Value of second derivative If $f (x)$ is twice differentiable function. Given that 
$$f(1)=1,f (2)=4,f (3)=9,$$
then is $f''(x) =2$ for all $x$ in $(1,3)$ or for some $x$ in $(1,3)$? 
I guessed the function to be $x^2$.So it's second derivative is always $2$. But my book says $f"(x)=2$ for some $x$ in $(1,3)$. Why is it so?
 A: We need not have $f(x) = x^2$ for all $x$. Think of a function $g$, which vanished outside the intervall $(1,2)$ say, then $f(x) = x^2+ g(x)$ has the same values as $x^2$ in $1$, $4$, $9$, but its derivative will not be $2$ troughout the interval. By polynomial interpolation you can - if you know interpolation - construct other examples, e. g. 
$$ f_2(x) = -\frac 83 x^3 + 17x^2 - \frac{88}3 x + 16 $$
also meets the conditions.
And: Just because $f_1(x) = x^2$ is the only function that comes to our mind, when reading the conditions $f(1) = 1$, $f(2) = 4$, and $f(3) = 9$, this does not mean, that it really is the only one. There are plenty of functions, and some of them perhaps also meet the given conditions, if you state that $f'' = 2$, you have to prove it. You cannot just say "It is the only one I know, hence I'm done.", that is not how math works. You have to start from the given conditions and prove your statement. 
To see that it is true for some $x$, consider $g \colon x \mapsto f(x) - x^2$. Then, as you noticed $g(1) = g(2) = g(3) = 0$ and $g$ is twice differentiable. Hence, by Rolle's theorem, there are $\xi_1 \in (1,2)$ and $\xi_2 \in (2,3)$ with $g'(\xi_1) = g'(\xi_2) = 0$. Another call to Rolle gives us an $x \in (\xi_1, \xi_2) \subseteq (1,3)$ with 
$$ 0 = g''(x) = f''(x) - 2. $$
A: Here is another example for the fun of it
$$ f= 6 - 3 \cos \left( \frac{\pi x}{3} \right) - \frac{7}{\sqrt{3}} \sin \left( \frac{\pi x}{3} \right) $$
it meets the conditions of $(1,1)$, $(2,4)$ and $(3,9)$ but $f'' = 2$ only at $x=2.03614537421\ldots$ within the range.
