What is the relation between the average rate of change and the derivative? A value in the range for any base polynomial function with a y-intercept of zero can be expressed as: $$f\left(x\right) = px$$ where $p$ is the average rate of change between $0$ and $x$. The average rate of change can be in turn expressed as $$p = \frac{ f'\left(0\right) + f'\left(x\right)}{2}$$ where $f'\left(0\right)$ is the instantaneous rate of change at $0$ (or in other words the derivative evaluated at zero) and $f'\left(x\right)$ is the instantaneous rate of change evaluated at $x$ (or in other words the derivative evaluated at the specific range value). In a base polynomial equation (or in other words a polynomial with one term of degree $d$ and leading coefficient $1$) with degree greater than 1 the derivative evaluated at zero can be further simplified to $0$. This leads to a final simplified equation: $$f(x) = \frac{f'(x)x}{2}$$
This final equation however fails to provide the range value for a polynomial above degree $2$. According to the power rule, if: $$f(x) = x^3$$ then: $$f'(x) = 3x^2$$ According to the above derived equation:
$$f(3) = \frac{f'(3)(3)}{2} = \frac{(3(3)^2)3}{2} = \frac{81}{2}$$
which is clearly the wrong answer. What is the flaw in the equation relating the average change of rate and the derivative?
 A: Unfortunately, $$p = \frac{ f'\left(0\right) + f'\left(x\right)}{2}$$ does not give the average rate of change. For example, try $f(x)=1-\cos x$. Your formula gives the average rate of change from $0$ to $\pi$ as $0$, when instead it should clearly be positive from examining the graph.
This is like saying that the average value of a function $f(x)$ on an interval $[a,b]$ is $\frac{f(a)+f(b)}{2}$. This is untrue for most functions. It happens to be true for linear functions, which are exactly what you get when taking the derivative of a quadratic polynomial, explaining why it works in that case.
A: There are two problems. First, you assume a value for $p$ in passing from the first equation to the second that is not correct.  You can define $p$ (which will be a function of $x$) by the first, but there is no reason it should agree with the second.  In practice, it will for polynomials up to second degree, but no further.   Second, you assume that for a function passing through the origin, $f'(0)=0$.  This is not true.  If $f(x)=5x, f'(0)=5$. 
