Analysis- if f assumes max/min for x in (a,b) prove f'(x) = 0

  • Thread starter sarahr
  • Start date
  • Tags
    Analysis
In summary, Sarah is trying to figure out what to do to prove that a differentiable function is continuous on a given interval. She is stumped and needs help.
  • #1
sarahr
13
0
hello,

i need to prove:
_______________________
if a differentiable function f:(a,b) ----> R (reals) assumes a max or a min at some x element of (a,b), prove that f'(x) = 0. why is this assertion false when [a,b] replaces (a,b)?
_______________________

-I'm stumped at where to start...because if i knew that f was continuous on [a,b] and f(a) = f(b), this would be Rolle's thm...which i think i could prove...

-or if f(a) did not equal f(b), then maybe i could get somewhere by mean value thm... (that is, if f was cont on [a,b])

-BUT i don't know that f is cont on [a,b]...

-the only thing i know given what I have is that since f is differentiable on (a,b) then its derivative f'(x) has the intermediate value property...can this get me anywhere?

-any help on how to get started on this proof would be greatly appreciated! thanks for reading through my lengthy post! :)
 
Physics news on Phys.org
  • #2
I don't understand the question. Are you saying that there is some x such that f(x)<=f(y) for all y in (a,b) (or >=). This does not imply f is constant. Just take f(x)=x^2 on (-1,1). Rather than try to guess what you mean, I'll let you clarify it.
 
  • #3
no, the question does not say that f(x)<=f(y) (or vice versa.) what is typed between the two lines is the exact question from my text. thanks!
 
  • #4
I'm sorry, I thought it meant f'(x)=0 for all x in (a,b), I didn't see x was the element where the max or min occurred. Then you can just use the definition of the derivative. You can show the left limit must be >=0 and the right must be <=0 (or the other way around, depending on if it's a max or a min), and so since these limits must be equal, they must be 0.
 
  • #5
Differentiable is stronger than continuous. (i.e. for a function to be differentiable, it must also be continuous.)

You might consider a nice simple example, like:
[tex]f(x)=x[/tex]
on
[tex](0,1)[/tex]
and
[tex][0,1][/tex]
 
  • #6
oh! this is very clear now. thanks so much!

sarah.
 

Related to Analysis- if f assumes max/min for x in (a,b) prove f'(x) = 0

1. What is the purpose of proving f'(x) = 0 in this scenario?

The purpose of proving f'(x) = 0 is to show that the function f has a critical point at x, meaning that the slope of the function is equal to 0 at that point. This can help us determine whether x is a maximum or minimum point for the function.

2. How does proving f'(x) = 0 prove that x is a maximum or minimum point?

If f'(x) = 0, this means that the slope of the function at x is equal to 0. This indicates that the function is either flat or has a horizontal tangent at x. Since a maximum or minimum point on a function occurs when the slope is equal to 0, proving that f'(x) = 0 at x confirms that x is a maximum or minimum point.

3. What is the significance of the interval (a,b) in this statement?

The interval (a,b) represents a range of values for x, where a < x < b. This means that the statement is specifically referring to a specific range of values for x and not just a single point.

4. Can this statement be applied to any type of function?

Yes, this statement can be applied to any differentiable function. If a function is differentiable, we can use the derivative to determine its critical points.

5. What other information do we need to prove that x is a maximum or minimum point?

In addition to proving that f'(x) = 0, we also need to check the concavity of the function at x. If the function is concave up (f''(x) > 0), then x is a minimum point. If the function is concave down (f''(x) < 0), then x is a maximum point. We can also use the second derivative test to confirm the nature of the critical point.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
798
  • Calculus and Beyond Homework Help
Replies
9
Views
730
  • Calculus and Beyond Homework Help
Replies
4
Views
937
  • Calculus and Beyond Homework Help
Replies
3
Views
694
  • Calculus and Beyond Homework Help
Replies
30
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
754
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
832
  • Calculus and Beyond Homework Help
2
Replies
36
Views
5K
Back
Top