Question

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

Explanation:

Step1: Recall continuity definition

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

Step2: Analyze $f(x)$

For $f(x)=

$$\begin{cases}x, & 0\leq x\\0, & x < 0\end{cases}$$

$, $\lim_{x
ightarrow0^{-}}f(x)=0$, $\lim_{x
ightarrow0^{+}}f(x)=0$, and $f(0) = 0$. So $f(x)$ is continuous at $x = 0$.

Step3: Analyze $g(x)$

For $g(x)=

$$\begin{cases}0, & 0\leq x\\2, & x < 0\end{cases}$$

$, $\lim_{x
ightarrow0^{-}}g(x)=2$, $\lim_{x
ightarrow0^{+}}g(x)=0$. Since $\lim_{x
ightarrow0^{-}}g(x)
eq\lim_{x
ightarrow0^{+}}g(x)$, $g(x)$ is not continuous at $x = 0$.

Step4: Analyze $f(-x)$

Let $y = f(-x)$. When $-x\geq0$ (i.e., $x\leq0$), $y=-x$; when $-x < 0$ (i.e., $x>0$), $y = 0$. $\lim_{x
ightarrow0^{-}}f(-x)=0$, $\lim_{x
ightarrow0^{+}}f(-x)=0$, and $f(0) = 0$. So $f(-x)$ is continuous at $x = 0$.

Step5: Analyze $|g(x)|$

$|g(x)|=

$$\begin{cases}0, & 0\leq x\\2, & x < 0\end{cases}$$

$, $\lim_{x
ightarrow0^{-}}|g(x)|=2$, $\lim_{x
ightarrow0^{+}}|g(x)|=0$. Since $\lim_{x
ightarrow0^{-}}|g(x)|
eq\lim_{x
ightarrow0^{+}}|g(x)|$, $|g(x)|$ is not continuous at $x = 0$.

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

$f(x)g(x)=0$ for all $x$. $\lim_{x
ightarrow0^{-}}f(x)g(x)=0$, $\lim_{x
ightarrow0^{+}}f(x)g(x)=0$, and $f(0)g(0)=0$. So $f(x)g(x)$ is continuous at $x = 0$.

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

For $x < 0$, $f(x)+g(x)=2$; for $x\geq0$, $f(x)+g(x)=x$. $\lim_{x
ightarrow0^{-}}[f(x)+g(x)]=2$, $\lim_{x
ightarrow0^{+}}[f(x)+g(x)]=0$. Since $\lim_{x
ightarrow0^{-}}[f(x)+g(x)]
eq\lim_{x
ightarrow0^{+}}[f(x)+g(x)]$, $f(x)+g(x)$ is not continuous at $x = 0$.

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

When $x < 0$, $f(x)=0$, then $g(f(x))=0$; when $x\geq0$, $f(x)=x\geq0$, then $g(f(x))=0$. $\lim_{x
ightarrow0^{-}}g(f(x))=0$, $\lim_{x
ightarrow0^{+}}g(f(x))=0$, and $g(f(0))=0$. So $g(f(x))$ is continuous at $x = 0$.

Answer:

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