Find the intervals in which the following functions are:
(i) increasing
.(ii)decreasing `f(x) = 2x^(2)-6x`

Answer

To find the intervals in which the function \( f(x) = 2x^2 - 6x \) is increasing or decreasing, we will follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^2 - 6x) \]
Using the power rule, we differentiate:

\[ f'(x) = 4x - 6 \]

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

\[ 4x - 6 = 0 \]

Solving for \( x \):

\[ 4x = 6 \\ x = \frac{6}{4} = \frac{3}{2} \]

### Step 3: Determine intervals
Now we will determine the intervals for increasing and decreasing by testing the sign of \( f'(x) \) around the critical point \( x = \frac{3}{2} \).

1. **Choose a test point in the interval \( (-\infty, \frac{3}{2}) \)**:
- Let’s choose \( x = 0 \):
\[ f'(0) = 4(0) - 6 = -6 \quad (\text{which is } < 0) \]
Therefore, \( f(x) \) is **decreasing** on \( (-\infty, \frac{3}{2}) \).

2. **Choose a test point in the interval \( (\frac{3}{2}, \infty) \)**:
- Let’s choose \( x = 2 \):
\[ f'(2) = 4(2) - 6 = 8 - 6 = 2 \quad (\text{which is } > 0) \]
Therefore, \( f(x) \) is **increasing** on \( (\frac{3}{2}, \infty) \).

### Step 4: Conclusion
Based on our analysis, we can conclude:
- The function \( f(x) \) is **decreasing** on the interval \( (-\infty, \frac{3}{2}) \).
- The function \( f(x) \) is **increasing** on the interval \( (\frac{3}{2}, \infty) \).

### Final Answer:
- Increasing Interval: \( \left( \frac{3}{2}, \infty \right) \)
- Decreasing Interval: \( \left( -\infty, \frac{3}{2} \right) \)

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