The interval on which the function `f(x)=2x^(2)-3x` is increasing or decreasing in :

Options

A. `[-oo,(3)/(4)]`
B. `[3,oo]`
C. `[(3)/(4),3]`
D. `[(3)/(4),oo]`

Answer

To determine the intervals on which the function \( f(x) = 2x^2 - 3x \) is increasing or decreasing, we will follow these steps:

### Step 1: Find the derivative of the function
To analyze the behavior of the function, we first need to find its derivative.

\[ f'(x) = \frac{d}{dx}(2x^2 - 3x) = 4x - 3 \]

### Step 2: Set the derivative equal to zero
Next, we find the critical points by setting the derivative equal to zero.

\[ 4x - 3 = 0 \]

Solving for \( x \):

\[ 4x = 3 \implies x = \frac{3}{4} \]

### Step 3: Determine the sign of the derivative
Now we will test the intervals around the critical point \( x = \frac{3}{4} \) to determine where the function is increasing or decreasing. The intervals to test are \( (-\infty, \frac{3}{4}) \) and \( (\frac{3}{4}, \infty) \).

1. **Choose a test point in the interval \( (-\infty, \frac{3}{4}) \)**:
- Let’s choose \( x = 0 \):
\[ f'(0) = 4(0) - 3 = -3 \quad (\text{which is } < 0) \]
Thus, \( f(x) \) is decreasing on the interval \( (-\infty, \frac{3}{4}) \).

2. **Choose a test point in the interval \( (\frac{3}{4}, \infty) \)**:
- Let’s choose \( x = 1 \):
\[ f'(1) = 4(1) - 3 = 1 \quad (\text{which is } > 0) \]
Thus, \( f(x) \) is increasing on the interval \( (\frac{3}{4}, \infty) \).

### Step 4: Conclusion
Based on our analysis, we can conclude:

- The function \( f(x) \) is **decreasing** on the interval \( (-\infty, \frac{3}{4}) \).
- The function \( f(x) \) is **increasing** on the interval \( [\frac{3}{4}, \infty) \).

### Summary of Intervals
- **Decreasing Interval**: \( (-\infty, \frac{3}{4}) \)
- **Increasing Interval**: \( [\frac{3}{4}, \infty) \)