Rolles Theorem/ Mean Value Theorem + First Derivative Test

In summary, the conditions for f'(c) = 0 for a < c < b are that f(x) must be continuous on [a,b] and differentiable on (a,b). This does not necessarily mean that f(a) = f(b) or that f has a relative extremum or point of inflection at x = c. It only guarantees the existence of a c such that f'(c) = 0.
  • #1
carlodelmundo
133
0

Homework Statement



Suppose that f(x) is a twice-differentiable function defined on the closed interval [a,b]. If f'(c) = 0 for a < c < b, which of the following must be true?

I. f(a) = f(b)

II. f has a relative extremum at x = c.

III. f has a point of inflection at x = c.

Homework Equations



Rolles Theorem states that there is a c such that f'(c) = 0 between [a,b] if f(x) is continuous on [a,b] and differentiable on (a,b).


The Attempt at a Solution



By Rolles Theorem, statement "I" must be correct. (That is, the endpoints are equal to each other). Is this correct?

I also put statement "II" as correct because since f'(c) = 0 for a < c < b... c must be a critical point or relative extremum. Question though: does this mean there are other points between a and b where f'(c) = 0 ?

For statement "III", I said this was incorrect. It's twice differentiable yes, but we don't know if f(x) has a point of inflection at x = c.

Thanks!
 
Physics news on Phys.org
  • #2
You said: "if f'(c) = 0 for a < c < b."
Did you mean for all c between a and b, or for some c between a and b?

Because in the first case, for example I is true (although I think it follows from the principal theorem of analysis) while in the second case it is not (you can easily think of a counter example for some specific a, b, c and function f).
 
  • #3
carlodelmundo said:
Rolles Theorem states that there is a c such that f'(c) = 0 between [a,b] if f(x) is continuous on [a,b] and differentiable on (a,b).

I think there is one more condition here that f(a) = f(b)
Otherwise there is the counter example of f(x)=ex
 
  • #4
Thank you both. aniketp, you're right. f(a) = f(b) is the last condition... I overlooked it... and since a < b, they can't be equal to each other.
 
  • #5
Are you saying that if a < b then f(a) cannot be equal to f(b) ?
 
  • #6
carlodelmundo said:

Homework Statement



Suppose that f(x) is a twice-differentiable function defined on the closed interval [a,b]. If f'(c) = 0 for a < c < b, which of the following must be true?

I. f(a) = f(b)

II. f has a relative extremum at x = c.

III. f has a point of inflection at x = c.

Homework Equations



Rolles Theorem states that there is a c such that f'(c) = 0 between [a,b] if f(x) is continuous on [a,b] and differentiable on (a,b).


The Attempt at a Solution



By Rolles Theorem, statement "I" must be correct. (That is, the endpoints are equal to each other). Is this correct?

I also put statement "II" as correct because since f'(c) = 0 for a < c < b... c must be a critical point or relative extremum. Question though: does this mean there are other points between a and b where f'(c) = 0 ?

For statement "III", I said this was incorrect. It's twice differentiable yes, but we don't know if f(x) has a point of inflection at x = c.

Thanks!
As for "I", you are confusing the theorem with its converse. Rolle's theorem says that if f(a)= f(b) (and other conditions) then there exist c such that f'(c)= 0. Saying that some f'(c)= 0 does NOT means that f(a) must equal f(b).

Consider [itex]f(x)= x^2[/itex], a= -1, b= 2, c= 0. f'(0)= 0 but [itex]f(-1)\ne f(2)[/itex]. Or [itex]g(x)= x^3[/itex], a= -1, b= 2. Again g'(0)= 0 but [itex]g(-1)\ne g(2)[/itex].

f has a relative extremum at 0 but not an inflection point. g has an inflection point at 0 but not a relative extremum. I would say none of those are necessarily true.
 
  • #7
Thank you all! And yes... I did confuse Rolle's Theorem. Thanks for the heads up
 

1. What is Rolle's Theorem?

Rolle's Theorem is a mathematical theorem that states that if a function is continuous on a closed interval and differentiable on the open interval, and the values of the function at the endpoints of the interval are equal, then there exists at least one point in the interval where the derivative of the function is equal to zero.

2. What is the Mean Value Theorem?

The Mean Value Theorem is a mathematical theorem that states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point in the interval where the derivative of the function is equal to the average rate of change of the function over the interval.

3. How are Rolle's Theorem and the Mean Value Theorem related?

Rolle's Theorem is a special case of the Mean Value Theorem where the average rate of change of the function over the interval is equal to zero. In other words, the Mean Value Theorem is a more general version of Rolle's Theorem.

4. What is the First Derivative Test?

The First Derivative Test is a method used to determine the relative extrema (maximum and minimum) of a function. It involves evaluating the sign of the derivative at critical points (where the derivative is equal to zero or undefined) to determine whether the function is increasing or decreasing at those points.

5. How can I use the First Derivative Test to find the relative extrema of a function?

To use the First Derivative Test, you must first find the critical points of the function by setting the derivative equal to zero and solving for x. Then, you can evaluate the sign of the derivative at these points. If the derivative is positive, the function is increasing and the critical point is a relative minimum. If the derivative is negative, the function is decreasing and the critical point is a relative maximum. If the derivative is zero, the critical point could be a maximum, minimum, or point of inflection. Further analysis is needed to determine which case it is.

Similar threads

Replies
1
Views
440
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
26
Views
2K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
220
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
167
  • Calculus and Beyond Homework Help
Replies
22
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
343
  • Calculus and Beyond Homework Help
Replies
2
Views
658
Back
Top