Differentiation and Taylor's Theorem

For $x\neq c$, write $f(x)-f(c)=\dfrac{f(x)-f(c)}{x-c}\,(x-c)$. As $x\to c$ the difference quotient tends to $f'(c)$ and $x-c$ tends to $0$, so the product tends to $f'(c)\cdot 0=0$ by the product law for limits. Hence $f(x)\to f(c)$, which is continuity at $c$.

Suppose $c$ is a local maximum, so $f(x)\le f(c)$ near $c$. For $x>c$ near $c$ the quotient $\frac{f(x)-f(c)}{x-c}\le 0$, and letting $x\to c^+$ gives $f'(c)\le 0$. For $x<c$ the quotient is $\ge 0$, and $x\to c^-$ gives $f'(c)\ge 0$. The two force $f'(c)=0$. A minimum is the maximum of $-f$.

By the extreme value theorem $f$ attains a maximum and a minimum on $[a,b]$. If both are attained at the endpoints, then since $f(a)=f(b)$ the maximum and minimum are equal, so $f$ is constant and $f'=0$ throughout $(a,b)$. Otherwise an extremum is attained at an interior point $\xi$, where Proposition 3 gives $f'(\xi)=0$.

Apply Theorem 4 to $g(x)=f(x)-f(a)-\dfrac{f(b)-f(a)}{b-a}(x-a)$, which is continuous on $[a,b]$, differentiable on $(a,b)$, and has $g(a)=g(b)=0$. The conclusion $g'(\xi)=0$ reads $f'(\xi)-\frac{f(b)-f(a)}{b-a}=0$, which is Equation (2).

Assume $x\neq a$; for $x=a$ both sides of Equation (3) equal $f(a)$ and the remainder vanishes, so the claim is trivial. Fix $x$ and define $F(t)=f(x)-\sum_{k=0}^n\dfrac{f^{(k)}(t)}{k!}(x-t)^k$, the error of the Taylor polynomial expanded at the moving base point $t$. The sum telescopes under differentiation. Write $T_k(t)=\frac{f^{(k)}(t)}{k!}(x-t)^k$, so $F=f(x)-\sum_{k=0}^n T_k$. For $k\geq 1$, $T_k'(t)=\frac{f^{(k+1)}(t)}{k!}(x-t)^k-\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}$, while $T_0'(t)=f'(t)$. Reindexing the second pieces by $j=k-1$ turns $\sum_{k=1}^n\frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}$ into $\sum_{j=0}^{n-1}\frac{f^{(j+1)}(t)}{j!}(x-t)^j$, which cancels all but the $k=n$ first piece, leaving

Now set $G(t)=F(t)-\Big(\dfrac{x-t}{x-a}\Big)^{n+1}F(a)$, which is well defined since $(x-a)^{n+1}\neq 0$. Then $G(a)=F(a)-F(a)=0$ and $G(x)=F(x)-0=0$, since $F(x)=0$. Because $f^{(n+1)}$ exists at every point of the interval (endpoints included), $f^{(n)}$ is differentiable and hence continuous there, so $F$, a combination of $f^{(0)},\dots,f^{(n)}$ with polynomial coefficients, and $G$ are continuous on the closed interval between $a$ and $x$ and differentiable on its interior. By Rolle's theorem there is $\xi$ strictly between $a$ and $x$ with $G'(\xi)=0$. Differentiating, $G'(t)=F'(t)+(n+1)\dfrac{(x-t)^n}{(x-a)^{n+1}}F(a)$, so $G'(\xi)=0$ gives

after substituting Equation (4). Cancelling $(x-\xi)^n$, which is nonzero because $\xi\neq x$, solves for $F(a)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$. Since $F(a)=f(x)-\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k$ by definition, rearranging is Equation (3).