Using rolle's theorem to show a function has only one real root

• November 3rd 2009, 06:46 AM
hazecraze
Using rolle's theorem to show a function has only one real root
$\longrightarrow$ Use rolle's theorem to show that $f(x)=3x-cosx-1$ has only one real root.

So, I know you use IVT to say it has at least one root and then contradict it with Rolle's, but in order to use the IVT you have to pick two points- one negative and one positive so there is point where it was zero. Is there any method for this or do you pick numbers and check to see if it works?
• November 3rd 2009, 07:24 AM
tonio
Quote:

Originally Posted by hazecraze
$\longrightarrow$ Use rolle's theorem to show that $f(x)=3x-cosx-1$ has only one real root.

So, I know you use IVT to say it has at least one root and then contradict it with Rolle's, but in order to use the IVT you have to pick two points- one negative and one positive so there is point where it was zero. Is there any method for this or do you pick numbers and check to see if it works?

1) Prove your function is always increasing (hint: check f'(x))

2) Check that [tex]f(0)f(1)<0[\math]

3) Now use the IVT and solve your problem.

Tonio
• November 3rd 2009, 11:09 AM
hazecraze
I'm really just concerned about finding the interval where f(c)=0.

I got stuck trying to find where it is increasing:
$3 + sin(x)=0$
$sin(x)=-3$
=

But it seems (how the book writes it) that you should just pick small numbers like 0 and 1 to see if it works. I guess in this case: 0 and pi works-(f(0) is negative and f(pi) is positive).
• November 3rd 2009, 11:14 AM
tonio
Quote:

Originally Posted by hazecraze
I'm really just concerned about finding the interval where f(c)=0.

I got stuck trying to find where it is increasing:
$3 + sin(x)=0$
$sin(x)=-3$
=

But it seems (how the book writes it) that you should just pick small numbers like 0 and 1 to see if it works. I guess in this case: 0 and pi works-(f(0) is negative and f(pi) is positive).

If you're doing calculus then you must know some basic trigonometry, don't you? Well, as $|\sin x|\leq 1\,\,\forall\,\,x\in \mathbb{R}$, then $f'(x)>0$ ALWAYS...right? And thus your functions is always strictly increasing and then it has at most one zero.

Now check point (2) in my first message and solve your problem.

Tonio
• November 3rd 2009, 11:33 AM
hjortur
There is no reason to use IVT, Rolle's theorem alone does the trick.

Assume that $f(x)=3x-cosx-1$ has at least two roots x=a and x=b.
Then f(a)=f(b)=0.

Now by Rolle's theorem there excists a c such that f'(c)=0.

But what is f'(x) ?

Hope that helps.