# Prove that the equation has at least a solution in the interval (0,1) 4x^3 + 3x^2 - 2x - 1 = 0

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### 2 Answers

We'll use the Rolle's theorem to prove that the given equation has a root over the interval (0,1).

Let's see how:

We'll choose a Rolle function f:[0,1]->R

f(x)=x^4+x^3-x^2-x

According to the Rolle's rule,

f(1)-f(0)=f'(c)(1-0)

where c belongs to (0,1).

If f(x) is a Rolle function, then f(1)=f(0).

f'(c)=0.

We'll differentiate Rolle's function and we'll get:

f'(x)=4x^3+3x^2-2x-1

**If f'(c)=0 ,then c is a root of f'(x), c belongs to (0,1). q.e.d.**

We know that if f(x) is a continuous function and f(a) and f(b) are of different sign, then f(x) should cross the X axis for some value c in the interval (a ,b).

So to have root for f(x) = 4x^3+3x^2-2x-1 , f(0) and f(1) are of different signs.

f(0) = 4*0^3 +3*0^2-2*0 -1 = -1 which is negative.

f(1) = 4*1^3+3*1^2-2*1 -1 = 4 ehich is positive.

So f(x) , being a continuous , should take all vaues between f(0) = -1 to f(1) = 4 for any x in the interval (0, 1).

Therefore there is one x = c for which f(c) = 0. This implies f(x) has a root in (0 , 1)