Question

let f(x) be the function given below.
f(x)=\begin{cases}\frac{1}{x}&\text{for }xgeq1\\3x + c&\text{for }x<1end{cases}
complete parts a and b below.
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 (xin(-infty,infty))?
if (c=square), then f(x) is continuous at for all (xin(-infty,infty)).

Explanation:

Step1: Recall the continuity condition

A function $y = f(x)$ is continuous at $x = a$ if $\lim_{x
ightarrow a^{-}}f(x)=\lim_{x
ightarrow a^{+}}f(x)=f(a)$. Here $a = 1$, $\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(3x + c)$ and $\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}\frac{1}{x}$.

Step2: Calculate the left - hand limit

$\lim_{x
ightarrow1^{-}}(3x + c)=3\times1 + c=3 + c$.

Step3: Calculate the right - hand limit

$\lim_{x
ightarrow1^{+}}\frac{1}{x}=\frac{1}{1}=1$.

Step4: Set the left - hand and right - hand limits equal

For the function to be continuous at $x = 1$, we set $3 + c=1$.

Step5: Solve for c

Subtract 3 from both sides of the equation $3 + c=1$. We get $c=1 - 3=-2$.

Answer:

$-2$