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.