Proving e^x ≥ 1 + x for All x ∈ [0,∞)

  • Context: Undergrad 
  • Thread starter Thread starter nuuc
  • Start date Start date
  • Tags Tags
    E^x
Click For Summary

Discussion Overview

The discussion revolves around proving the inequality e^x ≥ 1 + x for all x in the interval [0, ∞). Participants explore various approaches to establish this inequality, including the use of derivatives and function definitions.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Homework-related

Main Points Raised

  • One participant suggests defining the function f(x) = e^x - x - 1 to analyze the inequality.
  • Another participant proposes examining the derivative f'(x) to gain insights into the behavior of f(x).
  • There is a suggestion to apply the first derivative test to determine where f(x) is non-negative.
  • It is noted that the inequality holds for x = 0, but further analysis is required for x > 0.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the proof method, and multiple approaches are being discussed without resolution.

Contextual Notes

Some assumptions about the behavior of the function f(x) and its derivative are implied but not explicitly stated. The discussion does not resolve the mathematical steps necessary for a complete proof.

Who May Find This Useful

Readers interested in mathematical proofs, particularly those involving inequalities and calculus, may find this discussion relevant.

nuuc
Messages
2
Reaction score
0
how would i prove that e^x >= 1 + x for all x in [0,inf)?
 
Physics news on Phys.org
where are your thoughts?
hint: expand out to see.
 
Define f(x) = e^x - x - 1. What can f'(x) tell us?
 
morphism said:
Define f(x) = e^x - x - 1. What can f'(x) tell us?
like morphism stated, now what u have to do is just find f'(x)= (e^x-x-1)', what is f'(x) now?,
After that look at what interval the function f(x) is always greater or equal to zero, using the first derivative test,can you do it? and the result surely should be for any x greater than zero. It is obvious that for x=0 the >= sing is valid.
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
10
Views
2K