Let g(x)`=f(x)-2{f(x)}^2+9{f(x)}^3` for all `x in R` Then

Options

A. g(x) and f(x) increase and decrease together
B. g(x) increases whenever f(x) decreases and vice-versa
C. g(x) increases for all ` x in R`
D. g(x) decreases for all `x in R`

Answer

To solve the problem, we need to analyze the function \( g(x) = f(x) - 2(f(x))^2 + 9(f(x))^3 \) and determine when it is increasing or decreasing based on the derivative \( g'(x) \).

### Step-by-Step Solution:

1. **Differentiate \( g(x) \)**:
We start by differentiating \( g(x) \) with respect to \( x \):
\[ g'(x) = \frac{d}{dx}[f(x)] - 2 \frac{d}{dx}[(f(x))^2] + 9 \frac{d}{dx}[(f(x))^3] \]
Using the chain rule, we have:
\[ g'(x) = f'(x) - 2 \cdot 2f(x)f'(x) + 9 \cdot 3(f(x))^2f'(x) \]
Simplifying this gives:
\[ g'(x) = f'(x) - 4f(x)f'(x) + 27f(x)^2f'(x) \]

2. **Factor out \( f'(x) \)**:
We can factor \( f'(x) \) out of the expression:
\[ g'(x) = f'(x)(1 - 4f(x) + 27f(x)^2) \]

3. **Analyze the quadratic expression**:
Let \( h(f(x)) = 27f(x)^2 - 4f(x) + 1 \). We need to determine when \( h(f(x)) > 0 \) or \( h(f(x)) < 0 \).
- The coefficient of \( f(x)^2 \) is \( 27 \), which is positive.
- To find the roots of the quadratic, we calculate the discriminant:
\[ D = (-4)^2 - 4 \cdot 27 \cdot 1 = 16 - 108 = -92 \]
Since the discriminant is negative, the quadratic \( h(f(x)) \) does not cross the x-axis and is always positive (as the leading coefficient is positive).

4. **Conclusion about \( g'(x) \)**:
Since \( h(f(x)) > 0 \) for all \( f(x) \), we have:
\[ g'(x) = f'(x)(h(f(x))) > 0 \quad \text{if } f'(x) > 0 \]
\[ g'(x) < 0 \quad \text{if } f'(x) < 0 \]
Thus, \( g(x) \) is increasing when \( f(x) \) is increasing and decreasing when \( f(x) \) is decreasing.

### Final Answer:
The function \( g(x) \) is increasing when \( f'(x) > 0 \) and decreasing when \( f'(x) < 0 \).