# Math Help - Rolle's Theorem

1. ## Rolle's Theorem

This is the last part of a problem that I'm working through. The problem is on Rolle's theorem.

Using Rolles Theorem prove that for any real number λ: the function f(x) = x^3 - (3/2)x^2 + λ never has two distinct zeros in [0,1].

So I was thinking about ways I could do this: but when I calculated f(0) = λ, wheras f(1) = λ-1/2, but to use Rolle's theorem isnt f(a) = f(b) on the interval [a,b]?? This has got me a little confused.
Anyway I put that aside to try another way:
I thought perhaps I should assume for contradiction that I should assume there are two distinct zeros at c1 and c2. So f'(c1 = f'(c2) = 0. But then using Rolles Theorem there is a root of f between c1 and c2. But now I don't know where to go with this next! Please help!

2. Originally Posted by reeha
This is the last part of a problem that I'm working through. The problem is on Rolle's theorem.

Using Rolles Theorem prove that for any real number λ: the function f(x) = x^3 - (3/2)x^2 + λ never has two distinct zeros in [0,1].

So I was thinking about ways I could do this: but when I calculated f(0) = λ, wheras f(1) = λ-1/2, but to use Rolle's theorem isnt f(a) = f(b) on the interval [a,b]?? This has got me a little confused.
Anyway I put that aside to try another way:
I thought perhaps I should assume for contradiction that I should assume there are two distinct zeros at c1 and c2. So f'(c1 = f'(c2) = 0. But then using Rolles Theorem there is a root of f between c1 and c2. But now I don't know where to go with this next! Please help!
Yes, that second method is the way to go. But you are going the wrong way!
Rolles theorem does not say that if f'(c1)= f'(c2)= 0, then there is a zero of f between c1 and c2. It says the if f(c1)= f(c2)= 0, then there is a zero of f' between c1 and c2. Since $f(x)= x^3 - (3/2)x^2 + \lambda$, $f'(x)= 3x^2- 3x= 3x(x- 1)$. The zeros of that are x= 0 and x= 1. There is no 0 between such c1 and c2!