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:

$|R_{n}(f;x_{0})(x)|\leq \left(\sup _{t\in J}|f^{(n+1)}(t)|\right){\frac {|x-x_{0}|^{n+1}}{(n+1)!}}$

If $f^{(n+1)})$ is continuous on $J$, the $\sup$ can be replaced by $\max$:

$|R_{n}(f;x_{0})(x)|\leq \left(\max _{t\in J}|f^{(n+1)}(t)|\right){\frac {|x-x_{0}|^{n+1}}{(n+1)!}}$

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:

$|R_{n}(f;0)(x)|\leq \left(\sup _{t\in J}|f^{(n+1)}(t)|\right){\frac {|x|^{n+1}}{(n+1)!}}$

If $f^{(n+1)})$ is continuous on $J$, the $\sup$ can be replaced by $\max$:

$|R_{n}(f;0)(x)|\leq \left(\max _{t\in J}|f^{(n+1)}(t)|\right){\frac {|x|^{n+1}}{(n+1)!}}$