Extrema question - need more help please!!?

Question

Find the open interval on which the function is increasing or decreasing, and locate all relative extrema for the f(x)=x/(x+1) I posted the same question here http://answers.yahoo.com/question/index;_ylt=ArtVZyjuM0wXa4kJ.9lI9eEezKIX?qid=20061125130753AABrmR3 but I don't know how to respond to the people who answered. So I posted it again, because non of them were helpful and the last answer had a few errors (i.e., the derivative is 1/((x+1)^2) not -1/((x+1)^2) Usually, I'd set the derivative equal to zero, find the critical numbers and then find the relative min and max and see where the function is increasing/decreasing. But in this question, the derivative can never equal zero, and even though the derivative is undefined at -1, -1 is not defined for the function (so -1 can't serve as a critical number either). My question is, what do I do? What does this mean? That there's no relative extrema? And how would I figure out where it's inc/decreasing (cuz i cant sub -1in deriv.)

Answer 1

you are right, derivative is NEVER zero. In fact, it is always greater than zero. This means that the function is always increasing on the whole real line, minus the point x = -1 since the function is not defined there. there is no relative extrema since the function is monotone increasing and the interval are open intervals(no relative extrema at the starting and ending points of the interval)

Answer 2

Rewrite as f(x) = x(x+1)^(-1); at x --> -1, f(x) approaches an asymptote
The first derivative is f'(x) = (x+1)^(-1) - x(x+1)^(-2), by the chain rule
If you set this equal to 0, then solving for x involves the sqr(-1) = i, which has no meaning here; so indeed, you cannot do that trick.

What you can do is take the limit; this means, we know that there is a vertical asymptote at x = -1, so next we take the limit of f(x) as x -->-infinity and limit of f(x) as x -->+infinity.

lim(x-->-infinity)[f(x)] = lim[x/(x+1)] which I think is 1 and so is the limit as you approach + infinity; I could be wrong here about the limits, but probably not.

Answer 3

You're correct there are no relative extrema for this problem. Anytime the numerater in the derivative is unable to equal zero this means there isn't a relative min/max that exists.