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

Jan 5, 2017

In $\left[- 8 , 8\right] ,$ the absolute minimum is 0 at O. $x = \pm 8$ are the vertical asymptotes. So, there is no absolute maximum. Of course, $| f | \to \infty$, as $x \to \pm 8$..

#### Explanation:

The first is an overall graph.

The graph is symmetrical, about O.

The second is for the given limits $x \in \left[- 8 , 8\right]$

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,

$y = f \left(x\right) = 2 x + \frac{127}{2} \left(\frac{1}{x + 8} + \frac{1}{x - 8}\right)$, revealing

the slant asymptote y = 2x and

the vertical asymptotes $x = \pm 8$.

So, there is no absolute maximum, as $| y | \to \infty$, as $x \to \pm 8$.

$y ' = 2 - \frac{127}{2} \left(\frac{1}{x + 8} ^ 2 + \frac{1}{x - 8} ^ 2\right) = 0$, at $x = \pm 0.818 \mathmr{and} x = 13.832$,

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 $x = \pm 8$ ) is convex

in ${Q}_{2} \mathmr{and} c o n c a v e i b$Q_4#.

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