## University Calculus: Early Transcendentals (3rd Edition)

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 $(0,\infty)$.
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 $(0,\infty)$.