Originally Posted by

**acevipa** I'm not too sure how to do this question?

Suppose that $\displaystyle f$ is $\displaystyle (n+1)$-times differentiable on some open interval $\displaystyle I$ containing $\displaystyle 0$ and that $\displaystyle f^{(n+1)}$ is continuous on $\displaystyle I$. Suppose that $\displaystyle x\in I$ and recall that the remainder term in Taylor's theorem is given by:

$\displaystyle R_{n+1}(x)=\frac{1}{n!}\int_0^x f^{n+1}(t)(x-t)^n dt$

Use the mean value theorem for integrals to deduce that:

$\displaystyle R_{n+1}(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}$

where c is some real number between $\displaystyle 0$ and $\displaystyle x$