# Thread: Find intervals of f increasing/decreasing

1. ## Find intervals of f increasing/decreasing

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. 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. Originally Posted by 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. 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.
$\displaystyle \displaystyle f(x)=\frac{x^2}{x^2+3}$

$\displaystyle \displaystyle f'(x)=\frac{(x^2+3)(2x) - (x^2)(2x)}{(x^2+3)^2}$

$\displaystyle \displaystyle =\frac{6x}{(x^2+3)^2}$

like you said.

Set $\displaystyle \displaystyle f'(x) = 0$

So $\displaystyle \displaystyle 6x = 0 \implies x=0$

and so on..

3. Hi!

How can I solve (x^2+3)^2?
I cannot put x^2+3=0 because x^2=-3 does not exist.

4. Originally Posted by phillyolly
Hi!

How can I solve (x^2+3)^2?
I cannot put x^2+3=0 because x^2=-3 does not exist.
Why are you trying to solve $\displaystyle \displaystyle(x^2+3)^2=0$? Suppose we have two real numbers $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle b, b\ne0$, then $\displaystyle \displaystyle \frac{a}{b}=0 \iff a=0$.