Let $f:\mathbb{R}\to\mathbb{R}$, $f(x)=x|x-1|=\begin{cases} -x(x-1) &x\leq1 \\ x(x-1) & x>1 \end{cases}$.
Is this case distinction correct? Thank you!
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1Yes. The comment has a 15 character limit, so I had to say more XD – Joshua Tilley Jan 12 '23 at 10:31
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2This looks OK. You could also write it as $$x|x-1|=\begin{cases} -x(x-1) &x < 1 \ x(x-1) & x \geq 1 \end{cases}$$ or as $$x|x-1|=\begin{cases} -x(x-1) &x \leq 1 \ x(x-1) & x \geq 1 \end{cases}$$ since when $x=1$ both (sub)formulas evaluate to $0$ (i.e. both formulas agree when $x=1).$ – Dave L. Renfro Jan 12 '23 at 10:35