Derivative Proof: Prove f'(x)=lim x→0 (f(x+h)-f(x-h))/2h

In summary, the given problem states that if f' is continuous, then the limit of the difference quotient as x approaches 0 is equal to f'(x). The solution involves manipulating the difference quotient by replacing h with -h and taking the limit as h approaches 0, then using the definition of continuity and the limit equivalence of f(a+h) to solve for f'(x). It is also noted that the given problem may be solved with L'Hospital's theorem.
  • #1
v0id19
49
0

Homework Statement


if f' is continuous, show that:
[tex]\mathop{\lim}\limits_{x \to 0}(\frac{f(x+h)-f(x-h)}{2h})=f'(x)[/tex]
be sure to explain why f' must be continuous

Homework Equations


not really any equations, this is for AP Calc BC and we've just done L'Hospital's theorem and the derivatives/integrals of logs and inverse trig functions.

The Attempt at a Solution


I know that as x-->0 it becomes [tex]\frac{f(h)-f(-h)}{2h}[/tex] . I thought about proving that the difference quotient can be manipulated into the above formula, but haven't had any success.

Any pointers?
 
Physics news on Phys.org
  • #2
Are you certain that it's supposed to be as x -> 0 and not h? Aside from that, my only suggestion is to add and subtract f(x)/2h and apply some limit lemmas.
 
  • #3
yeah it says x-->0
what is a 'limit lemma'? o_O
 
  • #4
I think jgens is right, it's likely h -> 0.

Hmmm, I'm surprised this is Calc BC. I think I've seen this problem in Spivak before.

The trick is to consider the difference quotient and replace h with -h, and then take the limit as h -> 0. Then you should be able to work with the expression given.

EDIT: Ok, this is a slightly different problem. Hmmm, see if you can make use of the definition of continuity, and the equivalence of the statements lim x -> a f(x) and lim h -> 0 f(a+h).
 
Last edited:
  • #5
I think this approach may work:

Derivative of a function: y = f(x) than y - dy = f(x - h), hence dy/dx = lim h -> 0 (f(x) - f(x - h))/h

Given (f(x + h) - f(x - h))/2h = (f(x +h) - f(x) + f(x) - f(x - h))/2h. Take the limit as h -> 0 and I think that should yield a solution.

A limit lemma would be lim x-> c kf(x) = klim x -> c f(x) where k is a constant.
 
  • #6
thanks i'll give that a try--my only question is how do i get it from x-->0 to h-->0?
 
  • #7
So it turns out my math teacher actually did write the question wrong... it really is h-->0. it's solved with a simple l'hospital. thanks guys
 

Related to Derivative Proof: Prove f'(x)=lim x→0 (f(x+h)-f(x-h))/2h

What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is also known as the slope of the tangent line to the function at that point.

What is the purpose of a derivative proof?

The purpose of a derivative proof is to show that the formula for finding the derivative of a function, in this case f'(x)=lim x→0 (f(x+h)-f(x-h))/2h, is valid and can be used to calculate the derivative at any point.

How do you prove f'(x)=lim x→0 (f(x+h)-f(x-h))/2h?

The proof involves using the definition of a derivative, which is the limit of the difference quotient as h approaches 0. By manipulating the difference quotient and using algebraic techniques, the limit can be simplified to the given formula for f'(x).

Can the derivative proof be applied to all functions?

Yes, the derivative proof can be applied to all functions, as long as they are continuous and differentiable at the point of interest. However, some functions may require more complex proofs than others.

Why is the limit as h approaches 0 important in the derivative proof?

The limit as h approaches 0 is important because it represents the instantaneous rate of change of the function at a specific point. It allows us to calculate the slope of the tangent line to the function at that point, which is the definition of a derivative.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
282
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
933
  • Calculus and Beyond Homework Help
Replies
6
Views
726
  • Calculus and Beyond Homework Help
Replies
5
Views
310
  • Calculus and Beyond Homework Help
Replies
16
Views
2K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
6K
Back
Top