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The given piecewise function \( f(x) \) is defined as:

\[
f(x) = \begin{cases} 
-2x + 3 & \text{if } x < 1 \\
2x - 1 & \text{if } x \geq 1
\end{cases}
\]

We need to solve the following parts:

### (a) Find the domain of the function.
The domain of a function is the set of all possible input values (x-values) for which the function is defined. In this case, both pieces of the function are defined for all real numbers. The first piece \( -2x + 3 \) applies when \( x < 1 \), and the second piece \( 2x - 1 \) applies when \( x \geq 1 \).

Thus, the domain is all real numbers:  
\[
\text{Domain of } f(x): (-\infty, \infty)
\]

### (b) Locate any intercepts.
#### Finding the y-intercept:
The y-intercept occurs when \( x = 0 \). We use the first piece of the function since \( x = 0 < 1 \):

\[
f(0) = -2(0) + 3 = 3
\]

Thus, the y-intercept is \( (0, 3) \).

#### Finding the x-intercept(s):
For the x-intercept(s), set \( f(x) = 0 \) and solve for \( x \) in each piece.

- For \( x < 1 \), \( -2x + 3 = 0 \):
  \[
  -2x + 3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}
  \]
  Since \( \frac{3}{2} > 1 \), this x-intercept does not apply to the first piece.

- For \( x \geq 1 \), \( 2x - 1 = 0 \):
  \[
  2x - 1 = 0 \implies 2x = 1 \implies x = \frac{1}{2}
  \]
  However, \( \frac{1}{2} < 1 \), so this intercept only applies to the first

The function f is defined as a piecewise function: f(x) = -2x + 3 if x < 1 and f(x) = 2x - 1 if x >= 1. The tasks are to find the domain, locate intercepts, graph the function, and determine the range.

Analyze the piecewise function f(x) = -2x + 3 for x < 1 and f(x) = 2x - 1 for x >= 1. Solve for the domain, intercepts, graph, and range with detailed steps. Suitable for Grades 9-11.

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