Question

Let \displaystyle f(x) = \frac{x-6}{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 =

Answer 1

I know I have to get the derivative which is f'(x) = 8/(x+2)^2

Answer 2

\[f(x) = \frac{x-6}{x+2}\]

Answer 3

forget calc, just treat this rational function like you did in precalc class
vertical asymptote \(x=-2\) horizontal asymptote \(y=1\) and picture should be easy

Answer 4

assuming your derivative is right (i didn't check it) then since the numerator is 8 and the denominator is a square, that means the derivative is always positive (except at -2) so the function is always increasing

Answer 5

you remember graphing stuff like this before you took calc?
it is still the same

Answer 6

okay so what would I say for the increasing interval, cause it is not (-inf,inf) or (-inf,-2)U(-2,inf). So what would I say?
And I do remember those graphs.

Answer 7

i would say it is increasing everywhere except at -2 where it is undefined

Answer 8

if you want interval notation
\[(-\infty, -2)\cup (-2,\infty)\] which is a long winded way of saying "everything but -2"

Answer 9

there is clearly no local max or min
the derivative is never zero, the numerator of the derivative is a number

Answer 10

Okay, thanks, I got it now.

Answer 11

yw