Question

select the correct answer. which function is continuous across its domain? a. $f(x)=\

$$\begin{cases}x + 4, & -4\\leq x\\lt -2 \\\\ 0.5x^{2}, & -2\\leq x\\lt 4 \\\\ 25 - 3x, & 4\\leq x\\leq 8\\end{cases}$$

$ b. $f(x)=\

$$\begin{cases}x + 6, & -4\\leq x\\lt -2 \\\\ 0.5x^{2}, & -2\\leq x\\lt 4 \\\\ 20 - 3x, & 4\\leq x\\leq 8\\end{cases}$$

$ c. $f(x)=\

$$\begin{cases}x - 2, & -4\\leq x\\lt -2 \\\\ 0.5x^{2}, & -2\\leq x\\lt 4 \\\\ 25 - 3x, & 4\\leq x\\leq 8\\end{cases}$$

$ d. $f(x)=\

$$\begin{cases}x + 4, & -4\\leq x\\lt -2 \\\\ 0.5x^{2}, & -2\\leq x\\lt 4 \\\\ 20 - 3x, & 4\\leq x\\leq 8\\end{cases}$$

$

Explanation:

Step1: Check continuity at \(x = - 2\)

For a function \(y = f(x)\) to be continuous at \(x=a\), \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\)

• For \(x=-2\):
• \(\lim_{x

ightarrow - 2^{-}}f(x)\): When \(x
ightarrow - 2^{-}\), \(f(x)=x + 4\). So \(\lim_{x
ightarrow - 2^{-}}(x + 4)=-2 + 4=2\)

• \(\lim_{x

ightarrow - 2^{+}}f(x)\): When \(x
ightarrow - 2^{+}\), \(f(x)=0.5x^{2}\). So \(\lim_{x
ightarrow - 2^{+}}0.5x^{2}=0.5\times(-2)^{2}=2\)

• For options B: \(\lim_{x

ightarrow - 2^{-}}(x + 6)=-2+6 = 4
eq2\), for option C: \(\lim_{x
ightarrow - 2^{-}}(x - 2)=-2-2=-4
eq2\)

Step2: Check continuity at \(x = 4\)

• \(\lim_{x

ightarrow4^{-}}f(x)\): When \(x
ightarrow4^{-}\), \(f(x)=0.5x^{2}\). So \(\lim_{x
ightarrow4^{-}}0.5x^{2}=0.5\times16 = 8\)

• \(\lim_{x

ightarrow4^{+}}f(x)\):

• For option A: When \(x

ightarrow4^{+}\), \(f(x)=25-3x\), \(\lim_{x
ightarrow4^{+}}(25 - 3x)=25-12 = 13
eq8\)

• For option D: When \(x

ightarrow4^{+}\), \(f(x)=20-3x\), \(\lim_{x
ightarrow4^{+}}(20 - 3x)=20 - 12=8\)

Answer:

D. \(f(x)=

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

\)