Using Rolle's theorem to prove that the equation x ^ 3-3x + 1 = 0 has and only has one real root in (0,1) You have to use rolle

Using Rolle's theorem to prove that the equation x ^ 3-3x + 1 = 0 has and only has one real root in (0,1) You have to use rolle

Let f (x) = x ^ 3-3x + 1
Then, f (0) = 1 > 0
f(1)= -1＜0
According to the zero point theorem,
F (x) has at least one zero point in (0,1)
The following is to prove the uniqueness by using the counter evidence method:
Suppose that f (x) has at least two zeros a < B in (0,1),
Because f (a) = f (b) = 0
F (x) satisfies three conditions of Rolle's theorem on [a, b],
According to Rolle's theorem, there exists ξ ∈ (a, b)
Let: F & # 39; (ξ) = 0
f '(ξ)=3ξ^2-3=3(ξ^2-1)＜0
Therefore, F & # 39; (ξ) = 0 is not tenable and contradictory
So suppose f (x) has at least two zero errors in (0,1)
Therefore, f (x) has only one zero point in (0,1)
That is, the equation has only one real root in (0,1),