is it ok to use rolls theorem here??
question :
suppose that to the quadratic equation x^2+px+q=0
has two roots
prove that for
there are n different roots on the interval (x1,x2)
??
the prove:
by rolls theorem we have a point "c" on the interval of (x1,x2) for which
f'(c)=0
then they say
"that we can get another solution "g" on the interval of (x1,c) using rolls theorem"
but its not rolls theorem
we dont have two points for which there is a point "j" for which f'(j)=0
we have f(x1)=0 and f'(c)=0 (but its not f(c)=0 )
??
ok
my problem is very simple
if i have a function f(x)
and i have a point "a" for which f(a)=0
and i have point "b" for which f'(b)=0
is the a point "c" between [a,b] for which f''(c)=0
??
(i got one oint of f(x) and the other on f'(x) they are not on the same level)
so we can write .
Write that asprove that for
there are n different roots on the interval (x1,x2)
??
They aren't referring to f now, they are applying Rolle's theorem to f'.the prove:
by rolls theorem we have a point "c" on the interval of (x1,x2) for which
f'(c)=0
then they say
"that we can get another solution "g" on the interval of (x1,c) using rolls theorem"
but its not rolls theorem
we dont have two points for which there is a point "j" for which f'(j)=0we have f(x1)=0 and f'(c)=0 (but its not f(c)=0 )
??
It should be obvious that for , f' is also 0 at and . In fact that is true for all deritives up to the n-1 derivative.
Do you mean "is there a point "c" between a and b for which f"(c)= 0"?
Not necessarily. It depends on whether f'(a)= 0 also which is true for the polynomial in your first post.
For example if f(x)= (x- 1)(x- 2)= , f(1)= 0 and f(2)= 0 so there must exist "b" between 1 and 2 such that f'(b)= 2x- 3= 0. It is obvious that that point is b= 3/2 but f"(x)= 2 which is never 0.
But if f(x)= , then which is 0 at x= 1, x= 2 and at x= 3/2. Now applying Rolle's theorem to f', on the intervals from 1 to 3/2 and from 3/2 to 2 there must be a number b between 1 and 3/2 such that f"(b)= 0 and and a number c between 3/2 and 2 such that f"(c)= 0. That is, there are two values between 1 and 2 that make the second derivative 0 as the theorem says.
when you say
"on the intervals from 1 to 3/2 and from 3/2 to 1"
this is the same solution
you cant take the same interval but on the opposite direction
and say that its a different solution (its the same solution for the second derivative)
and when you say:
"there must be a number b between 0 and 3/2 such that f"(b)= 0 "
b need to be between the interval(between 1 to 3/2) not
"between 0 and 3/2"
??
Sorry, that was a typo. I meant "1 to 3/2 and 3/2 to 2".
I have edited my previous post.
this is the same solution
you cant take the same interval but on the opposite direction
and say that its a different solution (its the same solution for the second derivative)
and when you say:
"there must be a number b between 0 and 3/2 such that f"(b)= 0 "
b need to be between the interval(between 1 to 3/2) not
"between 0 and 3/2"
??
i am given that f(x)=x^2 + px + q
so i can present f(x) as f(x)=(x-x1)(x-x2)
so if we look at n=2 we get
f(x)=(x-x1)^2(x-x2)^2
f'(x)=2(x-x1)(x-x2)^2 + 2(x-x1)^2(x-x2)=2(x-x1)(x-x2)[x-x2+x-x1]=
=2(x-x1)(x-x2)[2x-x1-x2]
so x1 and x2 are solution on f'(x)
but this is a practical observation.
what do i need to say so for g(x)=[f(x)]^n
then x1,x2 are the roots till the n'th derivative??
If f(x)= (x-x1)^n(x-x2)^n then f'(x= n(x-x1)^(n-1)(x-x2)^n+ n(x-x1)^n(x-x2)^(n-1). f"= n(n-1)(x-x1)^(n-2)(x-x2)^n+ 2n^2(x-x1)^(n-1)(x-x2)^(n-1)+ n(n-1)(x-x1)^n(x-x2)^(n-2), etc. That is, the ith derivative is a sum of terms each including factors of (x-x1) and (x-x2) to at least the n-i power. As long as i< n, each term will include at least one factor of each of (x-x1) and (x-x2) and so will be 0 at x= x1 and x= x2.