[Math] the relation between the average rate of change and the derivative

calculusderivatives

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?

Best Answer

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.

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