What is the domain and range of #h(x)= 1 /x^2#?

1 Answer
Ratnaker Mehta
Jul 22, 2016

Domain of #h, D_h#, is #RR-{0}#.

Range of #h, R_h=(0,oo)#.

Explanation:

We know that Division by #0# is not permissible , so, we can not take x=0. Except this, x can take any value from #RR#, which means that the Domain of #h, D_h#, is #RR-{0}#.

Next, #AA x in D_h, x^2>0#, and so is, #1/x^2#.

Also let us note that, for each #x in D_h, h(x)!=0#, because, if for some #x in D_h, h(x)=1/x^2=0 rarr 1=0#, an impossible result.

Thus, Range of #h, R_h=(0,oo)#.

Related questions
See all questions in Domain and Range of a Function
Impact of this question
2891 views around the world
You can reuse this answer
Creative Commons License