Which is true about $f(x) $? Let $f$ be a twice differentiable function satisfying: $$ \cases{f(1) = 1 \\ f(2) =4\\ f(3) = 9  }$$  then :


*

*$ f''(x)= 2 $ for all $ x \in R $

*$f'(x) = 5 = f''(x) $ for some $ x\in (1,3)$

*There exists at least one $ x \in (1,3)$ such that $ f''(x)=2$


Then which of the following are true?
So what I basically tried is that I assumed that the function could be $ x^2$ But the answer I got from that was wrong. 
 A: It is true that $f(x)=x^2$ satisfies the requirements, but many other functions do as well. You can imagine putting squiggles on top of $x^2$ as long as you go through the points.  An example is below.   All the questions ask if they are true for all functions that satisfy the requirements, not just for one of them. My example shows that (1) can be false.  You should be able to modify it to find counterexamples to the second.  You do have $f'(x)=5$ for some $x$ because of the mean value theorem, but you might not have $f''(x)=5$ at that point.
A: We show that 3) is true.
Let $h(x)=f(x)-x^2$ then $h(1)=h(2)=h(3)=0$. Hence by the Mean Value Theorem, 
i) there is $x_1\in (1,2)$ such that $h'(x_1)=0$;
ii) there is $x_2\in (2,3)$ such that  $h'(x_2)=0$;
iii) there is $x_3\in (x_1,x_2)\subseteq (1,3)$ such that  $h''(x_3)=0$ (here the MVT is applied to $f'$).
It follows that $f''(x_3)=h''(3)+(x^2)''_{x=x_3}=0+2=2$.
A counterexample for 1) and 2) is $f(x)=x^2+\frac{1}{6}(x-1)(x-2)(x-3)$. Then
$$f'(x)=\frac{3x^2+11}{6} \quad\mbox{and}\quad f''(x)=x.$$
Hence $f''(x)=x=5\not\in (1,3)$.
