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f(x)=(\begin{cases}\frac{1}{x}&\text{for }xgeq1\\3x + c&\text{for }x<1e…

Question

f(x)=(\begin{cases}\frac{1}{x}&\text{for }xgeq1\\3x + c&\text{for }x<1end{cases}
complete parts a and b below.
is f(x) continuous for this choice of c? select the correct choice below.
a. the function f(x) is not continuous for this choice of c since the graph shows that the one - sided limits are not equal as x approaches 1.
b. the function f(x) is continuous for this choice of c since the graph shows that f(x) is defined at x = 1.
c. the function f(x) is not continuous for this choice of c since the graph shows that both one - sided limits do not exist as x approaches 1.
d. the function f(x) is continuous for this choice of c since the graph shows that both one - sided limits exist as x approaches 1.
(b) how must you choose c so that f(x) is continuous for all x∈(-∞,∞)?
if c = , then f(x) is continuous at for all x∈(-∞,∞).

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = a\), \(\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)\). Here \(a = 1\), \(\lim_{x
ightarrow 1^{-}}f(x)=\lim_{x
ightarrow 1^{-}}(3x + c)=3\times1 + c=3 + c\) and \(\lim_{x
ightarrow 1^{+}}f(x)=\lim_{x
ightarrow 1^{+}}\frac{1}{x}=1\).

Step2: Solve for c

Set the left - hand limit equal to the right - hand limit for continuity. So \(3 + c=1\). Solving for \(c\), we subtract 3 from both sides: \(c=1 - 3=- 2\).

Answer:

(b) If \(c=-2\), then \(f(x)\) is continuous at for all \(x\in(-\infty,\infty)\).