For the function $f(x) = x^{6}+x^{4}-1$ and the interval $[0,1]$, I need to find the number $\xi$ that occurs in the Mean Value Theorem: $\displaystyle \frac{f(b)-f(a)}{(b-a)} = f^{\prime}(\xi)$.
Plugging in all the given information, I get that $\displaystyle \frac{f(1) - f(0)}{1-0} = 6\xi^{5} + 4\xi^{3} \to \frac{(1^{6}+1^{4}-1)-(0^{6}-0^{4}-1)}{1} = 6\xi^{5} + 4\xi^{3} \to 2 =6\xi^{5} + 4\xi^{3} \to 1 = 3\xi^{5} + 2\xi^{3} \to 3\xi^{5} + 2\xi^{3} - 1= 0 $
So, this amounts to finding a root of the polynomial $3\xi^{5} + 2\xi^{3} - 1= 0 $ such that $0 < \xi < 1$
Using the rational roots theorem, the only possibility is that $\displaystyle \xi = \frac{1}{3}$; however, when I plug it into the equation $3\xi^{5} + 2\xi^{3} - 1= 0 $, I find that it does not give me a true statement.
Therefore, I am at an impasse to find the $\xi$ I'm looking for. Any assistance you could offer would be most welcomed - although the more detailed and concrete, the better. Thanks!