Prove that the equation 4x^3+3x^2=2x+1 have a solution 0<x<1.
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,
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:
If f'(c)=0 ,then c is a root of f'(x), c belongs to (0,1). q.e.d
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.