# 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)

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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.

Let f(x) = x^6 -2x^2+x

Then f(0) = *0^6-2*0^2+0 = 0

f(1) = 1^6-2*1^2+1 = 1-2+1 = 0.

Therefore f(0) = f(1) = 0.

f'(x) = d/dx{x^6-2x^2+x} = 5x^5 -4x+1.

f(x) is a continuous function in (0,1).

Therefore by Roll's theorem f'(x) = 0 in the interval (0,1) as f(x) is continous and differentiable in the interval and f(0)= f(1) = 0.

Therefore f'(x) = 5x^3-4x+1= 0 for some x = c in the interval (0,1). Therefore 5c^2-4c+1 = 0. That means c is root of 5x^3-4x+1 and c is in (0,1).