Question

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

Answer 1

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`}