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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 \(0 \le x \le 10.\)

 Apr 25, 2020
 #1
avatar+110189 
+2

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.

 Apr 25, 2020
 #2
avatar+135 
+2

thank you!

littlemixfan  Apr 25, 2020
 #3
avatar+135 
0

could you help me with part d? 

littlemixfan  Apr 25, 2020
 #4
avatar+110189 
+1

What answers do you have for b and c first.

OR what have you done to try and solve them?

I do not want to do all the work without you demonstrating that you are learning.

Melody  Apr 25, 2020
 #5
avatar+110189 
+1

Try these

 

\(f(0)=\\ f(1)=\\ f(2)=\\ f(2\frac{1}{2})=\\ f(3)=\\ f(3\frac{1}{3})=\\ f(3\frac{2}{3})=\\ f(4)=\\ f(4\frac{1}{4})=\\ f(4\frac{2}{4})=\\ f(4\frac{3}{4})=\\ f(5)=\\ f(5\frac{1}{5})=\\ etc\)

 

Try some values in between.

Can you see what is happening?

 Apr 25, 2020

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