For the function f(x) = 4 _ e (x - 1) - 2 x ^ 2 + 4x + 5, x > 1 ,...

Given $f\left(x\right) = 4\log_{e}\left(x - 1\right) - 2{x}^{2} + 4x + 5, x > 1$

${f}^{'}\left(x\right) = \dfrac{4}{x - 1} - 4\left(x - 1\right)$

For $1 0$

For $x > 2 ⇒ {f}^{'}\left(x\right) < 0$

Hence, $f\left(x\right)$ has a maxima at $x = 2$

$f\left(x\right) = - 1$ has two solution as the curve of $f\left(x\right)$ will cut it at exactly two points.

Now, $f\left(e\right) > 0$ and $f\left(e + 1\right) < 0$

i.e. $f\left(e\right) \cdot f\left(e + 1\right) < 0$

${f}^{'}\left(e\right) - {f}^{"}\left(2\right) = \dfrac{4}{e - 1} - 4\left(e - 1\right) + 8 > 0$