Answer
a) $f(x)$ is continuous on $R$.
b) $g(x)$ is continuous on $[0,\infty)$.
c) $h(x)$ is continuous on $(-\infty,0)\cup(0,\infty)$.
d) $k(x)$ is continuous on $(0,\infty)$.
Work Step by Step
a) $f(x)=x^{1/3}=\sqrt[3]x$
- Domain: $R$
- For all $x=c\in R$, we have $$\lim_{x\to c}f(x)=\sqrt[3]c=f(c)$$
Therefore, $f(x)$ is continuous on its entire domain, which is $R$.
b) $g(x)=x^{3/4}=\sqrt[4]{x^3}$
- Domain: $[0,\infty)$
- For all $x=c\in [0,\infty)$, we have $$\lim_{x\to c}g(x)=\sqrt[4]{c^3}=f(c)$$
- Since $g(x)$ is not defined on $(-\infty,0)$, $g(x)$ is not continuous on $(-\infty,0)$
Therefore, $g(x)$ is continuous on its entire domain, which is $[0,\infty)$.
c) $h(x)=x^{-2/3}=\sqrt[3]{x^{-2}}=\sqrt[3]{\frac{1}{x^2}}$
- Domain: $(-\infty,0)\cup(0,\infty)$
- For all $x=c\in (-\infty,0)\cup(0,\infty)$, we have $$\lim_{x\to c}h(x)=\sqrt[3]{\frac{1}{c^2}}=f(c)$$
- For $x=0$: Since $f(0)$ is not defined, $h(x)$ is not continuous at $x=0$.
Therefore, $h(x)$ is continuous on its domain, which is $(-\infty,0)\cup(0,\infty)$.
d) $k(x)=x^{-1/6}=\sqrt[6]{x^{-1}}=\sqrt[6]{\frac{1}{x}}$
- Domain: $(0,\infty)$
- For all $x=c\in (0,\infty)$, we have $$\lim_{x\to c}k(x)=\sqrt[6]{\frac{1}{c}}=f(c)$$
- Since $k(x)$ is not defined on $(-\infty,0]$, $k(x)$ is not continuous on $(-\infty,0]$
Therefore, $k(x)$ is continuous on its entire domain, which is $(0,\infty)$.
