Let f(x) be a periodic function of x with period 3 and f(x) = |2x − 1| + x for

−1 < x ≤ 2. Sketch the graph of the curve y = f(x) in the interval [−3, 4]

To sketch the graph of \( y = f(x) \) for the interval \( [-3, 4] \), let's break it down step by step, keeping in mind that \( f(x) \) is a periodic function with a period of 3 and its definition over the interval \( -1 < x \leq 2 \) is:

\[
f(x) = |2x - 1| + x
\]

### Step 1: Analyze \( f(x) \) over \( -1 < x \leq 2 \)

Let's first compute the values of \( f(x) = |2x - 1| + x \) for key points in the interval \( -1 < x \leq 2 \).

#### 1.1: Break into two cases for \( |2x - 1| \)

- For \( x \leq \frac{1}{2} \), \( 2x - 1 \leq 0 \), so \( |2x - 1| = -(2x - 1) = 1 - 2x \).
- For \( x > \frac{1}{2} \), \( 2x - 1 > 0 \), so \( |2x - 1| = 2x - 1 \).

Thus, we have two pieces:

1. \( f(x) = (1 - 2x) + x = 1 - x \) for \( -1 < x \leq \frac{1}{2} \).
2. \( f(x) = (2x - 1) + x = 3x - 1 \) for \( \frac{1}{2} < x \leq 2 \).

Now let's find the values of \( f(x) \) at key points:
- At \( x = -1 \), the interval is open, so no value is defined here.
- At \( x = 0 \), \( f(0) = 1 - 0 = 1 \).
- At \( x = \frac{1}{2} \), \( f\left( \frac{1}{2} \right) = 1 - \frac{1}{2} = \frac{1}{2} \).
- At \( x = 1 \), \( f(1) = 3(1) - 1 = 2 \).
- At \( x = 2 \), \( f(2) = 3(2) - 1 = 5 \).

### Step 2: Extend the function periodically

Since \( f(x) \) is periodic with a period of 3, this means we will repeat the pattern for intervals to the left and right of \( -1 < x \leq 2 \). Specifically:

- For \( -4 < x \leq -1 \), \( f(x) = f(x + 3) \).
- For \( 2 < x \leq 5 \), \( f(x) = f(x - 3) \).

### Step 3: Calculate values for the extended intervals

#### 3.1: For \( -4 < x \leq -1 \)
Since \( f(x) \) repeats every 3 units, calculate \( f(x) = f(x + 3) \):

- For \( x = -3 \), \( f(-3) = f(0) = 1 \).
- For \( x = -2 \), \( f(-2) = f(1) = 2 \).
- For \( x = -1 \), not defined, since the interval is open at \( -1 \).

#### 3.2: For \( 2 < x \leq 5 \)
Similarly, for \( 2 < x \leq 5 \), we use \( f(x) = f(x - 3) \):

- For \( x = 3 \), \( f(3) = f(0) = 1 \).
- For \( x = 4 \), \( f(4) = f(1) = 2 \).

### Step 4: Sketch the graph

1. From \( -3 \leq x \leq 0 \), the graph repeats the behavior from \( -1 < x \leq 2 \).
2. From \( 0 \leq x \leq 3 \), use the defined values for \( f(x) = |2x - 1| + x \).
3. From \( 3 \leq x \leq 4 \), repeat the earlier values using periodicity.

### Final Graph:
The graph will consist of repeating segments of a V-shaped curve with different slopes in each period.

Would you like assistance with a sketching tool or further clarification of the pattern?

Let f(x) be a periodic function of x with period 3 and f(x) = |2x − 1| + x for −1 < x ≤ 2. Sketch the graph of the curve y = f(x) in the interval [−3, 4].

Learn how to sketch the graph of the periodic function f(x) = |2x - 1| + x over the interval [-3, 4]. This problem covers key concepts such as piecewise functions, periodicity, and absolute value properties. Suitable for students in Grades 10-12.

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