how do you prove that e^x^>(x-1) ,for every x that belongs to IR?

Question

NOTE: e^x^ means that x is an exponent to the number e
proof can be done only by using calculus methods!

Answer 1

We shall prove an even stronger statement, that e^x≥x+1 for all x. It will easily follow from that that e^x > x-1 for all x.

Let f(x) = e^x - x - 1. Then f'(x) = e^x - 1. Obviously, f'(x) = 0 ⇔ e^x = 1 ⇔ x = 0. When x>0, f'(x) > 0 and when x < 0, f'(x) < 0. So suppose that there is a point x where f(x) < f(0). Then x is not itself 0, so either x>0 or x<0. If x>0, then by the mean value theorem, ∃c such that 0 0 for all c>0. Conversely, if x<0, then by the mean value theorem, ∃c such that x 0, contradicting the fact that f'(c) < 0 for all c < 0. Thus in either case, we get a contradiction, and thus ∀x, f(x) ≥ f(0) = 0.

Since e^x - x - 1 = f(x) ≥ 0, it follows immediately that e^x ≥ x+1 for all x. And we are done.

Answer 2

I suppose you mean to prove e^x >(x-1)

f(x) = e^x - x + 1
=> f'(x) = e^x - 1

Case 1: x > 0
=> f' (x) > 0 for x > 0
=> f(x) is an increasing function
So, for x > 0 => f(x) > f(0)

Case 2: x < 0
f'(x) = e^x - 1 < 0 for x < 0
=> f(x) is a decreasing function
So, for x < 0 => f(x) > f(0)

Thus, in both cases f(x) > f(0)
=> e^x -x + 1 > e^0 - 0 + 1
=> e^x > x + 1 > x - 1
=> e^x > x - 1

Answer 3

For x >0, apply Taylor formula to f(x)= e^x around x=0:

e^x = 1 + x + (x^2 e^a)/2! where a strictly is between 0 and x.

Since (x^2 e^a)/2! >= 0, you get

e^x >= 1 + x > x - 1, hence e^x > x - 1

For x =< 0, the statement is obvious, since e^x is strictly
positive and (x-1) is strictly negative.
(here >= means less or equal)

Answer 4

i'm assuming that N is the set of organic numbers (non-detrimental integers). Set f(x) = x^(a million/n). Then f(x) = exp[ ln(x^(a million/n)) ] = exp [(a million/n) * ln(x) ]. exp(x) is differentiable everywhere and ln(x) is differentiable at each x>0, so by employing the chain rule, exp [(a million/n) * ln(x) ] is differentiable at each x>0, yet it is in simple terms f(x) in cover, so we've that f is differentiable at each x>0. to coach that (x^a million/n)'=(a million/n)x^((a million/n)-a million), we've that log(x^(a million/n)) = (a million/n) ln(x). Differentiating the two aspects we've that (x^(a million/n))'/(x^(a million/n)) = (a million/n)(a million/x) by employing logarithmic differentiation. final, remedy for (x^(a million/n))' giving us (x^(a million/n))' = (a million/n)(a million/x)(x^(a million/n) = (a million/n)(x^-a million)(x^(a million/n) = (a million/N)(x^((a million/n)-a million)).

Answer 5

ln(x-1) lne^x=x
so e^x>x-1