# Given h(x)=2/x+1 find h^-1 and find the domain and range

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To find the inverse function we set it equal to y and solve for x.

`y=2/[x+1]=>x+1=2/y=>x=2/y-1=>x=[2-y]/y`

Hence the inverse function is `h^(-1)(x)=[2-x]/x`

**Domain**: The new function is not defined for x=0, thus the domain is D=`(-oo,0)U(0,oo)`

**Range**: Since h(x) is not defined for x=-1, then -1 can't be an element of the range of `h^(-1)(x)`

R=`(-oo,-1)U(-1,oo)`

The following graph of the inverse function confirm our findings.