Let m and t be real numbers, f (x) = (MX + T) / (x ^ 2 + 1), and the tangent slope of the image of F (x) at point m (0, f (0)) is 1 1. Find the value of real number M 2. If for any x ∈ [1,2], there is always t such that the inequality f (x) ≤ 2T holds, the value range of real number T is obtained 3. Let the two real number roots of the equation x & # 178; + 2tx-1 = 0 be A.B (a < b). If for any x ∈ [a, b], there always exists x1, X2 ∈ [a, b] such that f (x1) ≤ f (x) ≤ f (x2) is constant. Let g (T) = f (x2) - f (x1). When G (T) = 5, find the value of real number t

Let m and t be real numbers, f (x) = (MX + T) / (x ^ 2 + 1), and the tangent slope of the image of F (x) at point m (0, f (0)) is 1 1. Find the value of real number M 2. If for any x ∈ [1,2], there is always t such that the inequality f (x) ≤ 2T holds, the value range of real number T is obtained 3. Let the two real number roots of the equation x & # 178; + 2tx-1 = 0 be A.B (a < b). If for any x ∈ [a, b], there always exists x1, X2 ∈ [a, b] such that f (x1) ≤ f (x) ≤ f (x2) is constant. Let g (T) = f (x2) - f (x1). When G (T) = 5, find the value of real number t

Let m and t be real numbers, f (x) = (MX + T) / (x2 + 1), and the slope of the tangent of the image of F (x) at M (0, f (0)) is 1
Let m and t be real numbers, f (x) = (MX + T) / (x2 + 1), and the slope of the tangent of the image of F (x) at M (0, f (0)) is 1
(1) Find the value of real number M;
(2) If for any x ∈ [- 1,2], there is always t such that the inequality f (x) ≤ 2T holds, the value range of real number T is obtained
(2) ∵ for any x ∈ [- 1,2], there is always t such that the inequality f (x) ≤ 2T holds
For any x ∈ [- 1,2], there is always t such that the inequality t ≥ X / (2x ^ 2 + 1) holds, that is, t ≥ (x / (2x ^ 2 + 1)) min
I want to know how to get this step

(1) f'(x)=[m(x²+1)-(mx+t)*2x]/(x²+1)²
f'(0)=1
[m(0²+1)-(m*0+t)*2*0]/(0²+1)²=1
m=1
(2) f(x)=(x+t)/(x²+1)
-1=