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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?

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2 Answers

up vote 3 down vote accepted

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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I think you misunderstood my question. I'm specifically talking about base polynomial function with degree greater than $1$ and a leading coefficient $1$. Could you please provide an explanation for why the formula is invalid for these kind of functions? –  user26649 Jul 14 '12 at 3:47
Sorry, your original question only added that restriction at a later point. Hopefully my edit helps to clarify why this does not work. –  AMPerrine Jul 14 '12 at 3:57
Thank you, the edit really helped clear it up. –  user26649 Jul 14 '12 at 4:09
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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 $f(x)=5x, f'(0)=5$.

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I understand the second fallacy you pointed out. I edited the question to restrict this to polynomials with degree greater than one. However, I do not understand the second part of your answer - why is the formula for the average rate of change wrong? –  user26649 Jul 14 '12 at 3:36
If $f(x)=x^2+5x, f'(0)=5$. Your restriction that $f(0)=0$ (passing through the origin) works on the constant term, but not anything higher. There is no reason to think that the average rate of change is the average of the rate of change at the endpoints. One is a chord, the other is the average of two tangents. It is a feature of polynomials of degree 2 or less, not shared by most other functions. –  Ross Millikan Jul 14 '12 at 3:44
I see. In the example I mentioned above ($f(x) = x^3$), $f'(0) = 0$, so why would it fail in this scenario? And isn't the average rate the average of the infinite instantaneous rates in that interval (and in this scenario the average of the first and last rate should provide the rate of the whole interval)? If not, what is it the average of? –  user26649 Jul 14 '12 at 3:52
@FarhadYusufali: Because derivatives are a local thing and the average slope is global. You can make high order polynomials do anything you want locally, so we could have one that approximated a step function, with f(0)=0, f(1)=1 and f'(0)=f'(1)=0. There would be local squiggles, but it would fail your imagined relation that the average rate of change over (0,1) is the average of the derivatives at 0 and 1. –  Ross Millikan Jul 14 '12 at 3:57
Thank you for the help. –  user26649 Jul 14 '12 at 4:09
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