Radical Functions: Square Root and Cube Root

5. Radical function. Example: $f(x)=\sqrt[n]{x}$, $n\in N$.

1. Square root function. $f(x)=\sqrt{x}$, $x\ge0$ (since the square root of a negative number is not real, the domain of $\sqrt{x}$ is restricted to non-negative $x$) (Fig. 6.25). Domain: $[0,\infty)$; Range: $[0,\infty)$.

Note: (1) If $x$ is positive, it technically has two square roots; by convention $\sqrt{x}$ denotes the positive root and $-\sqrt{x}$ the negative one. (2) If $-4<x<9$, since $\sqrt{x}$ is only defined for $x\ge0$, we get $0\le\sqrt{x}<3$.

Ex. 6: Find the domain and range of $f(x)=\sqrt{9-x^2}$.

Solution: $f(x)$ is defined for $9-x^2\ge0$, i.e. $x^2-9\le0$, i.e. $(x-3)(x+3)\le0$. Therefore $[-3,3]$ is the domain (verify!).

To find the range, let $\sqrt{9-x^2}=y$. Since a square root is always non-negative, $y\ge0$ …(I). Also, squaring gives $9-x^2=y^2$. Since $-3\le x\le3$, i.e. $0\le x^2\le9$, i.e. $0\ge-x^2\ge-9$, i.e. $9\ge9-x^2\ge0$, i.e. $3\ge\sqrt{9-x^2}\ge0$, i.e. $3\ge y\ge0$ …(II). From (I) and (II), $y\in[0,3]$ is the range of $f(x)$.

2. Cube root function. $f(x)=\sqrt[3]{x}$ (Fig. 6.26). Domain: $R$; Range: $R$.

Note: if $-8\le x\le1$ then $-2\le\sqrt[3]{x}\le1$.

Ex. 7: Find the domain of $f(x)=\sqrt{x^3-8}$. …