（ln(1+x)）^2-X^2/(1+x)

（ln(1+x)）^2-X^2/(1+x)

Let t = x + 1
f(t)=(lnt)^2-(t-1)^2/t=(lnt)^2-(t-2+1/t)
Then f '(T) = 2 (LNT) / T-1 + 1 / T ^ 2 = 1 / T * [2lnt-t + 1 / T]
g(t)=2lnt-t+1/t
g'(t)=2/t-1-1/t^2=-(1/t-1)^2

[solution inequality] ln [(1 / 2) ^ X-1] < 1

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What's the difference between Cauchy's mean value theorem and Lagrange's
I feel like I just changed the right angle equation into the parameter equation. Everything else is the same

Cauchy mean value theorem is also called Cauchy mean value theorem. Let f (x), G (x) be continuous in [a, b], G '(x) ≠ 0 (x ∈ (a, b)), then there is at least one point, ξ ∈ (a, b), such that f' (ξ) / g '(ξ) = [f (b) - f (a)] / [g (b) - G (a)]
Edit the geometric meaning of this paragraph
If u = f (x), v = g (x), this form can be understood as a parametric equation, and [f (a) - f (b)] / [g (a) - G (b)] is the end slope of the connecting parametric curve. F '(ξ) / g' (ξ) represents the tangent slope at a point on the curve. Under the condition of the theorem, it can be understood as follows: there is at least one point on the curve represented by the parametric equation, and its tangent is parallel to the chord at the two end points, But Cauchy's mean value theorem is not only applicable to the curve represented by y = f (x), but also applicable to the curve represented by parameter equation. When G (x) = x in Cauchy's mean value theorem, Cauchy's mean value theorem is Lagrange's mean value theorem