I've got a question the linearization error proof part from the Book The calculus Lifesaver, we want to proof the error $g(a)=\frac{1}{2}{f}''(c)(x-a)^2$.
Original derivation progress:
The linearization equation: $L(x)=f(a)+{f}'(a)(x-a)$.
We set the error as $r(x)$, $r(x)=f(x)-L(x)=f(x)-f(a)-{f}'(a)(x-a)$.
Make $x$ as constant and $a$ as variable $t$: $g(t)=f(x)-f(t)-{f}'(t)(x-t)$, especially, $g(x)=0$.
So, ${g}'(t)=-{f}''(t)(x-t)$, especially, ${g}'(x)=0$.
Order $h(t)=g(t)-K(x-t)^2$, ${h}'(t)={g}'(t)+2K(x-t)$. We also know ${h}'(c) = \frac{h(x)-h(a)}{x-a}$, So we have
${g}'(c)+2K(x-c)=\frac{(g(x)-K(x-x)^2)-(g(a)-K(x-a)^2)}{x-a} = \frac{-g(a)+K(x-a)^2}{x-a}$, which is $g(a)-K(x-a)^2 = (x-a)(x-c)({f}''(c)-2K)$.
At here, it says as we can't handle $(x-c)$, the only way to eliminate it is to make the left part of the equation zero, which makes ${f}''(c)-\frac{2g(a)}{(x-a)^2}=0$, from which can get $g(a)=\frac{1}{2}{f}''(c)(x-a)^2$.
Question:
(1)Why we need to handle with $(x-c)$? Can't the form of $g(a)$ exists as having a $(x-c)$? (As is says in the book, we need to proof the form $g(a)=\frac{1}{2}{f}''(c)(x-a)^2$, and there's no $(x-c)$ in it. But that's because we already knew the equation. What if we don't know the equation in the first place? How do we deduce the equation then?)
(2)Why can just order $g(a)-K(x-a)^2 = 0$ to get $g(a)$? Is that because we can find an equation about the error g(a)?