Find intervals of f increasing/decreasing

[phillyolly]

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.

 

[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

 

[Mark44]

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.

If anybody can help me out understanding this, I will be very thankful.

 

[phillyolly]

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 $$f(x)\rightarrow 1$$?

 

[Mark44]

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.

 

[phillyolly]

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

 

[Dick]

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?

 

[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.

 

[Dick]

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.