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

We are given $$f(x) = 4\log_e(x-1) - 2x^2 + 4x + 5$$ for $$x > 1$$.

Find f'(x) and critical points.

$$f'(x) = \frac{4}{x-1} - 4x + 4 = \frac{4}{x-1} - 4(x-1) = \frac{4 - 4(x-1)^2}{x-1}$$

Setting $$f'(x) = 0$$: $$4 - 4(x-1)^2 = 0 \implies (x-1)^2 = 1 \implies x = 2$$ (since $$x > 1$$).

For $$x \in (1, 2)$$, $$(x-1)^2 < 1$$, so $$f'(x) > 0$$ (increasing).

For $$x \in (2, \infty)$$, $$(x-1)^2 > 1$$, so $$f'(x) < 0$$ (decreasing).

So Option A is correct.

Find f(2) and check Option B.

$$f(2) = 4\ln(1) - 8 + 8 + 5 = 5$$

Since $$f$$ increases on $$(1,2)$$ and decreases on $$(2,\infty)$$, the maximum is $$f(2) = 5$$. As $$x \to 1^+$$, $$f(x) \to -\infty$$, and as $$x \to \infty$$, $$f(x) \to -\infty$$. So $$f(x) = -1$$ (which is less than 5) has exactly two solutions (one on each side of $$x = 2$$). Option B is correct.

Compute f'(e) and f''(2) to check Option C.

$$f'(e) = \frac{4}{e-1} - 4(e-1) = \frac{4 - 4(e-1)^2}{e-1}$$

Since $$(e-1)^2 \approx (1.718)^2 \approx 2.952$$, we get $$f'(e) = \frac{4 - 11.81}{1.718} \approx -4.55$$.

$$f''(x) = -\frac{4}{(x-1)^2} - 4$$

$$f''(2) = -\frac{4}{1} - 4 = -8$$

$$f'(e) - f''(2) \approx -4.55 - (-8) = 3.45 > 0$$

Option C claims $$f'(e) - f''(2) < 0$$, which is false.

Check Option D.

$$f(e) = 4\ln(e-1) - 2e^2 + 4e + 5 = 4\ln(1.718) - 2(7.389) + 4(2.718) + 5 \approx 2.164 - 14.778 + 10.873 + 5 = 3.26 > 0$$

$$f(e+1) = 4\ln(e) - 2(e+1)^2 + 4(e+1) + 5 = 4 - 2(13.87) + 14.87 + 5 \approx -3.87 < 0$$

By the Intermediate Value Theorem, $$f(x) = 0$$ has a root in $$(e, e+1)$$. Option D is correct.

Answer: Option C is NOT correct.