Exponential Function Homework: Showing 0 ≤ e^x−1−x

[mtayab1994]

Homework Statement

k is a real number and $$f_{k}(t)=e^{t}-1-t-k\frac{x^{2}}{2}$$

1- Show that : $$(\forall x\epsilon\mathbb{R}):0\leq e^{x}-1-x$$

2- Show that : $$(\forall x>0)(\exists k\epsilon\mathbb{R}^{+})(\exists d\epsilon[0,x]):f(x)=f''(d)=0$$

3-Conclude that $$(\forall x\epsilon\mathbb{R}):|e^{x}-1-x|\leq\frac{x^{2}}{2}e^{|x|}$$

The Attempt at a Solution

For number 1 i said f(x) =e^x-1-x and f'(x)=e^x-1 so if x >0 than f in increasing and if x<0 f is a decreasing function so f(x)>f(0) in both cases so therefore: that 0<e^x-1-x.
Number 2 I don't know what to do can someone help please??

 

[mfb]

f_k(x)=0 for a specific x is sufficient to fix k, so you can show the existence of such a k first (using (1) and x^2/2 > 0).
You can calculate f'' and show the existence of such a d in a similar way afterwards.