The problem presents two functions, $g(x)$ and $f(x)$. $g(x) = -(x-1)^2 + 1 - \ln(x-1)$ $f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$ Part A asks to study the variations of $g(x)$ and calculate $g(2)$, then to prove $g(x) \ge 0$ on the interval $]1, 2]$ and $g(x) \le 0$ on $[2, +\infty[$. Part B deals with $f(x)$: determine the domain of definition $D_f$ and its limits at the boundaries, determine $f'(x)$ and verify that $f'(x) = \frac{g(x)}{(x-1)^2}$, draw the variation table of $f$, show that the line $y = -x + 3$ is an oblique asymptote, find the intersection of $C_f$ with the asymptote, and finally, with $h$ the restriction of $f$ on $I = ]2, +\infty[$, show that $h$ is bijective and calculate $h(3)$ and deduce $(h^{-1})'(\frac{\ln 2}{2})$.

1. Problem Description

The problem presents two functions, $g(x)$ and $f(x)$.

$g(x) = -(x-1)^2 + 1 - \ln(x-1)$

$f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$

Part A asks to study the variations of $g(x)$ and calculate $g(2)$, then to prove $g(x) \ge 0$ on the interval $]1, 2]$ and $g(x) \le 0$ on $[2, +\infty[$.

Part B deals with $f(x)$: determine the domain of definition $D_f$ and its limits at the boundaries, determine $f'(x)$ and verify that $f'(x) = \frac{g(x)}{(x-1)^2}$, draw the variation table of $f$, show that the line $y = -x + 3$ is an oblique asymptote, find the intersection of $C_f$ with the asymptote, and finally, with $h$ the restriction of $f$ on $I = ]2, +\infty[$, show that $h$ is bijective and calculate $h(3)$ and deduce $(h^{-1})'(\frac{\ln 2}{2})$.

2. Solution Steps

Part B:

1) Determine $D_f$ and limits at the bounds of $D_f$.

The function $f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$ is defined if $x-1 > 0$ and $x-1 \ne 0$.

Thus, $x > 1$.

$D_f = ]1, +\infty[$

$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-x + 3 + \frac{\ln(x-1)}{x-1})$

Let $u = x-1$, then as $x \to 1^+$, $u \to 0^+$.

$\lim_{x \to 1^+} f(x) = \lim_{u \to 0^+} (-(u+1) + 3 + \frac{\ln u}{u}) = \lim_{u \to 0^+} (-u + 2 + \frac{\ln u}{u}) = -\infty$ since $\lim_{u \to 0^+} \frac{\ln u}{u} = -\infty$.

$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} (-x + 3 + \frac{\ln(x-1)}{x-1})$

Since $\lim_{x \to +\infty} \frac{\ln(x-1)}{x-1} = 0$,

$\lim_{x \to +\infty} f(x) = -\infty$

2) Determine $f'(x)$ and verify that $f'(x) = \frac{g(x)}{(x-1)^2}$.

$f(x) = -x + 3 + \frac{\ln(x-1)}{x-1}$

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

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

5) b- Calculate $h(3)$ and deduce $(h^{-1})'(\frac{\ln 2}{2})$.

$h(x) = f(x)$ on $]2, +\infty[$

$h(3) = f(3) = -3 + 3 + \frac{\ln(3-1)}{3-1} = \frac{\ln 2}{2}$

Since $h(3) = \frac{\ln 2}{2}$, then $h^{-1}(\frac{\ln 2}{2}) = 3$.

We know that $(h^{-1})'(y) = \frac{1}{h'(h^{-1}(y))}$

So $(h^{-1})'(\frac{\ln 2}{2}) = \frac{1}{h'(h^{-1}(\frac{\ln 2}{2}))} = \frac{1}{h'(3)}$

$h'(3) = f'(3) = \frac{g(3)}{(3-1)^2} = \frac{g(3)}{4}$

$g(3) = -(3-1)^2 + 1 - \ln(3-1) = -4 + 1 - \ln 2 = -3 - \ln 2$

$h'(3) = \frac{-3 - \ln 2}{4}$

$(h^{-1})'(\frac{\ln 2}{2}) = \frac{1}{\frac{-3 - \ln 2}{4}} = \frac{4}{-3 - \ln 2} = -\frac{4}{3 + \ln 2}$

3. Final Answer

$h(3) = \frac{\ln 2}{2}$

$(h^{-1})'(\frac{\ln 2}{2}) = -\frac{4}{3 + \ln 2}$