Mistake in answer sheet? 
At $x = 3$, the function given by $f\left(x\right) =\begin{cases} x^2, & x < 3 \\ 6x-9, & x\geq 3 \end{cases}$ is?

$1$. undefined 
$2$. continuous but not differentiable 
$3$. differentiable but not continuous $\color{red} {Not\, possible}$
$4$. neither continuous nor differentiable
$5$. both continuous and differentiable
The book says $2$.  But I think it is wrong, because derivative existed for both functions and is the same at $x=3$.
$\dfrac{\mathrm{d}}{\mathrm{d}x} x^2\vert_{x=3} = 2\left(3\right) = 6$  and $\dfrac{\mathrm{d}}{\mathrm{d}x} 6x-9 = 6$
 A: The function is differentiable at $3$. The reason you give is intuitively reasonable. In principle it is not enough. Formally, you should prove that 
$$\lim_{x\to 3^-}\frac{f(x)-9}{x-3}=\lim_{x\to 3^+}\frac{f(x)-9}{x-3}.$$
A: $\dfrac{f(3+h)-f(3)}{h}=\begin{cases} 6+h, & h<0 \\ 6, & h>0 \end{cases}.$
Hence $\lim\limits_{h\to 0}\dfrac{f(3+h)-f(3)}{h}=6$, so $f'(3)$ exists and equals $6$.
However, the function $f'$ (which is defined everywhere) is not differentiable at $3$.
A: Edit: The book is not correct, and you most certainly are.
The derivative at $x=3$ does exist for both functions, though only one of the functions composing your piece-wise function is defined for that point (as the domain of the other does not include the point).
While the derivative at $x=3$ exists across the whole domain of $f(x)$, it has a different value for each constituent function of the total piece-wise function across most of its domain. Thankfully, at $x=3$, the derivative, as well as the function's value, for each function matches. Generally, the derivative of $x^2$ is $2x$, and the derivative of $6x-9$ is $6$. Evaluated at $x=3$, you will obtain identical values for the derivative. This is also the case when taking a limit analysis of $f(x)$, where the right hand limit equates to the left hand limit.
The piece-wise function as a whole has a continuous value and continuous derivative at $x=3$, so the function is both continuous and differentiable.
