Question 982214
We observe the leading term -2x<sup>6</sup>. The degree is the exponent 
6, an even number.  That tells us that the extreme left and right behaviors
are the same.  The negative coefficient -2 tells us that both extreme
left and right behaviors are downward.  Thus the function has a maximum 
value.  
</pre>
What is the value? What is the corresponding value of x?
<pre>
We can only answer those in reverse order. So we answer first:
</pre>
"At what value of x does this maximum value occur?"
<pre>
{{{y=-2x^6-32x^3-118}}}

We find the derivative

{{{"y'"=-12x^5-96x^2}}}

Set y' = 0

{{{-12x^5-96x^2=0}}}

We divide thru by the constant -12

{{{x^2(x^3+8)=0}}}

Factor the sum of two cubes:

{{{x^2(x+2)(x^2-2x+4)=0}}}

The real roots are x=0, x=-2  (the third factor yields only complex roots.)

We do a first derivative test of x=0 

intervals      |   x < -2 | -2 < x < 0 | x > 0
----------------------------------------------
test value t   |    -3    |     1      |   3
sign of y'(t)  |     +    |     -      |   -
direction      | upward   | downward   | downward
                 (incr.)     (decr.)     (decr.)
 
At -2 there is a change from increasing (upward) to decreasing (downward) 
Therefore there is a relative maximum pt. at x = -2.
At 0 there is no change from decreasing (downward), thus a horizontal
inflection point at x=0.

There are no other critical values of the derivative, so
the relative maximum pt. at x = -2, is an absolute maximum point.

Now we answer this question:
</pre>  
What is the maximum value reached at x = -2.
<pre>
We substitute in the original equation:

{{{y=-2x^6-32x^3-118}}}
{{{y=-2(-2)^6-32(-2)^3-118}}}
{{{y=-2(64)-32(-8)-118}}}
{{{y=10}}}

So the maximum value is 10.  Observe the maximum point (-2,10),
and also the horizontal inflection point at (0,-118)

{{{drawing(400,400,-4,4,-150,20,
locate(.1,-112,"(0,-118)"),
graph(400,400,-4,4,-150,20,-2x^6-32x^3-118), locate(-2.5,18,"(-2,10)") )}}}

Edwin</pre>