<s>[INST] Finding the intervals of increase and decrease and concavity of $y=\frac{\sqrt{1+|x-2|}}{1+|x|}$ I have the function $f(x)=\frac{\sqrt{1+|x-2|}}{1+|x|}$, which I need to analyze. I have already proven that $\lim\limits_{x \to +-\infty}{f(x)}=0$, therefore the function has only a horizontal asymptote at $y=0$. Then I looked at three cases for the derivative

*

*$x>2 \Longrightarrow f'(x)=\frac{1-3x}{2\sqrt{x-1}(1+x)^2}$

*$x \in[0,2]\Longrightarrow f'(x)=\frac{x-7}{x(1+x)^2\sqrt{3-x}}$

*$x<0 \Longrightarrow f'(x)=\frac{5-x}{2(1-x)^2\sqrt{3-x}}$
Now I need to look at the intervals of concavity of the function. I suppose I have to differentiate all derivatives for a second time and look at the results in each of the corresponding intervals. But this would require too much calculations. Isn't there a simpler way to find the intervals of increase/decrease and of concavity?
 [/INST] $$f(x)=\left\{
\begin{array}{ll}
 \frac{\sqrt{3-x}}{1-x} & x<0 \\
 \frac{\sqrt{3-x}}{x+1} & 0 \le x<2 \\
 \frac{\sqrt{x-1}}{x+1} & x\geq 2 \\
\end{array}\right.
$$
$$
f'(x)=
\left \{  
\begin{array}{ll}
 \frac{5-x}{2 \sqrt{3-x} (x-1)^2} & x<0 \\
 \frac{x-7}{2 \sqrt{3-x} (x+1)^2} & 0<x<2 \\
 \frac{3-x}{2 \sqrt{x-1} (x+1)^2} & x\ge2 \\
\end{array}\right.
$$
$$
f''(x)=
 \left\{ 
\begin{array}{ll}
 \frac{-3 x^2+30 x-59}{4 (3-x)^{3/2} (x-1)^3} & x<0 \\
 \frac{3 x^2-42 x+83}{4 (3-x)^{3/2} (x+1)^3} & 0<x<2 \\
 \frac{3 x^2-18 x+11}{4 (x-1)^{3/2} (x+1)^3} & x>2 \\
\end{array}\right.
$$
Plot below, for reference

$$...$$

</s>