Find intervals of f increasing/decreasing

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Homework Statement



(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)

Homework Equations


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.
 
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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
 
phillyolly said:

Homework Statement



(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)

Homework Equations





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.
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.
phillyolly said:
If anybody can help me out understanding this, I will be very thankful.
 
hunt_mat said:
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

1) How do you know it is an even function?
2) Why f(x)\rightarrow 1?
 
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.
 
So how can I find the intervals of concavity based on a second derivative?
 
phillyolly said:
So how can I find the intervals of concavity based on a second derivative?

Take the second derivative and figure out where it's positive and where it's negative?
 
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.
 
phillyolly said:
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.

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.
 
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