EVT and Fermats to prove f'(c)=0

[NWeid1]

Homework Statement

If f is differentiable on the interval [a,b] and f'(a)<0<f'(b), prove that there is a c with a<c<b for which f'(c)=0.

Homework Equations

The Attempt at a Solution

Well, I first tried to use IVT but I was having a hard time to I talked to my prof. and he said to use extreme value theorem and fermats theorem. So, by EVT, I know there will be an aboslute maximum and absolute minimum on [a,b]. By f'(a)<0<f'(b) I know that it will be decreasing and increasing and therefore will have a local miniimum on (a,b). Fermats theorem then says that if f(c) is a local extremum, then c must be a critical number of f. Which means the function WILL have a c to make f'(c)=0. I know this intuitively but I'm having a hard time rigorously proving it.

 

[NWeid1]

no one?

 

[Mark44]

Homework Statement

If f is differentiable on the interval [a,b] and f'(a)<0<f'(b), prove that there is a c with a<c<b for which f'(c)=0.

Homework Equations

The Attempt at a Solution

Well, I first tried to use IVT but I was having a hard time to I talked to my prof. and he said to use extreme value theorem and fermats theorem. So, by EVT, I know there will be an aboslute maximum and absolute minimum on [a,b]. By f'(a)<0<f'(b) I know that it will be decreasing and increasing and therefore will have a local miniimum on (a,b).

Please state the foregoing without "it". What you have written is very unclear. "It" could refer to f'(a), 0, or f'(b). I suspect that none of these is the antecedent.

Fermats theorem then says that if f(c) is a local extremum, then c must be a critical number of f. Which means the function WILL have a c to make f'(c)=0. I know this intuitively but I'm having a hard time rigorously proving it.

 

[NWeid1]

Well, I first tried to use IVT but I was having a hard time to I talked to my prof. and he said to use extreme value theorem and fermats theorem. So, by EVT, I know that f will have an aboslute maximum and absolute minimum on [a,b]. By f'(a)<0<f'(b) I know that f will be decreasing and increasing and therefore will have a local miniimum on (a,b). Fermats theorem then says that if f(c) is a local extremum, then c must be a critical number of f. Which means the function WILL have a c to make f'(c)=0. I know this intuitively but I'm having a hard time rigorously proving it.

 

[Mark44]

Can you say this more precisely?

By f'(a)<0<f'(b) I know that f will be decreasing and increasing

Where will f be decreasing?
Where will f be increasing?

If you haven't guessed, I'm trying to get you to make clear, unambiguous statements.

 

[NWeid1]

f will be decreasing on (a,o) and increasing on (0,b)

 

[gb7nash]

f will be decreasing on (a,o) and increasing on (0,b)

Not necessarily. We only know that the slope of f(x) at a is negative, and the slope of f(x) at b is positive. We don't know the exact intervals f is increasing or decreasing.

f'(a)<0<f'(b)

 

[Mark44]

f will be decreasing on (a,o) and increasing on (0,b)

In addition to what gb7nash said, you are assuming here that a < 0 and b > 0. All you have is the interval [a, b]. There is no indication of whether 0 is in the interval, to the left of it, or to the right of it. It's just some arbitrary interval.