$f(x) ={\frac{2}{1+x^2}}$
I need to find where this function is increasing, decreasing, concave and convex. I've found it's derivative:
$f'(x)=\frac{-4x}{(1+x^2)^2}$
Now you're supposed to make either $f'(x)>0$ when it's increasing and $f'(x)<0$ when it's decreasing, but that gives:
Increasing: $x<0$ Decreasing: $x>0$
But what does that actually mean? It's just confusing, usually when I solve these you get 2 solutions, so it's for example increasing on the interval of $(-2,2)$. What does this one tell me? What's the easiest way to find where this function is increasing and decreasing?
Then I also did the second derivative, which is:
$f''(x)= \frac{4(3x^2-1)}{(1+x^2)^3}$
How does this all help me find my solution?
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