Max-estimate version of Lagrange remainder formula

Statement

About a general point

Suppose $f$ is a function of one variable and $x_0$ is a point in the domain such that $f$ is $(n+1)$ times differentiable at $x_0$. Denote by $R_n(f;x_0)$ the function of $x$ given by $x\mapsto f(x)-P_n(f;x_0)(x)$, i.e., $R_n(f;x_0)$ is the remainder when we subtract from $f$ its $n^{th}$ Taylor polynomial at $x_0$.

For any $x$, let $J$ is the interval between $x$ and $x_0$ (it might be the interval $[x,x_0]$ or $[x_0,x]$ depending on whether $x<x_0$ or $x>x_0$). If $f$ is $n+1$ times differentiable everywhere on $J$, then we have:

$|R_n(f;x_0)(x)|\le ({max}_{t\in J}|f^{(n+1)}(t)|)\frac{|x-x_0{|}^{n+1}}{(n+1)!}$

About the point 0

Suppose $f$ is a function of one variable such that $f$ is $(n+1)$ times differentiable at $0$. Denote by $R_n(f;0)$ the function of $x$ given by $x\mapsto f(x)-P_n(f;0)(x)$, i.e., $R_n(f;0)$ is the remainder when we subtract from $f$ its $n^{th}$ Taylor polynomial at $0$.

For any $x$, let $J$ is the interval between $x$ and $0$ (it might be the interval $[x,0]$ or $[0,x]$ depending on whether $x<0$ or $x>0$). If $f$ is $n+1$ times differentiable everywhere on $J$, then we have:

$|R_n(f;0)(x)|\le ({max}_{t\in J}|f^{(n+1)}(t)|)\frac{|x{|}^{n+1}}{(n+1)!}$