# Homework Help: F(c) on a closed interval

1. Aug 10, 2008

### brizer

1. The problem: The function f is continuous for -3$$\leq$$c$$\leq$$5 and differentiable for -3<x<5. If f(-3)=6 and f(5)=-2, which of the following could be false?
(a) there exists c, where -3$$\leq$$c$$\leq$$5, such that f(c)$$\geq$$f(x) for all on the closed interval -3$$\leq$$x$$\leq$$5.
(b) There exists c, where -3$$\leq$$c$$\leq$$5, such that f'(c)=-1
(c) There exists c, where -3$$\leq$$c$$\leq$$5, such that f(c)=-1
(d) There exists c, where -3$$\leq$$c$$\leq$$5, such that f'(c)=0
(e) There exists c, where -3$$\leq$$c$$\leq$$5, such that f(c)=0

2. Relevant equations: IVT, mean value theorem

3. The attempt at a solution: If f(-3)=6 and f(5)=-2, then folloing the IVT, there is a c such that f(c)=-1 and f(c)=0. Following the mean value theorem, (-2-6)/(6--2)=-8/8=-1=f'(c). That leaves options (a) and (d). The only other theorem I can think of relevant to closed intervals of continuous, differentiable functions is Rolle's theorem which is not relevant. I feel like option (a) must be a theorem I can't remember, but I couldn't find it in my book, so I'm not sure.

2. Aug 10, 2008

### Dick

A continuous function on a closed and bounded interval has a absolute maximum and minimum. This is the absolute max case. I'm don't think this theorem has a specific name.

3. Aug 10, 2008

### HallsofIvy

One function that satisfies the given conditions is f(x)= 3- x

(b) There exists c, where -3[tex]\leq[tex]c[tex]\leq[tex]5, such that f'(c)=-1
Yes, that certainly could be false. In this example, f'(x)= -1 for all x

(d) There exists c, where -3[tex]\leq[tex]c[tex]\leq[tex]5, such that f'(c)=0
Yes, that certainly could be false.

The other three are, of course, true because of the intermediate value theorem (c and e) and the mean value theorem (a).

4. Aug 10, 2008

### brizer

I understand, there must be a maximum and minimum on a closed interval, so there must be some c that makes f(c) greater than all f(x) for real numbers. (a), then, must also be true. Thank you!

5. Aug 10, 2008

### carlodelmundo

It is called the Extreme Value Theorem (EVT).

6. Aug 10, 2008

Thank you!