The value range of the third power of the function y = (root x) + X is

The value range of the third power of the function y = (root x) + X is

First of all, the field 0 to positive infinity x cannot take negative numbers
So the range is 0 to positive infinity

(1) The value range of the root X-1 power of function y = 1 / 3

1) Radical X-1 ≥ 0
therefore
The power 0 of Y ≤ 1 / 3 = 1
Namely
The range of 0 is (0,1]
2）2x-x²=-（x²-2x）=-(x-1)²＋1≤1
therefore
The first power of Y ≥ 1 / 2 = 1 / 2
Namely
The range is [1 / 2, + ∞)

The value range of 1 / 1 of the function y = (2 to the power of X + 1) is

（0,1）

The value range of the function y = 3 (x power) / 3 (x power) + 1 is

Let t = 3 ^ x > 0
Then y = t / (T + 1) = (T + 1-1) / (T + 1) = 1-1 / (T + 1)
Because t + 1 > 0
So 0

Given the function f (x) = 1 - (2 to the x power + 1) of 2, calculate the range

f(x) = 1 - 2/(2^x+1)
0＜2^x＜+∞
1＜2^x+1＜+∞
0＜2/(2^x+1) ＜2
-1＜1 - 2/(2^x+1) ＜1
Range (- 1,1)

1. Judge the function f (x) = the x power of a + the x power of 1 / 1 a (a > 1). ① find the parity of the function; ② find the range of FX; ③ prove that f (x) is in the (- infinity, + infinity) is an increasing function

①f(x)=(a^x-1)/(a^x+1)
f(-x)=(a^-x-1)/(a^-x+1)
=(1-a^x)/(1+a^x)= -f(x)
So f (x) is an odd function
②f(x)=(a^x-1)/(a^x+1)
=1-(2/(a^x+1))
a^x＞0,
1-(2/(a^x+1))＞-1
So the range of F (x) is f (x) > - 1
Because a > 1
So a ^ x + 1 increases
SO 2 / (a ^ x + 1) decreases
Get 1 - (2 / (a ^ x + 1)) increment
Because x ∈ R
In other words, it is an increasing function on X ∈ (- infinity, + infinity)

Function f (x) = {3 X-1 power-2 (x ≤ 1), 3's 1-x power-2 (x > 1)} range

x≤1
x-1≤0
0<3^(x-1)≤1
-2x>1
1-x<0
0<3^(1-x)<1
-2 so the range (- 2,1]

The inverse function of F (x) = (10x power-10-x power) / (10x power + 10-x power)

Let t = 10 ˊ
Then f (T) = (t-1 / T) / (T + 1 / T) = (t-1) / (T + 1) = (T + 1-2) / (T + 1) = 1-2 / (T + 1)
Then f (T) = 1-2 / (T + 1) = y
Then 1-2 / (y + 1) = t
It can be replaced by T = 10 ˊ

How to find the value range of the first power of 2-x of y = 3

0

0

y=2^x -1/2^x +1
=1 - 2 / 2 ^ x + 1 (separate variable)
Now the range of 2 ^ x + 1 is [1 positive infinity]
So the range of 2 / 2 ^ x + 1 is (0 1]
Then the range of y = 1-2 / 2 ^ x + 1 is [0 1]
So the value range is [0 1]