Can the Extreme Value Theorem Apply to Function f on Interval [-3, 5]?

[brizer]

1. The problem: The function f is continuous for -3\leqc\leq5 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\leqc\leq5, such that f(c)\geqf(x) for all on the closed interval -3\leqx\leq5.
(b) There exists c, where -3\leqc\leq5, such that f'(c)=-1
(c) There exists c, where -3\leqc\leq5, such that f(c)=-1
(d) There exists c, where -3\leqc\leq5, such that f'(c)=0
(e) There exists c, where -3\leqc\leq5, such that f(c)=0


2. Homework 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.

 

[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.

 

[HallsofIvy]

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

(b) There exists c, where -3\leqc\leq5, such that f&amp;amp;amp;#039;(c)=-1&amp;amp;lt;br /&amp;amp;gt; Yes, that certainly could be false. In this example, f&amp;amp;amp;#039;(x)= -1 for all x&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; (d) There exists c, where -3\leqc\leq5, such that f&amp;amp;amp;amp;amp;amp;amp;#039;(c)=0&amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; Yes, that certainly could be false. &amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;gt; The other three are, of course, true because of the intermediate value theorem (c and e) and the mean value theorem (a).

 

[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!

 

[carlodelmundo]

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.

It is called the Extreme Value Theorem (EVT).

 

[Dick]

It is called the Extreme Value Theorem (EVT).

Thank you!