Certainly! To graph the function \(f(x) = x + 1\) for \(x < 2\), we can create a T-chart with different values of \(x\) and calculate the corresponding \(f(x)\) values. Since \(x < 2\), we can choose values like \(x = 0, 1, 2\) to illustrate the behavior of the function in this interval.
\[
\begin{array}c
\hline
x & f(x) = x + 1 \\
\hline
-1 & 0 \\
0 & 1 \\
1 & 2 \\
\hline
\end{array}
\]
Now, let's plot these points on a graph. Since \(x < 2\), we only need to consider the interval to the left of \(x = 2\).
```
|
3 | *
2 | * *
1 | * * *
| * * *
|-------------------
-1 0 1 2 3 4
```
The graph is a line starting from the point \((-1, 0)\) and increasing as \(x\) increases. The function \(f(x) = x + 1\) represents a line with a slope of 1 and a y-intercept at (0, 1). Since \(x < 2\), we only graph the portion of the line where \(x\) is less than 2.
To determine a solution within the given interval (\(x < 2\)), you can choose any \(x\) value from the T-chart. For example, \(x = -1\) is a solution within the interval, and its corresponding \(f(x)\) value is 0. Therefore, \((x, f(x)) = (-1, 0)\) is a solution within the given interval.