+0

+1
114
5
+135

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
+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.

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
+135
+2

thank you!

littlemixfan  Apr 25, 2020
#3
+135
0

could you help me with part d?

littlemixfan  Apr 25, 2020
#4
+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
+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