Htykuut

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

Find $r_2(x)$ and evaluate it in $x=-1$, i.e $r_2(-1)$.

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 $cin (-1,0)$ and
$$r_2(-1)=frac{1}{3(1-c)^3}.$$