SOLUTION: Quadratic equation if y=-2x^6-32x^3-118, does y have a maximum or a minimum value? What is the value? What is the corresponding value of x?

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Question 982214: Quadratic equation
if y=-2x^6-32x^3-118, does y have a maximum or a minimum value?
What is the value? What is the corresponding value of x?

Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
We observe the leading term -2x6. 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.

What is the value? What is the corresponding value of x?
We can only answer those in reverse order. So we answer first:

"At what value of x does this maximum value occur?"


We find the derivative



Set y' = 0



We divide thru by the constant -12



Factor the sum of two cubes:



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:

What is the maximum value reached at x = -2.
We substitute in the original equation:






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



Edwin

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