Caught Similarities

Let $g(x) = f(x) + f'(x) + f''(x)$ . Since $f(x)$ is a quadratic, so is $g(x)$ . We can then apply the fact that for any quadratic polynomial $g(x)$ , $g(x + 3) - 3g(x + 2) + 3g(x + 1) - g(x) = 0$ for all $x$ .

Setting $x = 0$ , we find $g(3) = 3g(2) - 3g(1) + g(0) = 3 \cdot 23 - 3 \cdot 13 + 12 = 42.$

For $f(x)=ax^2+bx+c$ ,

$\begin{matrix} (\alpha) \; \; f(0)+f'(0)+f''(0) \! \! \! \! \! &=2a+b+c \! \! \! \! \! &= 12\\ (\beta) \; \; f(1)+f'(1)+f''(1) \! \! \! \! \! &=5a+2b+c \! \! \! \! \! &= 13\\ (\gamma) \; \; f(2)+f'(2)+f''(2) \! \! \! \! \! &=10a+3b+c \! \! \! \! \! &= 23 \end{matrix}$

Doing $\dfrac{\alpha + \gamma - 2 \beta}{2}$ , we find out $a=\dfrac{9}{2}$ . Doing $\beta - \alpha$ , we have $b = -\dfrac{25}{2}$ , and finally $c=\dfrac{31}{2}$ .

Because $f(3)+f'(3)+f''(3) =17a+4b+c$ , we have that our desired value is $\boxed{42.}$

i'm not sure how i got this but for each x in f(x) the resulting summation of the function, first and second derivatives is part of a sequence. From x=0 to x=1 we get a difference of 1 (13-12) and from x=1 to x=2 you get a difference of 10 (23-13). So the difference between these two differences (10-1) is 9 so the next summation for x=3 should be 10+9= 19 more than the sum of 23 for x=2. So i added 23 and 19 and got 42 in my head. Can anyone help me understand if there is a logical explanation to this extrapolation? Is it because some property of a parabola? Let me know what you did

We let $g(x)=f(x)+f'(x)+f''(x)$ . Since deg $(f(x))=2$ , we must have deg $(f'(x))=1$ and deg $(f''(x))=0$ and thus deg $(g(x))=$ max $($ deg $(f(x))$ , deg $(f'(x))$ , deg $(f''(x)))=2$ . Thus, $\exists a,b,c \in \Bbb R$ such that $g(x)=ax^{2}+bx+c$ . Hence we have $3$ variables and $3$ equations. Solving, we have $a=\frac{9}{2},b=-\frac{7}{2}$ and $c=12$ . Thus $g(3)=\frac{9}{2}\times 3^{2}-\frac{7}{2}\times 3 + 12=42.$