Question

Calculus help!!

Use MVT to prove that \(\forall x \in\mathbb R\) , \(e^x>1+x\) \(x\neq 0\)
Please, help.

Answer 1

I would help if I could, but that looks like college homework... :) I have studied that... so sorry..

Answer 2

haven't

Answer 3

we let \[f(x)=e^x-(1+x)\]
we have \[f'(x)=e^x-1\] f'(x) depends on the sign of x
if x>0 then f'(x)>0 and the same for the for x<0
using mvt we can have 0<c<x for some \[x\ne0\]
such that \[f(x)=f(0)+f'(c)x\]
we need to show that f'(c)x>0 , f(0)=0

Answer 4

if we can either take x <c<0 or 0<c<x
the result will be the same

Answer 5

mvt states that if f is continues at [a,b] and differentiable (a,b)
then there exist a c in [a,b] such that \[\large f'(c)=\frac{f(b)-f(a)}{b-a}\]
equivalent to \[f(b)=f(a)+f'(c)(b-a)\]

Answer 6

i choose and interval [0,x] to use the MVT and we have all what we need!

Answer 7

since x>0 and f'(x) depends x
you can show that f'(c)x>0 and done!