It is proved that the function f (x) = X-1 / X decreases monotonically on (1, positive infinity)

It is proved that the function f (x) = X-1 / X decreases monotonically on (1, positive infinity)

Let x1, X2 ∈ (1, + ∞), and x1 ＜ X2, so: 1 ＜ x1 ＜ X2, so: x1-1 ＞ 0, x2-1 ＞ 0, x2-x1 ＞ 0, because f (x1) - f (x2) = 1 / (x1-1) - 1 / (x2-1) = (x2-x1) / [(x1-1) (x2-1)] > 0, so: function y = 1 / (x-1) is a monotone decreasing function in the interval (1, + ∞)

The function f (x) is a monotone decreasing function on [0, + ∞), f (x) ≠ 0 and f (2) = 1. The monotonicity of function f (x) = f (x) + 1F (x) on [0,2] is obtained

F ′ (x) = f ′ (x) - F ′ (x) f 2 (x) = f ′ (x) (1 − 1F 2 (x)); ∵ f (x) is a monotone decreasing function on [0, + ∞); ∵ f ′ (x) ＜ 0; if f (2) = 1, X ∈ [0, 2], f (x) ≥ 1, ∵ 1 − 1F 2 (x) ≥ 0; ∵ f ′ (x) ≤ 0; ∵ f (x) monotonically decreases on [0, 2]

Given the function f (x) = the square of X + (A-3) + 4, if f (x) is an even function, find the value of real number a, if f (x) is a decreasing function in the interval (negative infinity, 2)
It's fast to find the value range of real number a

A = 3 and a is less than or equal to 1