how to count the roots of a function for $ f(x) = \frac{(2x+1)}{x^2*(x+1)^2}$

Question

When constructing the sign table for the function underneath, the signs of the second derivative should be:

(sorry, don't know how to fix tables...)

x -1 0

f"(x) + | not defined | -

However, the roots should be calculated twice, each, no? So then I don't get why f" gets positive for x < -1. Because on the right side, it's -, then changes to - for -1

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

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

Thx in advance!

babi

Answer 1

Note that there is a sign problem, the second derivative is $-\frac{2x+1}{x^2(x+1)^2}$.

Now since the denominator is always $\ge 0$, the sign of $f''(x)$ is not hard to compute. The only places it can change sign are at points where the numerator is $0$. This only happens at $x=-1/2$.

Because of the singularities at $x=0$ and $x=-1$, it is probably best to say that $f''(x)\gt 0$ in $(-\infty,-1)$, also in $(-1,-1/2)$, and that $f''(x)\lt 0$ in $(-1/2,0)$ and in $(0,\infty)$.

It would be misleading to say, for example, that the second derivative is positive for $x\lt -1/2$.

Answer 2

It suffices to count the numerator's roots, since those will make the fraction zero. So,

\begin{align*} f''(x) = 0 &\iff 2x+1=0 \\ &\iff x=-\frac{1}{2} \end{align*}

It can be shown that this is not an extraneous root by checking the denominator.