Q.Find the domain and range of the function $f(x)=\sqrt{16-x^2}$.

A radical function has the form $f(x)=\sqrt[n]{x}$. The square root function $f(x)=\sqrt{x}$ needs a non-negative input (the square root of a negative number isn't real), so its domain is $[0,\infty)$; by convention $\sqrt{x}$ denotes only the non-negative root even though $x>0$ technically has two square roots ($\pm\sqrt x$). Its range is likewise $[0,\infty)$. The cube root function $f(x)=\sqrt[3]{x}$ behaves completely differently: since a negative number does have a real cube root, it accepts every real input, giving domain $R$ and range $R$ — this is the key structural fact that makes cube roots (unlike square roots) never need a domain restriction when inverting a cubic. …