Question

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

Explanation:

Step1: Recall continuity condition

A piece - wise 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 a function $f(x)$ with $f(x)=g(x)$ for $x\lt - 2$ and $f(x)=h(x)$ for $x\geq - 2$, we check $\lim_{x
ightarrow - 2^{-}}g(x)=\lim_{x
ightarrow - 2^{+}}h(x)$.
For option A:
$\lim_{x
ightarrow - 2^{-}}(x - 2)=-2-2=-4$
$\lim_{x
ightarrow - 2^{+}}0.5x^{2}=0.5\times(-2)^{2}=2$
Since $-4
eq2$, option A is not continuous at $x = - 2$.
For option B:
$\lim_{x
ightarrow - 2^{-}}(x + 4)=-2 + 4=2$
$\lim_{x
ightarrow - 2^{+}}0.5x^{2}=0.5\times(-2)^{2}=2$

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

For $f(x)$ with $f(x)=0.5x^{2}$ for $x\lt4$ and $f(x)=25 - 3x$ for $x\geq4$
$\lim_{x
ightarrow4^{-}}0.5x^{2}=0.5\times4^{2}=8$
$\lim_{x
ightarrow4^{+}}(25 - 3x)=25-3\times4=25 - 12 = 13$
Since $8
eq13$, option B is not continuous at $x = 4$.

Step4: Check continuity at $x = 4$ for option C

For $f(x)$ with $f(x)=0.5x^{2}$ for $x\lt4$ and $f(x)=20 - 3x$ for $x\geq4$
$\lim_{x
ightarrow4^{-}}0.5x^{2}=0.5\times4^{2}=8$
$\lim_{x
ightarrow4^{+}}(20 - 3x)=20-3\times4=20 - 12 = 8$

Step5: Check continuity at $x=-2$ for option C

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

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

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

Answer:

C. $f(x)=

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

$