Question

Let $f\left(x\right)=[x^2/3]$. Find f(s) if

1. $S=\{-2,-1,0,1,2\}$
2. $S=\left\{0,1,2,3,4,5\right\}$
   
3. $S=\left\{1,5,7,11\right\}$
   
4. $S=\left\{2,6,10,14\right\}$
   

Short answer

1. $f\left(s\right)=\left\{0,1,3\right\}$
   
2. $f\left(S\right)=\{0,1,3,5,8\}$
   
3. $f\left(S\right)=\{0,8,16,4\}$
   
4. $f\left(s\right)=\{1,12,,33,65\}$
   

Step-by-step solution

Step: 1

Given$f\left(x\right)=[x^2/3]$

a.

If x = - 2 then $f\left(x\right)=[4/3]$

If x = - 1 then $f\left(x\right)=[1/3]$

If x = 0 then $f\left(x\right)=\left[0\right]$

If x = 1 then $f\left(x\right)=[1/3]$

If x = 2 then $f\left(x\right)=[4/3]$

If x = 3 then $f\left(x\right)=[9/3]$

Therefore $f\left(s\right)=\left\{0,1,3\right\}$

Step: 2

b.

If x = 0 then$f\left(x\right)=\left[0\right]$

If x = 1 thenrole="math" localid="1668420397075" $f\left(x\right)=[1/3]$

If x = 2 then$f\left(x\right)=[4/3]$

If x = 3 then$f\left(x\right)=[9/3]$

If x = 4 then$f\left(x\right)=[16/3]$

If x = 5 then$f\left(x\right)=[25/3]$

Therefore$f\left(S\right)=\{0,1,3,5,8\}$

Step: 3

c.

If x = 1 then$f\left(x\right)=[1/3]$

If x = 5 then$f\left(x\right)=[25/3]$

If x = 7 then$f\left(x\right)=[49/3]$

If x = 11 then$f\left(x\right)=[121/3]$

Therefore $f\left(S\right)=\{0,8,16,4\}$

Step: 4

d.

If x = 2 then$f\left(x\right)=[4/3]$

If x = 6 then$f\left(x\right)=[36/3]$

If x = 10 then$f\left(x\right)=[100/3]$

If x = 14 then$f\left(x\right)=[196/3]$

Therefore $f\left(s\right)=\{1,12,33,65\}$