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nuuc
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how would i prove that e^x >= 1 + x for all x in [0,inf)?
like morphism stated, now what u have to do is just find f'(x)= (e^x-x-1)', what is f'(x) now?,morphism said:Define f(x) = e^x - x - 1. What can f'(x) tell us?
e^x is a mathematical constant, also known as Euler's number, which is approximately equal to 2.71828. It is used in the proof as it is the base of the natural logarithm, ln, which is a fundamental function in calculus.
Proving e^x ≥ 1 + x for all x ∈ [0,∞) is important because it is a fundamental inequality in mathematics and has many applications in fields such as calculus, physics, and economics. It also helps to understand the behavior of exponential functions.
The proof uses mathematical induction, which is a technique for proving a statement for all natural numbers. It starts by showing that the statement is true for the first natural number, and then assuming it is true for n, it is proved for n+1. This process is repeated until the statement is proved for all natural numbers.
Yes, the proof can be extended to other intervals or values of x. However, it may require additional steps or adjustments to the initial proof. For example, if we want to prove e^x ≥ 1 + x for all x ∈ [-1, ∞), we would need to consider a different base case and adjust the inductive step accordingly.
Yes, there are many real-life applications of this proof. For example, in economics, the proof can be used to show that the continuous compounding of interest will always result in a higher return compared to simple interest. In physics, it is used to model exponential growth and decay in natural phenomena. It is also used in computer science and engineering to analyze algorithms and the complexity of problems.