Let $f:\mathbb{R} \to \mathbb{R}$ be a function, $f(x)=\ln(1-x)$.
Let $T_{2,f,0}$ be the MacLaurin polynomial of second degree for $f$ and $r_2$ it's remainder.
Find $r_2(x)$ and evaluate it in $x=-1$ i.e $r_2(\frac{1}{2})$.
My first thought was to use Lagrange's form of remainder...
$r_2(x)=\frac{f^{'''}(c)}{3!}x^3=\frac{\frac{-2}{(1-c)^3}}{3!}=\frac{-2}{3(1-c)^3}x^3$
Now, I don't understand that
"evaluate it in $x=-1$ i.e $r_2(\frac{1}{2})$"
I anyway evaluated it in $x=-1$ and $x=\frac{1}{2}$:
When $x=1$: $r_2=\frac{2}{3(1-c)^3}$
When $x=\frac{1}{2}$ $r_2=\frac{-1}{12(1-c)^3}$
Which none of them gives me the correct answer, which apparently is
$r_2=\frac{1}{3(1-c)^3}$ with $c \in (-1,0)$.
Do I have to find the remainder in another way?
I think that it should be
By the the way, your computation is almost correct, you just missed a factor $2$.
The remainder of second degree is $$r_2(x)=f(x)-T_{2,f,0}(x)=\frac{f'''(c)}{3!}x^3=\frac{-2}{3!(1-c)^3}x^3=\frac{-x^3}{3(1-c)^3}$$ where $c$ is some point between $x$ and $0$ and $$f'(x)=-(1-x)^{-1}\,,\,f''(x)=-(1-x)^{-2}\,,\,f'''(x)=-2(1-x)^{-3}.$$ Therefore if $x=-1$ then $c\in (-1,0)$ and $$r_2(-1)=\frac{1}{3(1-c)^3}.$$