# Find intervals of f increasing/decreasing

1. Jul 9, 2010

### phillyolly

1. The problem statement, all variables and given/known data

(a) Find the intervals on which f is increasing or decreasing.
(b) Find the local maximum and minimum values of f.
(c) Find the intervals of concavity and the inflection points.

x2/(x2+3)

2. Relevant equations

3. The attempt at a solution

a) I find that
f'(x)=6x/(x2+3)2

I am stuck because (x2+3)2 has no solutions and I don't know how to define x in order to proceed further.
If anybody can help me out understanding this, I will be very thankful.

2. Jul 9, 2010

### hunt_mat

Look at the function to see what it's it does. as $$x\rightarrow\pm\infty$$, $$f(x)\rightarrow 1$$. Also note that f(x) is an even function. Your calculation shows that the one and only minimum is at x=0. This should allow you to answer the question.

Mat

3. Jul 9, 2010

### Staff: Mentor

Why are you just looking at the denominator? Since x2+3 has no real solutions, this means that the domain of f is the entire real line.

f is increasing on intervals for which f'(x) > 0, and is decreasing on intervals for which f'(x) < 0.

You're going to need the second derivative, too, in this problem.

4. Jul 9, 2010

### phillyolly

1) How do you know it is an even function?
2) Why $$f(x)\rightarrow 1$$?

5. Jul 9, 2010

### Staff: Mentor

1) Because f(-x) = f(x) for all x
2) Because x2/(x2 + 3) = 1 - 3/(x2 + 3). As x --> inf, f(x) --> 1. As x --> -int, f(x) --> 1.

6. Jul 11, 2010

### phillyolly

So how can I find the intervals of concavity based on a second derivative?

7. Jul 11, 2010

### Dick

Take the second derivative and figure out where it's positive and where it's negative?

8. Jul 11, 2010

### phillyolly

I did so, and in this case, x=1. however, the graph is concave down at 0 and it has nothing to do with a point 1.

9. Jul 12, 2010

### Dick

x=1 isn't an 'interval of concavity'. It's just a point. If you mean it's an inflection point, yes it is. But it's not the only one.