• Support PF! Buy your school textbooks, materials and every day products Here!

MVT prove question

  • #1
1,395
0
suppose f and g are continues on [0,1]
and differentiable on (0,1)
and f'(x)g(x) differs f(x)g'(x)
for every x existing in (0,1)
prove that there is a point c in [0,1] so g(c)=0

??

for what purpose do i need to know that
"f and g are continues on [0,1]
and differentiable on (0,1)
??"

g(c)=0 is the use of MVT

i dont know how to merge the peaces
??
 

Answers and Replies

  • #2
33,477
5,167
Your problem as presented doesn't make any sense to me, since it gives information about the derivative of (fg)(x), but then doesn't ask you to do anything with that information.

As written, it's not possible to prove what you're asked to prove. Here's a counterexample.
Let f(x) = x + 1, and g(x) = x^2 + 1
Both functions are continuous for all reals, so a fortiori they are continuous on [0, 1].
Both functions are differentiable everywhere, which certainly implies they are differentiable on (0, 1).

(fg)'(x) = d/dx[x^3 + x^2 + x + 1] = 3x^2 + 2x + 1 is positive for all reals.

g(x) >= 1 for all reals, which implies there does not exist a number c in [0, 1] so that g(c) = 0.
 
  • #3
1,395
0
for what purpose do i need to know that
"f and g are continues on [0,1]
and differentiable on (0,1)
??"
 
  • #4
33,477
5,167
for what purpose do i need to know that
"f and g are continues on [0,1]
and differentiable on (0,1)
??"
I have no idea.

Also, one regular on this forum noticed that my counterexample doesn't work. I took the condition that f'(x)g(x) not equal to f(x)g'(x) to mean that (fg)'(x) was not zero. I'll have to think about this problem some more.

This condition does mean that the numerator in the result from the quotient rule is not zero, but how that figures into the problem I haven't the faintest idea.

Transgalactic, are you sure that what you posted is the same as what's in your book or that your were given?
 
  • #5
1,395
0
this is the question:

suppose f and g are continues on [0,1]
and differentiable on (0,1)
and [tex]f'(x)g(x)\neq f(x)g'(x)[/tex]for every x existing in (0,1)
prove that there is a point c in [0,1] so g(c)=0

??
 
  • #6
1,395
0
the solution starts with
"suppose there is no point c on [0,1]then we define a new function
[tex]T(x)=\frac{f(x)}{g(x)}[/tex] ..."

why are they defining a new function how is that linked with not having a point??
 
Last edited:
  • #7
33,477
5,167
Now it starts to make sense. The conclusion of the statement you're trying to prove is "for some number c in [0, 1], g(c) = 0.

The proof you are looking at, or the suggestion for a start is a proof by contradiction, for which the hypothesis part of your statement is assumed to be true, but the conclusion is assumed to be false. In other words, the conclusion is now "for every number c in [0, 1], g(c) is not 0."

The function T is a quotient, and to find T'(x), you need the quotient rule. As I mentioned in an earlier post, this problem seemed to have something to do with the quotient rule, but I didn't at the time know what it was, since you hadn't presented this new information.

There are certain conditions that have to be met before you can use the quotient rule. If you don't know what they are, look at how the quotient rule is defined.
 
  • #8
1,395
0
here is the full prove :
suppose there is no point c on [0,1] so g(c)=0 then we define a new function
[tex]T(x)=\frac{f(x)}{g(x)}[/tex]
so because its an elementary function so its differentiable on [0,1] interval.
and because we assumed that g(x) differs zero on [0,1]
[tex]
T'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}
[/tex]

and because we are given
[tex]
f'g(x)\neq f(x)g'(x)
[/tex]
from that equation we can get
[tex]
f'g(x)-f(x)g'(x)\neq0
[/tex]
now because we are given that [tex]g(x)\neq0[/tex]
then for all [0,1] interval [tex]T'(x)\neq0[/tex]

now we look at the ends of the interval
x=0
[tex]T(0)=\frac{f(0)}{g(0)}=0[/tex] its true because we assume that [tex]g(x)\neq0[/tex]
and f(0)=0 are given
the same thing for T(1)=0
so there is a point "c" on T for which T'(c)=0
and that contradicts our assumption that [tex]g(x)\neq0[/tex]



i cant see how it contradicts
in the end we talk about T(x) not g(x)

how its a contradiction??

i see that we proved that on one hand we cant have extreme point on T(x)
but on the other hand we have to because of rolls theorem

but its a contradiction about T(x) not g(x)

??
 
Last edited:
  • #9
33,477
5,167
Go back and look at your first post. What are you (or they) trying to prove in that post?
 
  • #10
1,395
0
we need to prove that there is a point in g(x) for which g(c)=0

thats what i told before

i cant see how its a contradiction
 
  • #11
33,477
5,167
So they assumed that for all numbers c in [0, 1] g(c) != 0, then then set up the function T(x) = f(x)/g(x). The work you showed shows that with the original hypotheses, it must be that g(c) = 0. That's the contradiction.
 
  • #12
1,395
0
thanks :)
 

Related Threads on MVT prove question

  • Last Post
Replies
3
Views
965
  • Last Post
Replies
0
Views
912
  • Last Post
Replies
1
Views
2K
Replies
1
Views
3K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
1K
Top