Analysis: Theorem 5.2 Apostol Analysis

Question

theorem 5.2

The theorem as I read it states if $f(x)$ is differentiable at $c$ then $f(x)-f(c)=(x-c)f'(c)$ voor $x \in (a,b)$

But of course this is only true for $x\to c$. Can someone explain how I misinterpret this theorem.

Answer 1

$f^*$ is different with the derivate of $f$.

$f^*$ is a function where the value for $x\neq c$ is $\frac{f(x)-f(c)}{x-c}$, and $f'(c)$ at $x=c$.

Answer 2

Theorem states, that $f^*(c) = f'(c)$ and not that $f^* \equiv f'$.

Answer 3

Let [...] be: "$\forall x\in(a,b),f(x)-f(c)=(x-c)f^*(x)$ and $f^*(c)=f'(c)$", i.e.

$\forall x\in(a,b)\setminus\{c\},f^*(x)=\frac{f(x)-f(c)}{x-c}$ and $f^*(c)=f'(c).$

In Apostol's statement, you can replace equivalently "there is a function $f^*$ which is continuous at $c$ and which satisfies [...]" by:

"the function $f^*$ defined by [...] is continuous at $c$",

i.e.

$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=f'(c).$