The interval in which `y= (1)/(4x^(3)-9x^(2) + 6x)` is increasing is

Options

A. `(-oo,oo)`
B. `(0,1//2)`
C. `(1//2,1)`
D. `(1, oo)`

Answer

To determine the interval in which the function \( y = \frac{1}{4x^3 - 9x^2 + 6x} \) is increasing, we need to analyze the behavior of the denominator \( f(x) = 4x^3 - 9x^2 + 6x \). The function \( y \) is increasing when \( f(x) \) is decreasing.

### Step-by-Step Solution:

1. **Identify the function**:
We have \( f(x) = 4x^3 - 9x^2 + 6x \).

2. **Find the derivative**:
We need to find the derivative \( f'(x) \) to determine where \( f(x) \) is decreasing.
\[ f'(x) = \frac{d}{dx}(4x^3 - 9x^2 + 6x) = 12x^2 - 18x + 6. \]

3. **Factor the derivative**:
We can factor out the common term:
\[ f'(x) = 6(2x^2 - 3x + 1). \]
Now, we need to factor the quadratic \( 2x^2 - 3x + 1 \):
\[ 2x^2 - 3x + 1 = (2x - 1)(x - 1). \]
Thus, we have:
\[ f'(x) = 6(2x - 1)(x - 1). \]

4. **Find critical points**:
Set \( f'(x) = 0 \):
\[ 6(2x - 1)(x - 1) = 0. \]
This gives us the critical points:
\[ 2x - 1 = 0 \implies x = \frac{1}{2}, \quad x - 1 = 0 \implies x = 1. \]

5. **Determine the sign of \( f'(x) \)**:
We will test the intervals determined by the critical points \( x = \frac{1}{2} \) and \( x = 1 \):
- For \( x < \frac{1}{2} \) (e.g., \( x = 0 \)):
\[ f'(0) = 6(2(0) - 1)(0 - 1) = 6(-1)(-1) = 6 \quad (\text{positive}). \]
- For \( \frac{1}{2} < x < 1 \) (e.g., \( x = 0.75 \)):
\[ f'(0.75) = 6(2(0.75) - 1)(0.75 - 1) = 6(1.5 - 1)(-0.25) = 6(0.5)(-0.25) = -0.75 \quad (\text{negative}). \]
- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f'(2) = 6(2(2) - 1)(2 - 1) = 6(4 - 1)(1) = 6(3)(1) = 18 \quad (\text{positive}). \]

6. **Conclusion**:
The function \( f(x) \) is decreasing in the interval \( \left(\frac{1}{2}, 1\right) \). Therefore, the function \( y \) is increasing in this interval.

### Final Answer:
The interval in which \( y \) is increasing is \( \left(\frac{1}{2}, 1\right) \).