How would one begin to simplify the following?
$$ \frac{2x-x\lvert x-1\rvert+x\lvert x\rvert+5}{\lvert x\rvert+1}$$
I have found the domain as
$$\begin{cases} \dfrac{x+5}{1-x}, & \ {(-\infty,0]} \\[4pt] \dfrac{2x^2+x+5}{x+1}, & \ {[0,1)} \\[4pt] \dfrac{5x+5}{x+1}, & \ {[1,+\infty)} \\ \end{cases}$$
Thank you for any help!
Hints
When dealing with something like |r|, you want to consider two cases:
$r < 0$ and $r \geq 0.$
Here you have 3 conditions to worry about.
Is $(x - 1) \geq 0$ or is $(x - 1) < 0.$
Is $(x) \geq 0$ or is $(x) < 0.$
Is $|x| + 1 = 0.$ If possible, this would make the denominator $= 0,$ which is not allowed. However, this situation is impossible because $|x|$ can not equal $-1$.
Try to "collapse" the cases that represent points 1 and 2 above into 3 distinct intervals. Then, for each interval, create a totally distinct function that applies only to that interval.
I overlooked that the OP has shown work and that therefore, it is okay to complete the problem.
The original function is
$$ f(x) = \frac{2x-x\lvert x-1\rvert+x\lvert x\rvert+5}{\lvert x\rvert+1}$$
The three intervals will be
Interval 1: $x < 0.~$ The specific function will be $f_1(x).$
$$ f_1(x) = \frac{2x ~- ~[(x)(1 - x)] ~+~ [x(-x)] ~+~ 5}{(-x) ~+ ~1}.$$
Interval 2: $0 \leq x < 1.~$ The specific function will be $f_2(x).$
$$ f_2(x) = \frac{2x ~- ~[(x)(1 - x)] ~+~ [x(x)] ~+~ 5}{(x) ~+ ~1}.$$
Interval 3: $1 \leq x.~$ The specific function will be $f_3(x).$
$$ f_3(x) = \frac{2x ~- ~[(x)(x - 1)] ~+~ [x(x)] ~+~ 5}{(x) ~+ ~1}.$$
At this point, none of $~f_1(x), f_2(x),~$ or $~f_3(x)$ are employing any absolute value signs. Now, you can consider whether there is any simplification that is common to all three functions $~f_1(x), f_2(x),~$ and $~f_3(x).$
From my perspective, other than multiplying the numerators out, I see no other possible simplification that can be applied to all three functions.