Questions: Determine the domain and range. 2. f(x) = 2x-5 if x>1 4x-3 if x ≤ 1

Transcript text: e domain and range. 2. $f(x)=\left\{\begin{array}{l}2 x-5 \text { if } x>1 \\ 4 x-3 \text { if } x \leq 1\end{array}\right.$

Solution

Solution Steps

Step 1: Identify the piecewise function

The given function is: \[ f(x) = \left\{ \begin{array}{ll} 2x - 5 & \text{if } x > 1 \\ 4x - 3 & \text{if } x \leq 1 \end{array} \right. \] This is a piecewise function with two cases: one for $ x > 1 $ and another for $ x \leq 1 $.

Step 2: Graph the function for $ x > 1 $

For $ x > 1 $, the function is $ f(x) = 2x - 5 $. This is a linear function with a slope of $ 2 $ and a y-intercept at $ (0, -5) $. However, since $ x > 1 $, the graph starts at $ x = 1 $ (exclusive) and extends to the right.

Step 3: Graph the function for $ x \leq 1 $

For $ x \leq 1 $, the function is $ f(x) = 4x - 3 $. This is another linear function with a slope of $ 4 $ and a y-intercept at $ (0, -3) $. The graph includes $ x = 1 $ and extends to the left.

Step 4: State the domain and range

The domain of $ f(x) $ is all real numbers, $ (-\infty, \infty) $, because the function is defined for all $ x $.
The range of $ f(x) $ depends on the outputs of both cases. For $ x > 1 $, $ f(x) = 2x - 5 $ produces values greater than $ -3 $. For $ x \leq 1 $, $ f(x) = 4x - 3 $ produces values less than or equal to $ 1 $. Thus, the range is $ (-\infty, 1] \cup (-3, \infty) $.