Proving Rolle's Theorem: Proving Zero's of f(x) in [0,1]

  • Thread starter Thread starter nlews
  • Start date Start date
  • Tags Tags
    Theorem
nlews
Messages
11
Reaction score
0
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 isn't 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!
 
Physics news on Phys.org
Rolle's theorem states that if a function is identical on two distinct points between which it's defined and has a derivative, then there is a point between them where the derivative vanishes.
So after making the assumption, you have f(c1)=f(c2)=0 rather than f'(c1)=f'(c2)=0 (as you wrote), which means there is a point c where f'(c)=0, and 0<c<1. Is this possible?
 
Moderator's note:

Thread moved from "Calculus & Analysis".

Homework assignments or any textbook style exercises are to be posted in the appropriate forum in our https://www.physicsforums.com/forumdisplay.php?f=152" area. This should be done whether the problem is part of one's assigned coursework or just independent study.
 
Last edited by a moderator:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top