Can f(x) ever be greater than e^x?

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    Calculus Calculus 1
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Discussion Overview

The discussion revolves around the question of whether a function f(x) can be greater than or equal to the exponential function e^x for all x ≥ 0. The conversation includes aspects of calculus, particularly focusing on derivatives and the behavior of functions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant initially claims that f(x) > e^x for every x ≥ 0, based on the condition f(0) = 1 and f'(x) ≥ f(x).
  • Another participant challenges this claim by pointing out that f(0) = e^0, suggesting that the initial assertion is incorrect.
  • A later post corrects the initial claim to f(x) ≥ e^x for every x ≥ 0, indicating a shift in the argument.
  • Participants discuss the requirement to show work in the context of homework policy, emphasizing the importance of understanding the concepts involved.
  • One participant suggests using the Mean Value Theorem as a potential approach to the problem, indicating that they have access to basic theorems from their calculus course.
  • Another participant proposes considering the case where f(x) = e^x, suggesting that proving equality could lead to insights about the inequality.
  • There is a suggestion that if equality holds on the interval, then the theorem may be true, referencing the properties of the exponential function's derivative.

Areas of Agreement / Disagreement

Participants express differing views on the initial claim regarding the relationship between f(x) and e^x. There is no consensus on whether f(x) can be greater than e^x, and the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants reference various mathematical concepts and theorems, but there are limitations in the discussion regarding the assumptions made about f(x) and the specific conditions under which the claims are evaluated.

shgidi
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we have f(0)=1, f '(x)>=f(x)

we shall prove f(x)>e^x for every x>=0

thanx to the solvers
 
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what? that's not true f(0) = e^0
 
sorry, i had a smal mistake:

we have f(0)=1, f '(x)>=f(x)

we shall prove f(x)>=e^x for every x>=0

thanx to the solvers
 
Ok. Welcome to Physicsforums shgidi, I hope you enjoy it here =]

Please read our homework policy (there's a sticky at the top of this sections page), you must post your homework in the homework section! Also, we require you must show us any work that you have attempted (and yes you must attempt whatever you know) because otherwise its not helping you, its just doing your homework!

It is only because I am so VERY nice that I will give you a hint :P
In the interval we are interested in, x \geq 0, the function is always increasing because f(0) = 1, so f'(0) is great than 1, and they both just keep increasing. Since they are both positive, divide by f(x) on both sides. Now take an integral of that inequality from 0 to t. I basically did it for you :(
 
OK, I promise for the next time to look and find the homework section.
I've posted my question here, because I saw here many questions in the same style, so I thought it would fit.

anyway, 10x for your solution, but it doesn't quiet help me, because Integrals is not part of the material of my current course.
 
shgidi said:
OK, I promise for the next time to look and find the homework section.
I've posted my question here, because I saw here many questions in the same style, so I thought it would fit.

anyway, 10x for your solution, but it doesn't quiet help me, because Integrals is not part of the material of my current course.
And that's one of the reasons we ask that people show what they have tried: we have no idea what concepts you have available to do this.

You might try the Mean Value Theorem. I hope you know what that is!
 
Ok, Ill try again... this is a question from a test I had. The course is calculus 1, and I have for use all of the basic theorems, including mean value, lagrange etc..

I tried some variations of those theorems, but I still have no solution.
Ill be REALLY glad to see a solution.

thanks.
 
Are you able to treat the case = instead of >= ?

Do that and then try to think about > .
 
epenguin said:
Are you able to treat the case = instead of >= ?

Do that and then try to think about > .

you don't need to prove the >, if = is satisfied on the interval then the theorem is true. how is this not simply true definition since e^x's derivative is itself hence g(x)=e^x=g'(x)
 

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