Proving e^x ≥ 1 + x for All x ∈ [0,∞)

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SUMMARY

The inequality e^x ≥ 1 + x holds true for all x in the interval [0, ∞). To prove this, define the function f(x) = e^x - x - 1. By calculating the first derivative, f'(x) = e^x - 1, we can analyze the behavior of f(x). The first derivative test indicates that f'(x) is non-negative for x ≥ 0, confirming that f(x) is always greater than or equal to zero in this interval.

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  • Understanding of calculus, specifically derivatives
  • Familiarity with exponential functions
  • Knowledge of the first derivative test
  • Basic concepts of function behavior and intervals
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  • Learn about the first derivative test in detail
  • Explore the concept of limits and continuity in calculus
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nuuc
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how would i prove that e^x >= 1 + x for all x in [0,inf)?
 
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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.
 

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