I am trying to solve this & not sure if it is the right answers, the question is:
Given function f(x) is $3$ times derivable at $0$, and the Maclaurin series of the function f(x) up to the order $2$ is :
$f(x) = ax + bx^2 + R_2(x)$.
- Does $f''(0)=b$?
My answer is no because $f''(0)=2b$, is that correct?
Given $a=1$, and -
$$g(x) = \begin{cases} \frac{f(x)}{x} & \text {if $x\ne0$} \\ 1 & \text {if $x=0$} \end{cases}$$
Does $g'(0)=b$?
the answer is yes, does my answer is correct?
Your answers for the first one is right; however, I'm pretty sure the second one is false. Indeed, differentiability at $x_0$ implies $g$ is continuous at $x_0$. So I'll show $g$ is not continuous at $x=0$.
Indeed, we shall compare $\lim_{x \rightarrow 0} f(x)/x$ with $\lim_{x \rightarrow 0} 0$. If $g$ is continuous, they should agree.
As $f(x) = x + bx^2 + R_2(x)$, we see $\frac{f(x)}{x} = 1 + bx + R_2(x)/x$. By definition of Taylor Series, $R_2(x) = O(x^3)$, so we see $R_2(x)/x = O(x^2)$; in particular $\lim_{x \rightarrow 0} R_2(x)/x = 0$. So $\lim_{x \rightarrow 0} \frac{f(x)}{x} = 1 \neq 0$. So $g$ is not continuous at $x=0$, so it is not differentiable at $x=0$, so $g'(0)$ is undefined.