Let $f(x)=\ln(3x+1)$. I computed the third order Taylor polynomial of $f$ around $0$:
$$T_3(x)=3x-\frac{9}{2}x^2+9x^3. $$
So, I need to evaluate an approximation of $\ln(1.03)$ and $\ln(0.97)$ using this, and I need to prove that the error in both cases is less that $10^{-6}.$
$\ln(1.03)=f(0.1).$So, the approximation is $T_3(0.1)=0.264$
$\ln(0.97)=f(-0.1)$. So, the approximation is $ T_3(-0.1)=-0.264$
My problem is the proof about the error.
I want to use the Lagrange's formula for the error:
$$R_3(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$$ for some $c\in(0,x)$ in the first case and $c\in(x,0)$ in the second case.
In the first case, I tried to use the maximum of $f^{(n+1)}$ in the interval $(0,0.1).$
$$f^{(4)}(x)=-\frac{486}{(3x+1)^4} $$
The function $f^{(4)}$ is increasing on the interval $(0,0.1),$ so an upper bound for it is $$f^{(4)}(0.1)=-170.162109\dots\geq-171.$$
So, if $M=-171$,
$$R_3(x)\leq \frac{M}{4!}(0.1)^4=-0.000709008\dots>10^{-4}>10^{-6}.$$
So, I couldn't prove what I want.
What can I do?