Expert: Scotto - 2/19/2009

Question
1.) Determine the following about the graph of the equation y= (8/x^3)-(6/x).
b.) Find the x-coordinates of each point at which y is the relative maximum and each point at which y is the relative minimum. Justify your answer.
c.) Find the x-coordinate of each point of inflection. Justify your answer.

So I found f'(x) and it is y'=(24x/x^6)-(12x/x^2) and I tried to set it = 0 and solve but I'm having trouble setting it to zero because there's an horizontal asymptote there. I know the relative max=(-2,2) & relative min=(2,-2) but how do I get there?
For (c) my f''=[(24x^6-144x)/x^12]-[(12x^2-14x)/x^4] and I still cant set it =0 and solve. Can you help me?

Answer
I won't go over the answers given since a simpler approach would be to say the 8/x^3 = 8x^(-3) and 6/x = 6x^(0-1).

1b) The derivative of the function is -24/x^4 + 6/x^2.
This can be factored into (6/x²)(-4/x²+1).

Note that the first term is always non zero.

The second term has zeros at x² = 4
{since it's -1+1 at these points}.

The points are x=2 and x=-2.
Put these back into the function to get y.
Take the second derivative and put in 2 and -2
to see if they are max (negative 2nd derivative)
or min (positive second derivative).

c) Inflection points are where the second derivative is 0.

The first derivative was -24/x^4 + 6/x^2, so knowing this is
-24x^(-4) + 6x^(-2), the second derivative is -96/x^5 + 12/x^3,
since it still the power rule.  

The 12/x^3 can be factored out, leaving (12/x^3)(-8/x^2 + 1).
The only place this is zero is at -8/x² + 1 = 0.
This occurs when x² = 8, or x=±2√2.

=====================================================================

As a note for these types of problems, the answer for the first
derivative and what is used to find the next derivative can
sometimes be different.  What I mean is to express the first
derivative, it should be factored.  To get the second derivative,
use the unfactored form of the first derivative.