$\require{color}$
Given that $$ f(x) = \begin{cases} \arctan(3x), & x<0 \\[2ex] \ \left(x+\frac32\right)\ln\left(2x+1\right), & x\geq0 \end{cases}$$ find where the function is increasing or decreasing.
My solution:
Differentiating gives
$$ f'(x) = \begin{cases} \ \frac3{1+9x^2}, & x<0 \\[2ex] \ \ln\left(2x+1\right)+\frac{2x+3}{2x+1}, & x\geq0 \end{cases}$$
$\circ$ For $x<0:$
$$f'(x)>0\implies\frac3{1+9x^2}>0\implies \{\forall x\in \mathbb{R}\mid \frac3{1+9x^2}>0\}.$$
$\color{red} (?) $ For $0\leq x<\frac12:$
$$f'(0)=3,\quad f'\left(\frac12\right)=\ln(2)+2,\quad f''(x)=\frac{4x-2}{(1+2x)^2}$$
Since the endpoints of the interval are positive and there aren't any negative minimums in such interval, then we can imply that $f'(x)$ is positive in $0\leq x<\frac12$ and therefore $f(x)$ is increasing. By this, $f(x)$ is increasing
$\circ$ For $x\geq\frac12:$
$$f'\left(\frac12\right)=\ln(2)+2,\quad f''(x)=\frac{4x-2}{(1+2x)^2}>0\implies x\geq\frac12.$$
Since $f''(x)$ is positive for $x>1/2$ then $f'(x)$ is increasing. Knowing that $f'(1/2)$ is positive implies that $f'(x)$ is always positive in the given interval and therefore $f(x)$ is increasing
This shows that $f(x)$ is always increasing.
My questions arises on the $\color{red} (?) $ sections. I am not sure if my implication with the minimums is correct. If not, is there any other way to prove this? I also want to add that I am not allowed to use Newton's Method or any other root approximating methods.
For $x\ge 0$ you have already computed $$f'(x) = \ln(2x+1)+\frac{2x+3}{2x+1}$$ and both terms are positive. Therefore $f$ is strictly increasing for $x\ge 0.$