It is proved that when x ≠ 0, ex ＞ 1 + X, f (x) = ex-1-x

It is proved that when x ≠ 0, ex ＞ 1 + X, f (x) = ex-1-x

This proposition is wrong. It holds only when x > 0
Let f (x) = e ^ x-1-x
F '(x) = e ^ X-1 > 0 (when x > 0)
Therefore, f (x) increases singly on (0, + ∞)
f(0)=0
So there is always e ^ x ＞ 1 + X on (0, + ∞)

Prove whether the Rolle theorem f (x) = IX |, [- 1,1] is satisfied

In fact, in this interval, there is derivative or equal to 1
Or equal to - 1

Proof of Lagrange mean value theorem
Please be more detailed

Constructing auxiliary function
It's easy to know
1.g(a)=g(b)=0;
2. G (x) is continuous in [a, b];
3. G (x) is derivable in (a, b)
According to Rolle's theorem, there exists ξ in (a, b) such that G & # 39; (ξ) = 0, that is, there exists
F & # 39; (ξ) (B-A) = f (b) - f (a) & nbsp;, it is proved