Posted by **Matrix School 2** on Saturday, July 11, 2009 at 1:43pm.

consider the relation R=[(1,2),(2,5),(3.10),(4,17),(5,26)]. find the

i)range ii)domain iii)inverse

- Set theory -
**MathMate**, Saturday, July 11, 2009 at 4:21pm
Assuming the relation is

f(x)=y=x^{2}+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

- Set theory -
**MathMate**, Saturday, July 11, 2009 at 4:29pm
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=x^{2}+1,

the domain is

{1,2,3,4,5}

the range is

{2,5,10,17,26}

The inverse is as above.

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