Give an alternative proof of Taylor's Theorem with a single application of Rolle's Theorem by proceeding as follows. Let the notation and hypothesis be as in the statement of Taylor Theorem (Proposition $4.23$ ). Also, as in the proof of Taylor's Theorem, for $x \in[a, b]$, let
$$
P(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots+\frac{f^{(n)}(a)}{n !}(x-a)^{n} .
$$
Define $g:[a, b] \rightarrow \mathbb{R}$ by
$$
g(x)=f(x)+f^{\prime}(x)(b-x)+\frac{f^{\prime \prime}(x)}{2 !}(b-x)^{2}+\cdots+\frac{f^{(n)}(x)}{n !}(b-x)^{n}+s(b-x)^{n+1},
$$
where $s=[f(b)-P(b)] /(b-a)^{n+1} .$ Show that $g(a)=g(b)=f(b) .$ Apply Rolle's Theorem to $g$ to deduce Taylor's Theorem.