I'm dealing with absolute value inequalities. I've realized it would be so much easier to confirm the solution sets I'm obtaining with a curve of the function.
Currently I'm doing $\left|x^2-2x\right|\lt x$. Now this can be done pretty easily to get the solution set as $x\in (1,3)$. Is there a way to construct a graph for this function namely $f(x)=\left|x^2-2x\right|-x$ without using any sort of calculator whatsoever? Also what should be the general strategy to tackle graphs of this kind?
I can easily graph $f(x)=\left| x^2-2x\right| $ but the presence of that $x$ is causing trouble in achieving the graph. Thanks.
The key is to find the values of $x$ for which the expression inside the absolute value signs changes sign.
In your example, $x^2-2x=x(x-2)$ so it changes $+,-,+$ on the intervals $(-\infty,0),\,(0,2),\,(2,\infty)$, respectively. Thus
$f(x)=\begin{cases}x^2-3x&\text{ on }(-\infty,0)\\ x-x^2&\text{ on }(0,2)\\ x^2-3x&\text{ on }(2,\infty)\end{cases}$