) #

Explanation:

I took 30' for this answer. I should have a break. Please make another answer, without attempting to edit my answer. After some time, I would edit my answer, myself.. I have to repeat this because some rush to supply typo omissions of characters like #. I would do it myself. Logical and computational errors can be pointed out in the comment column and not by editing my answer. .

graph{(x-4)^2/(x^2-4) [-30, 30, -15, 15]}

x-intercept ( y = 0) : 4

y-intercept ( x = 0 ) : `-4`

By actual division and resolving into partial fractions,

`y=f = 1+1/(x-2)-9/(x+2)`

As `x to 2 and -2, f to +-oo`

`f'=-1/(x-2)^2+9/(x+2)^2`

`=8((x-1)(x-4))/(x^2-4)^2`

#=0, when x = 1 and 4,

`f darr` intervals:

`<0, x in (1, 2) and (2, 4)` and

`f uarr` intervals :

`>0, x in (-oo, -2), (-2, 1) and (4, oo)`

Local minimum``f(4)=0#

Local maximum : `f(1)=-3`

`f''=2/(x-2)^3-18/(x+2)^3=0`, when `9^(1/3)(x-2)=(x+2)`, giving

`x=2(9^(1/3)+1)/(9^(1/3)-1)=5.7034, nearly`

`f'''= -6/(x-2)^4+54/(x+2)^4 ne 0, at x = 5.034`, and here y = 0.05,

nearly

So, the POI is at x = 5.034, nearly