Find the interval on which the function is increasing Find the interval on which the function $f(x) = 3 / (x^4 + 1)$ is increasing. 
I draw a graph and got the answer the interval should be $(-2, 3) \cup (2, 3)$, wonder why it is not correct though?
Thanks and Sorry for not formatting the question very well. 
 A: The function is as follows: $f(x)=\frac {3}{x^4+1}$.
Taking the derivative of $f(x) = y$ with respect to its argument x, and setting $\frac {dy}{dx}=0$, gives us information regarding where $f(x)$ reaches its local extremum/slope of $0$ (local maximum or minimum). 
So find such a point where the function achieves a zero rate, or constant slope, set it's derivative to equal zero and solve for x.
Thus we have: 
$$f(x)=\frac {3}{x^4+1}\Rightarrow f'(x)=\frac {-12x^3}{(x^4+1)^{-2}}=0\Rightarrow x=0$$
This implies that our function $f(x)$ achieves a slope of zero at $x=0$. Plugging this value back into our original function gives a corresponding point of $(x,y)=(0,3)$.
To determine the intervals of increase and decrease, we simply create a deleted neighborhood of $f(x), s.t. x\in[-\infty,0]\cup[0,\infty].$ Test points within these intervals (into $f'(x)$), to determine whether the function is increasing or decreasing for that given bound.
Finally, we see the interval of increase is $x\in(-\infty,0]$ and the interval of decrease is $x\in[0,\infty)$. Please note the point $0$ is included in these intervals, as a slope of zero is mathematically defined as a function which both increases and decreases.
A: Hint: What do derivatives tell you? When is $-12x^3\geq 0$? 
A: HINT
You have
$$
f(x) = 3\left(x^4+1\right)^{-1}
$$
so you can compute $f'(x)$ by chain rule and see where $f'(x) > 0$.
A: Simply diffrenciate the $$ f(x) = \frac{3}{(x^4 + 1)}$$ 
to get $$ f^\prime(x) = \frac{-12 x^3}{(x^4 + 1)^2}$$
and for increasing conditions,  $$ f^\prime(x)>0$$
which gives you basically,
$$ x^3<0$$
thus, any $$x \in R: x<0$$
or in human languages, any Real negative number will make the function increasing.  Cheers!
