Obtain an upper bound for the error incurred in approximating $f(0.1)$ for $f(x)=\frac{1}{1-x}$ by $P_5(0.1)$
My approach: Find Maclaurin series $P_5(x)=1+x+x^2+x^3+x^4+x^5$
According to remainder theorem $|R_5(0.1)|\leq \frac{M\cdot(0.1)^6}{6!}$. But how can I get $M$ because I didn't know what $M$ is the upper bound for $f^{(n+1)}(x)$ in the $x\in[0,0.1]$. Besides can anyone help me to understand general concept to find $M$. Because it's easy to get $M$ for $\sin(x),\cos(x),e^x$ function but I don't know how to find it for other function.
Any hint or solution will be appreciated. Thanks in advance.
Hint. Note that in this case $$f(x)-P_5(x)=\frac{1}{1-x}-\frac{1-x^6}{1-x}=\frac{x^6}{1-x}.$$ P.S. By the way $f^{(n)}(x)=\frac{n!}{(1-x)^{n+1}}$ so what is $M=\max_{x\in [0,0.1]}|f^{(6)}(x)|$?