Question

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

Answer 1

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.