Question 356042
a) The domain of {{{sqrt(x)}}} is {{{x>=0}}} since you can't take the square root of a negative number. So the domain of {{{j(x)=(sqrt(x))^2}}} is also {{{x>=0}}}



b) To find the range, you can either graph {{{j(x)=(sqrt(x))^2}}} and note the possible y values, or you can find the inverse function and find the domain of that inverse function. Either way, you'll find that the range is {{{j(x)>=0}}}