Question

Let f(x)= 4-(8/x)+(8/x^2). Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).

1. f is increasing on the intervals
2. f is decreasing on the intervals
3. The relative maxima of f occur at x =
4. The relative minima of f occur at x =

The derivative is f'(x) = (8 (-2 + x))/x^3 and x = 2
This is as far as I got. Help please.

Answer 1

let me check the derivative, then we can complete this

Answer 2

did you get
\[\frac{8(x-2)}{x^3}\]

Answer 3

yup, that is it.

Answer 4

so only zero is at \(x=2\)

Answer 5

the derivative changes sign twice, at \(x=0\) and at \(x=2\)

Answer 6

function itself is undefined at \(x=0\) that is a vertical asymptote

Answer 7

can you help with one person? @satellite73

Answer 8

derivative is positive, then negative, then positive
function is increasing, then decreasing, then increasing
that makes the point at \(x=2\) a local min

Answer 9

and x = 0 would be a local max, right?

Answer 10

oh no!

Answer 11

function is not defined at \(x=0\)
you have a vertical asymptote there

Answer 12

okay, so no local maximum.

Answer 13

no not at all
this is the same as
\[f(x)=\frac{4 (x^2-2 x+2)}{x^2}\]\
another regular rational function

Answer 14

Okay. Thanks again. Your help is appreciated.

Answer 15

yw

Answer 16

when you do these on a computer, you should cheat so see what you are working with

Answer 17

http://www.wolframalpha.com/input/?i=+4-%288%2Fx%29%2B%288%2Fx^2%29