For the function `f(x)=4 log_e (x - 1) – 2x^2 + 4x +5, x gt 1`, which one of the following is NOT correct ?

Options

A. A f is increasing in (1, 2) and decreasing in `(2, oo)`
B. f (x) = -1 has exactly two solutions
C. `f'(e) – f "(2) < 0 `
D. f(x)=0 has a root in the interval (e, e +1)

Answer

To solve the problem, we need to analyze the function \( f(x) = 4 \log_e (x - 1) - 2x^2 + 4x + 5 \) for \( x > 1 \) and determine which of the given statements is NOT correct.

### Step 1: Find the first derivative \( f'(x) \)

To analyze the increasing and decreasing behavior of the function, we first compute the first derivative:

\[ f'(x) = \frac{d}{dx} \left( 4 \log_e (x - 1) - 2x^2 + 4x + 5 \right) \]

Using the derivative of logarithm and polynomial functions, we have:

\[ f'(x) = \frac{4}{x - 1} - 4x + 4 \]

### Step 2: Set \( f'(x) = 0 \)

To find critical points, we set the first derivative equal to zero:

\[ \frac{4}{x - 1} - 4x + 4 = 0 \]

Multiplying through by \( x - 1 \) (valid since \( x > 1 \)):

\[ 4 - 4x(x - 1) + 4(x - 1) = 0 \]

This simplifies to:

\[ 4 - 4x^2 + 4x + 4x - 4 = 0 \]

Thus:

\[ -4x^2 + 8x = 0 \]

Factoring out \( -4x \):

\[ -4x(x - 2) = 0 \]

The critical points are \( x = 0 \) and \( x = 2 \). Since we are only interested in \( x > 1 \), we consider \( x = 2 \).

### Step 3: Analyze intervals for increasing/decreasing behavior

We can analyze the sign of \( f'(x) \) in the intervals \( (1, 2) \) and \( (2, \infty) \):

- For \( x \in (1, 2) \):
Choose \( x = 1.5 \):
\[ f'(1.5) = \frac{4}{1.5 - 1} - 4(1.5) + 4 = \frac{4}{0.5} - 6 + 4 = 8 - 6 + 4 = 6 > 0 \]
Thus, \( f(x) \) is increasing in \( (1, 2) \).

- For \( x \in (2, \infty) \):
Choose \( x = 3 \):
\[ f'(3) = \frac{4}{3 - 1} - 4(3) + 4 = \frac{4}{2} - 12 + 4 = 2 - 12 + 4 = -6 < 0 \]
Thus, \( f(x) \) is decreasing in \( (2, \infty) \).

### Step 4: Determine the maximum

Since \( f(x) \) is increasing on \( (1, 2) \) and decreasing on \( (2, \infty) \), we conclude that \( x = 2 \) is a local maximum.

### Step 5: Check the statements

1. **Statement 1**: \( f(x) \) is increasing on \( (1, 2) \) and decreasing on \( (2, \infty) \) - **Correct**.
2. **Statement 2**: \( f(x) = -1 \) has exactly two solutions - **Incorrect** (since it can only intersect the horizontal line twice).
3. **Statement 3**: \( f'(e) < f''(2) \) - **Needs verification**.

### Conclusion

The statement that is NOT correct is **Statement 2**.