[Math] How to find where the function is decreasing/increasing/concave/convex $f(x) ={\frac{2}{1+x^2}}$

derivatives

$f(x) ={\frac{2}{1+x^2}}$

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?

Best Answer

  1. The endpoints $\pm\infty$ are always present. Your results say that $f$ is increasing on $(-\infty, 0)$ and decreasing on $(0, \infty)$.
  2. For convexity / concavity you need to do the same thing to $f''$, $f'' > 0$ implies convexity and $f'' < 0$ implies concavity. Note that the numerator is the only interesing component of $f''$ because the denominator is always $\ge 1$, so it doesn't change the sign. Now to find the critical points, solve for $$4(3x^2-1) = 0 \Leftrightarrow x^2-\frac13=0 \Leftrightarrow x = \pm \frac1{\sqrt3}$$
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