Question

2 select the correct answer. which function is continuous across its domain? a. $f(x)=\begin{cases}x + 4, &-4leq xlt - 2\\0.5x^{2},&-2leq xlt4\\25 - 3x,&4leq xleq8end{cases}$ b. $f(x)=\begin{cases}x + 4, &-4leq xlt - 2\\0.5x^{2},&-2leq xlt4\\20 - 3x,&4leq xleq8end{cases}$ c. $f(x)=\begin{cases}x - 2, &-4leq xlt - 2\\0.5x^{2},&-2leq xlt4\\25 - 3x,&4leq xleq8end{cases}$ d. $f(x)=\begin{cases}x + 6, &-4leq xlt - 2\\0.5x^{2},&-2leq xlt4\\20 - 3x,&4leq xleq8end{cases}$

Explanation:

Step1: Recall continuity condition

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. We need to check the continuity at $x=- 2$ and $x = 4$ for each piece - wise function.

Step2: Check continuity at $x=-2$ for option A

For $x
ightarrow - 2^{-}$, $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(x + 4)=-2 + 4=2$. For $x
ightarrow - 2^{+}$, $\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(0.5x^{2})=0.5\times(-2)^{2}=2$.

Step3: Check continuity at $x = 4$ for option A

For $x
ightarrow4^{-}$, $\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{-}}(0.5x^{2})=0.5\times4^{2}=8$. For $x
ightarrow4^{+}$, $\lim_{x
ightarrow4^{+}}f(x)=\lim_{x
ightarrow4^{+}}(25 - 3x)=25-3\times4=13$. Since $8
eq13$, option A is not continuous.

Step4: Check continuity at $x=-2$ for option B

For $x
ightarrow - 2^{-}$, $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(x + 4)=-2 + 4=2$. For $x
ightarrow - 2^{+}$, $\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(0.5x^{2})=0.5\times(-2)^{2}=2$.

Step5: Check continuity at $x = 4$ for option B

For $x
ightarrow4^{-}$, $\lim_{x
ightarrow4^{-}}f(x)=\lim_{x
ightarrow4^{-}}(0.5x^{2})=0.5\times4^{2}=8$. For $x
ightarrow4^{+}$, $\lim_{x
ightarrow4^{+}}f(x)=\lim_{x
ightarrow4^{+}}(20 - 3x)=20-3\times4 = 8$. Also, $f(4)=20 - 3\times4=8$. So the function in option B is continuous at $x=-2$ and $x = 4$.

Step6: Check continuity at $x=-2$ for option C

For $x
ightarrow - 2^{-}$, $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(x - 2)=-2-2=-4$. For $x
ightarrow - 2^{+}$, $\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(0.5x^{2})=0.5\times(-2)^{2}=2$. Since $-4
eq2$, option C is not continuous.

Step7: Check continuity at $x=-2$ for option D

For $x
ightarrow - 2^{-}$, $\lim_{x
ightarrow - 2^{-}}f(x)=\lim_{x
ightarrow - 2^{-}}(x + 6)=-2 + 6=4$. For $x
ightarrow - 2^{+}$, $\lim_{x
ightarrow - 2^{+}}f(x)=\lim_{x
ightarrow - 2^{+}}(0.5x^{2})=0.5\times(-2)^{2}=2$. Since $4
eq2$, option D is not continuous.

Answer:

B. $f(x)=

$$\begin{cases}x + 4, &-4\leq x\lt - 2\\0.5x^{2},&-2\leq x\lt4\\20 - 3x,&4\leq x\leq8\end{cases}$$

$