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how would i prove that e^x >= 1 + x for all x in [0,inf)?
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.
PREREQUISITESStudents of calculus, mathematics educators, and anyone interested in understanding inequalities involving exponential functions.
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?