What is the domain and range of this function and its inverse #f(x) = sqrt(x + 7)#?

1 Answer
Apr 19, 2015

Domain of f(x)= {x#in#R, #x>= -7#}, Range= {y#in#R, y#>=0#}
Domain of #f^-1 (x)#= { x#in#R}, Range = {y#in#R, , #y>= -7#}

The domain of the function would be all x, such that #x+7>=0#, or #x>= -7#. Hence it is {x#in# R, #x>=-7#}

For range, consider y=#sqrt(x+7)#. Since#sqrt(x+7) # has to be #>=0#, it is obvious that #y>=0#. Range would be {y#in#R, y#>=0#}

The inverse function would be #f^-1 (x)#= #x^2 -7#.

The domain of the inverse function is all real x that is { x#in#R}

For the range of the inverse function solve y= #x^2#-7 for x. It would be x= #sqrt(y+7)#. This clearly shows that #y+7>=0#. Hence Range would be {y #in#R, #y>= -7#}