+226 Let \(f(x) = \lfloor x \lfloor x \rfloor \rfloor\) for \(x \ge 0.\)
(a) Find all \(x \ge 0\) such that \(f(x) = 1. \)
(b) Find all \(x \ge 0\) such that \(f(x) = 3. \)
(c) Find all \(x \ge 0\) such that \(f(x) = 5.\)
(d) Find the number of possible values of\( f(x)\) for \(0 \le x \le 10.\)
+118673 I have not seen a question like this before but this looks kind of like a step broken parabola to me.
As x increases so does f(x) but it happens in steps.
part a)
If x=1 then f(1)=1
If x=2 then f(2)=4
If x=1.99 then f(x) = floor of 1.99*1 = 1 this will be true of any x value between 1 and 2
so
for \(1\le x<2\qquad f(x)=1\)
likewise
for \(0\le x<1\qquad f(x)=0\)
I have not really through much further than that but that gives you a good start for your own thought process.
Please be careful, it is tricky. Try drawing the graph to help you.
I'm going for a walk. You can show me what you have done when I get back if you want to. I'll be happy to help more.
