Question

suppose that the function \\(f\\) is defined, for all real numbers, as follows.

\\f(x)=\

$$\begin{cases} -2x+3 & \\text{if } x<2 \\\\ x-3 & \\text{if } x \\ge 2 \\end{cases}$$

\\

graph the function \\(f\\). then determine whether or not the function is continuous.

Explanation:

Graph the first piece

Using the Piecewise Linear Functions knowledge point:
For \(x < 2\), the function is \(f(x) = -2x + 3\).
At \(x = 0\), \(f(0) = 3\).
At \(x = 2\), \(f(2) \to -1\) (open circle at \((2, -1)\)).

Graph the second piece

Using the Piecewise Linear Functions knowledge point:
For \(x \ge 2\), the function is \(f(x) = x - 3\).
At \(x = 2\), \(f(2) = -1\) (solid circle at \((2, -1)\)).
At \(x = 5\), \(f(5) = 2\).

Determine continuity

The left-hand limit is \(\lim_{x \to 2^-} f(x) = -1\).
The right-hand limit is \(\lim_{x \to 2^+} f(x) = -1\).
Since \(\lim_{x \to 2} f(x) = f(2) = -1\), the function is continuous.

Answer:

The graph consists of two linear segments meeting at the point \((2, -1)\):

• A ray starting from an open circle at \((2, -1)\) and passing through \((0, 3)\) to the left.
• A ray starting from a solid circle at \((2, -1)\) and passing through \((5, 2)\) to the right.

Because the two pieces meet at the same value, there is no break in the graph.

The function is continuous.