(i) Determine the natural domain of the function $f(x)=\frac{\sqrt x}{\sqrt{ 5-x}}$.
Whatever is under a radical must be non-negative. And the denominator must be non-zero.
Using these rules I got the domain to be = [0,5).
(ii) Let $f$ above have codomain ℝ. Is $f$ surjective? Justify your answer.
I know that for a function to be surjective, the function maps an element x to every element y. However I do not know how I would find this in this case?
Any help is appreciated!
Best Answer
(i) Notice that you wanted $5-x>0$ and $\sqrt{5-x}\neq 0$. But the second statement is equivalent to $5-x>0$, since the square root function is already non-negative. And from $5-x>0$ can you conclude that $5>x$ or equivalently $x<5$. On the other hand do you have $\sqrt{x}$ in the numerator and thus you also want $x\geq 0$.Thus the natural domain is $[0,5)$.
(ii) The fact that the square root function is non-negative implies that $f$ always attains positive values and thus any negative value won't be reached by $f$.