i am given that
$\displaystyle
\frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0 \\
$
prove that there is a solution for 0<x<1
using roll theorem
$\displaystyle
a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\
$
i am given that
$\displaystyle
\frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0 \\
$
prove that there is a solution for 0<x<1
using roll theorem
$\displaystyle
a_0 x^n + a_1 x^{n - 1} + .. + a_n = 0 \\
$
so you take a function
$\displaystyle
f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x\\
$
which is an anti derivative of my given equation that i need find solution to.
$\displaystyle
f'(x)=a_0 x^n + a_1 x^{n - 1} + .. + a_n
$
so the extreme point of f(x) is the solution of f'(x)
thanks