Let f(x)=ln(1-3x). Find a quadratic polynomial p(x) so that p(0)=f(0), p'(0)=f'(0) and p''(0)=f''(0).
I don't know how to do this :/
First a quick note if you know how to find a Taylor series centered at x=0 you are done. Just take the first 3 terms.
So lets suppose you don't so we want a quadratic function
$\displaystyle p(x)=ax^2+bx+c$ and we want
$\displaystyle f(0)=p(0) \iff ln(1)=c \iff c=0$
So now we have $\displaystyle p(x)=ax^2+bx \implies p'(x)=2x+b$
$\displaystyle f'(0)=\frac{-3}{1-3(0)}=2(0)+b=p'(0) \implies b=-3$
so now we have $\displaystyle p''(x)=2a$
$\displaystyle f''(0)=\frac{-9}{(1-3(0))^2}=2a \implies a=-\frac{9}{2} $
So we get
$\displaystyle p(x)=-\frac{9}{2}x^2-3x$