As I understand it, the mean value theorem is where $${f}'(c)=\frac{f(b)-f(a)}{b-a}$$ if $f$ is continuous on the open interval (a, b) and differentiable on the closed interval [a,b].

A problem in the current homework set on WebAssign has me confused. Given $f(x)=x^{7}$ on the closed interval [0, 1], determine whether the MVT can be applied to the closed interval [a,b].

Since $f(x)= x^{7}$ has a similar profile to a cubic function graph, and it is differentiable to ${f}'(x)= 7x^{6}$, it passes two criteria for the MVT.

Now, solving the MVT formula:

$${f}'(x)=\frac{f(b)-f(a)}{b-a} \Rightarrow \frac{[1^{7}]-[0^{7}]}{1-0} \Rightarrow \frac{1-0}{1-0} \Rightarrow \frac{1}{1}= 1$$

Now, I need to find a number $c$ between 0 and 1 that f'(c)=1. However, the only whole number possiblities from the [0, 1] interval produce $${f}'(0)= 7(0)^{6}= 0 \neq 1$$ $${f}'(1)= 7(1)^{6}= 7 \neq 1$$

Am I missing something here? The question has two parts: identify whether the MVT is applicable, and find the numbers $c$ that fit the theorem on the interval. The closest number for $c$ that I've found that works is 0.724, which gives a value of 1.00815, but it doesn't match 1 perfectly.


Your statement of the Mean Value Theorem is somewhat misleading, and has the intervals flipped. What the Mean Value Theorem states is that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$. (Continuity on the larger interval, not the smaller one!)

As to your question: the interval $[0,1]$ includes all real numbers between $0$ and $1$, and not only the whole numbers. The solution will be some real number $c$, $0\lt c \lt 1$, such that $f'(c) = 1$.

In this case, you want a value of $c$ for which $1 = f'(c) = 7c^6$. So you want to find a value of $c$ for which $c^6 = \frac{1}{7}$. You can write down an expression that tells you exactly what $c$ is, though you won't be able to find any finite decimal expression that gives you $c$, because $c$ is an irrational number (it's not even a fraction). Nonetheless, you can write it down. (Just like you can write down a real number $d$ such that $d^2 = 2$: namely, $d=\sqrt{2}$; voila!).

Added. The intuition of the Mean Value Theorem is:

(i) If the function is "nice", then there is a point between $a$ and $b$ where the tangent is parallel to the line joining $(a,f(a))$ and $(b,f(b))$. Note that "nice" means continuous, has tangents everywhere, and you consider all points between $a$ and $b$.

Perhaps more useful for intuition:

(ii) If you think of $f(x)$ as being position, so that $f'(x)$ is velocity, then $\frac{f(b)-f(a)}{b-a}$ is the average velocity over the entire trip. What the Mean Value Theorem tells you is that there has to be at least one point in time during which your instantaneous velocity was equal to your average velocity. E.g., if you traveled 100 miles in two hour, then there had to be some instant at which you were traveling at 50 miles per hour. But it doesn't say that it had to be either when you were starting, an hour into the trip, or when you finished (the "whole numbers" on $[0,2]$). Not even that it had to be at some "nice time" (12:30, or 1:45, or 12:37). Some time; maybe you cannot even write it down exactly, but at some time.


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