Known function f (x) = x2-1 x，x∈（1，2]， (I) judge the monotonicity of F (x) and prove your conclusion by definition; (II) find the range of F (x)

Known function f (x) = x2-1 x，x∈（1，2]， (I) judge the monotonicity of F (x) and prove your conclusion by definition; (II) find the range of F (x)

(Ⅰ) f (x) is an increasing function on (1,2]. It is proved as follows: let X1 and X2 be any two real numbers on interval (1,2] and X1 < X2, then f (x1) - f (x2) = x12-1x1-x22 + 1x2 = (x1-x2) (x1 + x2) - x2-x1x1x1x2 = (x1-x2) (x1 + x2 + 1x1x2) ? 1 < X1 < x2 ≤ 2 ? X1 + x2 + 1x1

Monotonicity of function f (x) = (1 / 3) ^ (x ^ 2-2x), and find the value range

Let t (x) = x ²- 2x=(x-1) ²- one
F (T) = (1 / 3) ^ t is the subtraction function
When (- ∞, 1], t (x) is the subtractive function, f (T) = (1 / 3) ^ t is the subtractive function, so f (x) is the increasing function
When [1, + ∞), t (x) is an increasing function, f (T) = (1 / 3) ^ t is a decreasing function, so f (x) is a decreasing function
Therefore, the function increases the interval (- ∞, 1) and decreases the interval [1, + ∞)
When x = 1, f (1) = 3
Value range (0,3]

Given that f (x) = 2 ^ X-1 / 2 ^ x + 1. (1) find the value range of the function; (2) Let g (x) = x ^ 2 / 2F (x) judge the parity of the function, and prove it Given that f (x) = 2 ^ X-1 / 2 ^ x + 1. (1) find the value range of the function; (2) Let g (x) = x ^ 2 / 2F (x) judge the parity of the function, and prove it

F (x) = (2 ^ x + 1-2) / (2 ^ x + 1) = 1-2 / (2 ^ x + 1) ∵ 2 ^ x is the increasing function on R, the value range is (0, + 00) ∵ 2 ^ x + 1 is the increasing function on R, the value range is (1, + 00) ∵ 1 / (2 ^ x + 1) is the decreasing function on R, the value range is (0,1) ∵ - 2 / (2 ^ x + 1) is the increasing function on R, and the value range is (- 2,0) ∵ 1-2 / (2 ^ x + 1) is the increasing function on R

A proof of function limit If X - > + ∞ and X - > - ∞, the limits of function f (x) exist and are equal to a, Prove Lim X - > ∞ f (x) = a

For any given positive number, C > 0
Because when X - > + ∞, the limit of F (x) exists and is equal to a,
Then it is known from the definition of limit that for C, there is a positive number X1 > 0, so that when x > x1, there is a constant | f (x) - a | - ∞, the limit of F (x) exists and is also equal to a, so there is a positive number x2 > 0, so that X

Let the function z = arctanx / y, find the total differential DZ

zx=1/(1+(x/y) ²) * 1/y =y/(x ²+ y ²)
zy=1/(1+(x/y) ²) * (-x/y ²) =- x/(x ²+ y ²)
therefore
dz=y/(x ²+ y ²) dx-x/(x ²+ y ²) dy

How to prove that the function y = 1 / X is unbounded on (0,2)

Counter evidence:
Suppose 1 / X is bounded on (0,2), let m > 2
Then | 1 / X | ≤ M
Take x = 1 / (2m)
Then | 1 / X | = 2m > m
Contradicting assumptions,
Therefore, the hypothesis is not tenable
So the function y = 1 / X is unbounded on (0,2)