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

hazecraze · #1 $Old$ November 3rd, 2009, 08:46 AM

$\longrightarrow$ 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?

tonio · #2 $Old$ November 3rd, 2009, 09:24 AM

Quote:

Originally Posted by hazecraze $View Post$

$\longrightarrow$ 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 [math]f(0)f(1)<0[\math]

3) Now use the IVT and solve your problem.

Tonio

hazecraze · #3 $Old$ November 3rd, 2009, 01:09 PM

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).

tonio · #4 $Old$ November 3rd, 2009, 01:14 PM

Quote:

Originally Posted by hazecraze $View Post$

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

hjortur · #5 $Old$ November 3rd, 2009, 01:33 PM

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.