# Prove that the equation 4x^3+3x^2=2x+1 have a solution 0<x<1.

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### 2 Answers

To prove that the given equation has a root in the set (0,1), we'll apply Rolle's theorem.

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

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

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

According to the Rolle's theorem:

where c belongs to (0,1).

If f(1)=f(0) (condition valid only for a Rolle function)=> f'(c)=0

We'll differentiate the Rolle 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

4x^3+3x^2= 2x+1.

To show that there is a root between the 0<x<1.

We rewrite the equation by subtracting 2x+1 from both sides:

4x^3+3x^2-2x-1 = 0.

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

f(x) = 4x^3+3x^2-2x+1 is a obviously continuous function in (0 ,1) for all values of x.

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

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

Therefore f(x) being a continous function should take all the values between f(0) = -1 and f(1) = 4 for 0 < x <1.

So zero being a value between -1 and 4, f(x) = 0 for some x in the interval (0 , 1).

Therefore 4x^3+3x^2-2x-1 = 0 for some x for which 0 < x < 1.

That proves 4x^3+3x^2 =2x+1 has a solution c , for which 0 < c< 1.