Question

consider the relation R=[(1,2),(2,5),(3.10),(4,17),(5,26)]. find the
i)range ii)domain iii)inverse

Answer 1

Assuming the relation is
f(x)=y=x²+1 where x ∈ ℕ (i.e. x is a member of natural numbers), then the domain is
{x:[1,5] ∈ ℕ}
Range is
{y:[2,26] ∈ ℕ}
Inverse:
f^-1(y) = sqrt((y-1)) : f^-1(y)>0

Answer 2

Correction for range, which is the set of all values produced by the function within the given domain. For the above relation
f(x)=y=x²+1,
the domain is
{1,2,3,4,5}
the range is
{2,5,10,17,26}
The inverse is as above.