math question

[computerex]

Can someone prove: (f'(b)+f'(a))/2+f(a) = f(b) given |b-a| = 1 ?

 

[Izack]

Not at 12AM I can't...
Maybe in the morning...

 

[statickid]

why the heck would you want to prove that??

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hahah ok with that out of the way, are you trying to figure something out and are not sure if it can be proven or not,

OR

have you been told that it can be proven->as in an acedemic problem to test your proving ability?

YOUR NOT CHEATING ARE YOU???:lol:

 

[computerex]

http://www.physicsforums.com/showthread.php?t=400938

 

[cjp]

Can someone prove: (f'(b)+f'(a))/2+f(a) = f(b) given |b-a| = 1 ?

Try this example:
f(x) = x^3
a = 3
b = 4

Observe that |b-a| = 1

f(a) = 27
f(b) = 64
f'(x) = 3*x^2
f'(a) = 27
f'(b) = 48
(f'(b)+f'(a))/2+f(a) = (48+27)/2+27 = 64.5

So, for this example, the equation is not true.
So, it is not true in general.
So, it can not be proved.

 

[computerex]

Thank you for providing a useful response cjp. I was thinking the same thing. I was puzzled as to why it worked for the problem in class. If you didn't read the physics forum thread, I'll post the problem here:

The slope of f(x) is twice the x value. Given that f(2) = 3, what is f(3)?

I solved this simply by integrating 2x and using the initial condition to find the function, then plugging 3 to get the answer 8.

The class however did (f'(3)+f'(2))/2+f(2) to get eight as well.
The responder in the physics forum thread says that the statement in question is true for all quadratic. Is there a proof for that?

 

[cjp]

I solved this simply by integrating 2x and using the initial condition to find the function, then plugging 3 to get the answer 8.

How can you integrate if you don't know what the function f(x) is?

The responder in the physics forum thread says that the statement in question is true for all quadratic. Is there a proof for that?

That may be possible. Am I doing your homework?
If it is true for all quadratics, then it shouldn't be too hard to find a proof. Please try to do it yourself first, and then if you fail, explain what you tried, and where you got stuck. Then I'll give you a small hint to get you going again.

 

[computerex]

This is a AP Calculus BC course, we don't do proofs. The AP exam is actually in roughly one and a half hour, so no, this isn't a homework assignment either
http://www.collegeboard.com/student/testing/ap/cal/cal2.html

This was a problem in a sample exam:

The slope of f(x) is twice the x value. Given that f(2) = 3, what is f(3)?

Here is my solution:

f'(x) = 2x [since the slope is twice the x value]
f(x) = integral (2x dx)
= x^2 + c
Use the initial condition f(2) = 3 to solve for c:

4+c=3
c=-1

so the function is f(x) = x^2 - 1
You plug three in it: f(3) = 9-1 to get 8.

I got the right answer. The majority of the class got the right answer as well, but they did by taking the average rate of change between x = 3 and x = 2 then adding it to the initial value of 3 to get eight. I thought that this didn't make sense. It's an intuitive uneasiness. So I consulted the physics forum/this forum about the question. Responders in the physics forum stated that generally speaking the method the class used doesn't hold, but it holds for quadratics. I have already tried proving this myself. I am not getting any good results. So I asked whether someone else can provide the proof.

What I came up with makes no sense. Considering the fact that this is a high school elementary Calculus course, and that I haven't taken a class that deals with formally proving things, I don't expect my proofs to be rigorous and thus are just wrong... I explicitly stated the problem and a solution for it so that I don't have people accusing me of having them do my homework. So two irrelevant replies, one slightly useful, and the other accusatory. Brilliant.

 

[cjp]

The slope of f(x) is twice the x value.

OK, so I read over that. So, you are right, you have f'(x) = 2*x, and you can integrate that to get the answer. Sorry.

Now, proving that (f'(b)+f'(a))/2+f(a) = f(b) given |b-a| = 1 holds for all quadratic functions shouldn't be too hard, as I said before.

My first hint:

• What is the generic form of quadratic functions?
• What is the derivative of that generic form?
• Just substitute these into your formula, and substitute b = a+1 or b = a-1.

---------- Post added at 01:18 PM ---------- Previous post was at 01:06 PM ----------

I found that it is only true for all quadratic functions if b-a = 1, but not for b-a = -1, and not for any other value of b-a.

 

[computerex]

f(x) = ax^2 + bx + c
f'(x) = 2ax + b

I will try much luck later, it is 7:21 am and I have to get to school by 7:45