Question

If $A=\{-1,0,2,5,6,11\},B=\{-2,-1,0,18,28,108\}$ and $f\left(x\right)=x^2-x-2,$ find $f\left(A\right).$ Is $f\left(A\right)=B?$

Answer 1

Given, $f\left(x\right)=x^2-x-2$

Now, $f(-1)=(-1{)}^2-(-1)-2=0,f(0)=0^2-0-2=-2[1]$

$f\left(2\right)=2^2-2-2=0,f\left(5\right)=5^2-5-2=18$

$f\left(6\right)=6^2-6-2=28$

and $f\left(11\right)=(11{)}^2-11-2=108[1]$

$\therefore$ $f\left(A\right)=\left\{f\right(x):x\in A\}=\left\{f\right(-1),f(0),f(2),f(5),f(6),f(11\left)\right\}$

$=\{0,-2,0,18,28,108\}=\{-2,0,18,28,108\}\left[1\right]$

We observe that $-1\in B,$ but $-1\notin f\left(A\right)$

So, $f\left(A\right)\ne B\left[1\right]$