Problem+12+Max+and+Mike+J.

Problem 12:

For the function : f(x)= x²/x²+3, find a) The intervals over which f is increasing and decreasing b) The local maximum(s), minimum(s) values of f c) The intervals of concavity and inflection

a) Using the quotient rule, we can find the first derivative.

f'(x) = 6x/(x²+3)²

From the equation of the first derivative, and from looking at the graph of the original function we know that f(x) is decreasing over the interval (-∞,0) and from (0,∞). Although the slopes are positive on the right hand side of the graph, the slopes are decrease in magnitude as they approach 0. Therefore, both sides of the graph are decreasing, so there are no points where f(x) is increasing.

Ms. Sweeney comment: Well, you want to be using calculus and derivatives to be garnering information about the graph without looking at it, so the rules of the game is really that you don't make conclusions based on looking at the original function (which is of course reasonable in any other situation -- right now we want to practice using the derivative!). When you say "slopes" what do you mean? If you mean the first derivative, then say it, since it appears that you want to use the First Derivative Test! Secondly, if we are looking for where the function increases and where it decreases, our only question is whether or not the first derivative is positive or negative -- we don't really care whether they increase or decrease. So....Mike I will leave it up to you to fix this. Where is the first derivative positive? Where is it negative? What does this mean about where the function increases and where it decreases? b) Local max's and min's occur at critical values which are defined as points where the derivative does not exist or is equal to zero. Our function a cusp opening up from the origin, therefore our function has a local minimum at the point( (0,0). There are no local maximums.

Actually a local max or min occurs at a critical value but REALLY at a critical value at which the function changes from increasing to decreasing or vice versa -- in other words, where the first derivative changes from positive to negative, or vice versa. Where does this happen? 

c) The simplest way to determine concavity is to look at the graph of the original function.  The function is clearly concave down over the intervals : (-∞,0) and (0,∞). There is also another way to determine concavity. If the second derivative is negative the graph is concave down, which is the case for our function. There is no point of inflection because there is no point where the graph of f(x) changes from positive to negative or vice versa.

What is you are not able to look at the graph of the function -- if you had a "no calculator" test, for example??? Really, you would need to rely on the second derivative. What IS it in this case? I actually suspect that there IS an inflection point -- in other words, that there is a value (or two) at which the second derivative DOES change signs!  