About $f(x)$ satisfying $f(0) = 0, f(1) =1, f(2) = 2, f(3) = 4$, and $f'(x)$ is differentiable on $\mathbb{R}$. let $f(x)$ be a function such that 

$$f(0) = 0$$

$$f(1) =1$$ $$f(2) = 2$$ $$f(3) = 4$$ $$f'(x) \text{is differentiable on } \mathbb{R}$$ Prove that there is a number in the interval $(0,3)$ such
that $0 &lt; f''(x)&lt;1$

I'm really stuck.
thanks
Hint: Show that a some point $x \in (0,1)$, $f'(x) = 1$. Then for some $y \in (2,3)$, $f'(y) = 2$. Now use that to deduce the result by applying the MVT again.

Added:

Having found such $x$ and $y$, there must be a $z \in (x,y)$ such that

$$f''(z) = \frac{f'(y) - f'(x)}{y - x}$$

Now think about how you can bound that expression above and below.
I would suggest as a first attempt, curve fitting a polynomial. Let's fit to a cubic, since the general cubic equation has four parameters (and you have four known points).
$$f(x) = ax^3+bx^2+cx+d$$
$$f(0) = a*0^3+b*0^2+c*0+d = 0 \rightarrow d = 0$$
$$f(1) = a*1^3+b*1^2+c*1+d = 1 \rightarrow a+b+c=1$$
$$f(2) = a*2^3+b*2^2+c*2+d = 2 \rightarrow 8a+4b+2c = 2$$
$$f(3) = a*3^3+b*3^2+c*3+d = 4 \rightarrow 27a+9b+3c = 4$$
Solving the last three equations simultaneously gives us:
$$a=-\frac{1}{2}, b=\frac{3}{2}, c = 0$$
Therefore, the actual cubic equation that will pass through all these points is:
$$f(x)=\frac{-x^3}{2}+\frac{3x^2}{2}$$
Since we have the actual function, let's take the derivative:
$$f'(x)=\frac{-3x^2}{2}+3x$$
$$f''(x)=-3x+3$$
Since we have the second derivative being linear, we can figure out where the point $f(x)=0.5$ is. If that point lies inside $(0,3)$, then we have proven the function.
$$f''(x)=3-3x=0.5 \rightarrow x=\frac{-2.5}{-3}=\frac{5}{6}$$
This is inside the domain of the function, so we are good.

Alternatively, we could use the MVT like others have done and say:
$$f''(0)=3$$
$$f''(3)=-6$$
Since the second derivative is continuous (a straight line), by the MVT there exists a value in $(0,3)$ such that $0&lt;f''(x)&lt;1$.