To determine if the function [tex]\(f(x)\)[/tex] is continuous at [tex]\(x = 1\)[/tex], we need to check if the left-hand limit, right-hand limit, and the value of the function at [tex]\(x = 1\)[/tex] are all equal.
The piecewise function [tex]\(f(x)\)[/tex] is defined as:
[tex]\[
f(x) =
\begin{cases}
5 + x & \text{for } x < 1 \\
4 & \text{for } x = 1 \\
7x - 1 & \text{for } x > 1
\end{cases}
\][/tex]
To ensure continuity at [tex]\(x = 1\)[/tex]:
1. Left-hand limit as [tex]\(x\)[/tex] approaches 1:
[tex]\[
\lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} (5 + x) = 5 + 1 = 6
\][/tex]
2. Right-hand limit as [tex]\(x\)[/tex] approaches 1:
[tex]\[
\lim_{x \to 1^{+}} f(x) = \lim_{x \to 1^{+}} (7x - 1) = 7 \cdot 1 - 1 = 6
\][/tex]
Both the left-hand limit and the right-hand limit as [tex]\(x\)[/tex] approaches 1 are equal to 6.
For [tex]\(f(x)\)[/tex] to be continuous at [tex]\(x = 1\)[/tex], the value of the function at [tex]\(x = 1\)[/tex] must also equal the limit as [tex]\(x\)[/tex] approaches 1 from both sides. Therefore,
[tex]\[
f(1) = \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{+}} f(x) = 6
\][/tex]
Given the options:
a) 4
b) 6
c) 98
d) 10
[tex]\(f(x)\)[/tex] will be continuous at [tex]\(x = 1\)[/tex] when [tex]\(f(1) = 6\)[/tex].
Thus, the correct answer is [tex]\(\boxed{6}\)[/tex].
