select the correct answer. which function is continuous across its domain? a. $f(x)=\\begin{cases}x + 4, &…

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

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

Answer

Explanation:

Step1: Check continuity at (x = - 2)

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

  • For (x=-2):
    • (\lim_{x\rightarrow - 2^{-}}f(x)): When (x\rightarrow - 2^{-}), (f(x)=x + 4). So (\lim_{x\rightarrow - 2^{-}}(x + 4)=-2 + 4=2)
    • (\lim_{x\rightarrow - 2^{+}}f(x)): When (x\rightarrow - 2^{+}), (f(x)=0.5x^{2}). So (\lim_{x\rightarrow - 2^{+}}0.5x^{2}=0.5\times(-2)^{2}=2)
  • For options B: (\lim_{x\rightarrow - 2^{-}}(x + 6)=-2+6 = 4\neq2), for option C: (\lim_{x\rightarrow - 2^{-}}(x - 2)=-2-2=-4\neq2)

Step2: Check continuity at (x = 4)

  • (\lim_{x\rightarrow4^{-}}f(x)): When (x\rightarrow4^{-}), (f(x)=0.5x^{2}). So (\lim_{x\rightarrow4^{-}}0.5x^{2}=0.5\times16 = 8)
  • (\lim_{x\rightarrow4^{+}}f(x)):
    • For option A: When (x\rightarrow4^{+}), (f(x)=25-3x), (\lim_{x\rightarrow4^{+}}(25 - 3x)=25-12 = 13\neq8)
    • For option D: When (x\rightarrow4^{+}), (f(x)=20-3x), (\lim_{x\rightarrow4^{+}}(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})