Let $f(x),\; a<x<b$ be a real-valued, $n$ -times differentiable function, and let $a<x_0<b$ be a fixed base-point. We will show that for all $x\neq x_0$ in the domain of the function, there exists a $\xi$ , strictly between $x_0$ and $x$ such that $$f(x) = \sum_{k=0}^{n-1} f^{(k)}(x_0)\, \frac{(x-x_0)^k}{k!} + f^{(n)}(\xi)\,\frac{(x-x_0)^n}{n!}.$$

Fix $x\neq x_0$ and let $R$ be the remainder defined by $$f(x) = \sum_{k=0}^{n-1} f^{(k)}(x_0)\, \frac{(x-x_0)^k}{k!} + R\,\frac{(x-x_0)^n}{n!}.$$ Next, define $$F(\xi) = \sum_{k=0}^{n-1} f^{(k)}(\xi)\, \frac{(x-\xi)^k}{k!} + R\,\frac{(x-\xi)^n}{n!},\quad a<\xi<b.$$ We then have