What are the absolute extrema of #f(x)=(2x^3-x)/((x^2-64)# in #[-8,8]# ?

1 Answer
Jan 5, 2017

Answer:

In #[-8, 8],# the absolute minimum is 0 at O. #x = +-8# are the vertical asymptotes. So, there is no absolute maximum. Of course, #|f| to oo#, as #x to +-8#..

Explanation:

The first is an overall graph.

The graph is symmetrical, about O.

The second is for the given limits #x in [-8, 8]#

graph{((2x^3-x)/(x^2-64)-y)(y-2x)=0 [-160, 160, -80, 80]}

graph{(2x^3-x)/(x^2-64) [-10, 10, -5, 5]}

By actual division,

# y = f(x) = 2x +127/2(1/(x+8)+1/(x-8)) #, revealing

the slant asymptote y = 2x and

the vertical asymptotes #x = +-8#.

So, there is no absolute maximum, as #|y| to oo#, as #x to +-8#.

#y'=2-127/2(1/(x+8)^2+1/(x-8)^2)=0#, at #x = +-0.818 and x=13.832#,

nearly.

#y'= 127((2x^3+6x)/((x^2-64)^3)#, giving x= 0 as its 0. f''' is #ne# at

x=0. So, origin is the point of inflexion (POI). In #-8, 8]#, with respect to the

origin, the graph ( in between the asymptotes #x = +-8# ) is convex

in #Q_2 and concave ib #Q_4#.

So, the absolute minimum is 0 at the POI, O.