Use the fact that d/dx(x^6 - 2x^2 + x) = 6x^5 - 4x + 1 to show that the equation: 6x^5 - 4x + 1 = 0 has at least one solution in the interval (0,1)

We have the function f(x) = x^6 - 2x^2 + x. f'(x) = 6x^5 - 4x +
1.

Now we see that f(1) = 1^6 - 2*1^2 + 1 = 1 -2 +1 =
0

And f(0) = 0 + 0 +0 = 0.

Therefore we
can use Rolle's Theorem which states that if a function f(x) is continuous in the closed interval
[ a, b] and differentiable at every point in (a,b) and if f(a) = f(b) = 0, then there is at least
one number c between a and b at which f'(c) =0.

Now the function we
have is f(x) = x^6 - 2x^2 + x. This is continuous in [ 0,1] and differentiable at all points in
(0 ,1).

As f(0) = f(1) = 0 , there lies a point between 0 and 1
where f'(x) = 6x^5 - 4x + 1 = 0.

This is the solution for the
equation 6x^5 - 4x +1 = 0.