Is it known or where does this lead to? I am  eleventh class student, recently I started learning calculus. I was experimenting on various things, and found a new thing. It is as follows. Let us consider a function $f(x)$which is continuous. So we have derivative $f^{\prime}(x)$ of that function, and if we figure out all the values of the function, and the derivative there is an interesting linking between both of them. 
That is  $$f(b)=f(b-1)+\left\lceil\dfrac{f^{\prime}(b)+f^{\prime}(b-1)}{2}\right\rceil$$ where $\lceil K\rceil$ is the ceil function of K. For example, let us consider $f(x)=x^3,f^{\prime}(x)=3x^2$ and we have $f(1)=1$, so $$f(2)=f(1)+\left\lceil\dfrac{f^{\prime}(2)+f^{\prime}(1)}{2}\right\rceil$$
$$f(2)=1+\left\lceil\dfrac{12+3}{2}\right\rceil=8=2^3$$
Some of my teachers told that its already known and it's called as mean value theorem, but 
I don't know whether it's already there or it's a new one, and what is the intuitive explanation for that ? It is useful in finding the next immediate point, to a given point. Generally we know that $f(a+h) \approx f(a) +f^{\prime}(a).h$, but how is this different from that ? 
If it is a trivial question for experts present here, please do excuse me, but I will be happy in knowing the reason. 
Thank you ! 
 A: One little remark: the statement " Let us consider a function $f(x)$ which is continuous. So we have derivative $f'(x)$  of that function..." is not true. Continuity of a function $f$ in a given point $x$ is not a sufficient condition for the differentiability of $f$ in that point. 
Take for example  $f(x)=|x|$; it is continuous at $x=0$ but not differentiable at that point, as the left derivative  and right derivative are different (in sign), as you can quickly check just using  definitions.
A: It doesn't work in general. Try $f(x)=\sqrt[3]{x}$ and $b=2$; or $f(x)=ax$ for some non-integer $a$, with any $b$.
Also, it turns out that not every continuous function has a derivative everywhere (in fact, some don't have a derivative anywhere), but you'll learn more about this as you progress.
A: Or try $f(x) = \frac{1}{2} x$, b = 1.
I would like to point out why this formula can't be possibly right:
$$f(x)=f(x-1)+\left\lceil\dfrac{f^{\prime}(x)+f^{\prime}(x-1)}{2}\right\rceil$$
it says that $f(x)-f(x-1)$ is integer. That would be very sad if we could only change our functions by integer value if we move by one in $x$-axis.
A: As Hagen pointed out above, this is not true even for the example you gave - you have a computational error.
It is true that if $f(x)$ is a cubic polynomial, then $$f(b)=f(b-1)+\frac{f'(b)+f'(b-1)-a}{2}$$ where $a$ is the coefficient of $x^3$ in the polynomial.
In particular, if $f(x)=x^3$ and $b$ is an integer, it thus means that we can show that:
$$f(b)=f(b-1)+\left\lfloor\frac{f'(b)+f'(b-1)}{2}\right\rfloor$$
This formula fails to work for general polynomials of degree higher than $3$, or for non-polynomial functions $f$.
