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

1 Answer
bp
Apr 19, 2015

Domain of f(x)= {xinR, x>= -7}, Range= {yinR, y>=0}
Domain of f^-1 (x)= { xinR}, Range = {yinR, , y>= -7}

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

For range, consider y=sqrt(x+7). Sincesqrt(x+7) has to be >=0, it is obvious that y>=0. Range would be {yinR, 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 { xinR}

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 inR, y>= -7}

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