consider the functions\nf(x)=\begin{cases}2&0leq x\\0&x < 0end{cases}\nand\ng(x)=\begin{cases}0&0leq x\\2&x…

consider the functions\nf(x)=\begin{cases}2&0leq x\\0&x < 0end{cases}\nand\ng(x)=\begin{cases}0&0leq x\\2&x < 0end{cases}\nin each part, is the given function continuous at (x = 0)? enter \yes\ or \no\.\n(a) (f(x))\n(b) (g(x))\n(c) (f(-x))\n(d) (|g(x)|)\n(e) (f(x)g(x))\n(f) (f(x)+g(x))\n(g) (g(f(x)))\nnote: this question wont show you which parts are right and wrong until all of them are correct.\nnote: you can earn partial credit on this problem.

consider the functions\nf(x)=\begin{cases}2&0leq x\\0&x < 0end{cases}\nand\ng(x)=\begin{cases}0&0leq x\\2&x < 0end{cases}\nin each part, is the given function continuous at (x = 0)? enter \yes\ or \no\.\n(a) (f(x))\n(b) (g(x))\n(c) (f(-x))\n(d) (|g(x)|)\n(e) (f(x)g(x))\n(f) (f(x)+g(x))\n(g) (g(f(x)))\nnote: this question wont show you which parts are right and wrong until all of them are correct.\nnote: you can earn partial credit on this problem.

Answer

Explanation:

Step1: Recall continuity condition

A function $y = h(x)$ is continuous at $x = a$ if $\lim_{x\rightarrow a^{-}}h(x)=\lim_{x\rightarrow a^{+}}h(x)=h(a)$.

Step2: Analyze $f(x)$

For $f(x)=\begin{cases}2&0\leq x\0&x < 0\end{cases}$, $\lim_{x\rightarrow0^{-}}f(x)=0$, $\lim_{x\rightarrow0^{+}}f(x)=2$, $f(0) = 2$. Since $\lim_{x\rightarrow0^{-}}f(x)\neq\lim_{x\rightarrow0^{+}}f(x)$, $f(x)$ is not continuous at $x = 0$.

Step3: Analyze $g(x)$

For $g(x)=\begin{cases}0&0\leq x\2&x < 0\end{cases}$, $\lim_{x\rightarrow0^{-}}g(x)=2$, $\lim_{x\rightarrow0^{+}}g(x)=0$, $g(0)=0$. Since $\lim_{x\rightarrow0^{-}}g(x)\neq\lim_{x\rightarrow0^{+}}g(x)$, $g(x)$ is not continuous at $x = 0$.

Step4: Analyze $f(-x)$

$f(-x)=\begin{cases}2&0\leq -x\0&-x < 0\end{cases}=\begin{cases}2&x\leq0\0&x>0\end{cases}$. $\lim_{x\rightarrow0^{-}}f(-x)=2$, $\lim_{x\rightarrow0^{+}}f(-x)=0$, $f(-0) = 2$. Since $\lim_{x\rightarrow0^{-}}f(-x)\neq\lim_{x\rightarrow0^{+}}f(-x)$, $f(-x)$ is not continuous at $x = 0$.

Step5: Analyze $|g(x)|$

$|g(x)|=\begin{cases}0&0\leq x\2&x < 0\end{cases}$. $\lim_{x\rightarrow0^{-}}|g(x)|=2$, $\lim_{x\rightarrow0^{+}}|g(x)|=0$, $|g(0)| = 0$. Since $\lim_{x\rightarrow0^{-}}|g(x)|\neq\lim_{x\rightarrow0^{+}}|g(x)|$, $|g(x)|$ is not continuous at $x = 0$.

Step6: Analyze $f(x)g(x)$

$f(x)g(x)=\begin{cases}2\times0 = 0&0\leq x\0\times2=0&x < 0\end{cases}=0$ for all $x$. $\lim_{x\rightarrow0^{-}}f(x)g(x)=0$, $\lim_{x\rightarrow0^{+}}f(x)g(x)=0$, $f(0)g(0)=0$. So $f(x)g(x)$ is continuous at $x = 0$.

Step7: Analyze $f(x)+g(x)$

$f(x)+g(x)=\begin{cases}2 + 0=2&0\leq x\0 + 2=2&x < 0\end{cases}=2$ for all $x$. $\lim_{x\rightarrow0^{-}}(f(x)+g(x))=2$, $\lim_{x\rightarrow0^{+}}(f(x)+g(x))=2$, $(f(0)+g(0))=2$. So $f(x)+g(x)$ is continuous at $x = 0$.

Step8: Analyze $g(f(x))$

When $x\geq0$, $f(x)=2$, then $g(f(x))=g(2)=0$. When $x < 0$, $f(x)=0$, then $g(f(x))=g(0)=0$. $\lim_{x\rightarrow0^{-}}g(f(x))=0$, $\lim_{x\rightarrow0^{+}}g(f(x))=0$, $g(f(0))=0$. So $g(f(x))$ is continuous at $x = 0$.

Answer:

(a) no (b) no (c) no (d) no (e) yes (f) yes (g) yes