Find the range of the square of the function y = (the fourth power of X + the square of X + 5) / (the square of X + 1)

Find the range of the square of the function y = (the fourth power of X + the square of X + 5) / (the square of X + 1)

Separation after deployment
y=(x^4+x^2+5)/(x^4+x^2+1)=1+4/(x^4+x^2+1)
Change yuan
Let t = x ^ 2, then y = 1 + 4 / (T ^ 2 + T + 1) (t ∈ R)
Then range: (0,19 / 3]

Find the value range of (x square-4x) power of function y = 1 / 3, X ∈ [0,5)

Y = (1 / 3) ^ (x? - 4x) = (1 / 3) ^ ((X-2) 2x - 4) because the base number a = 1 / 3, let u = (X-2) 2 - 4, y = (1 / 3) ^ u decrease with the increase of U, and because x ∈ [0,5), u ∈ [- 4,5)
So y ∈ (1 / 243,81]

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

x^2+1≥1
Zero

Given the function y = (1 / 3 of (x square + 2x + 5 power)), find its monotone interval and range of values The square of the exponent is 2x x + 1

x^2+2x+5=(x+1)^2+4>=4
Base 1 / 3 greater than or less than 1
So (1 / 3) ^ x is a minus function
x^2+2x+5>=4
So (1 / 3) ^ (x ^ 2 + 2x + 5) 0
So the range (0,81)
x^2+2x+5=(x+1)^2+4
The opening is upward, and the axis of symmetry is x = - 1
So X-1, increasing
Base 1 / 3 greater than or less than 1
So (1 / 3) ^ x is a minus function
So the monotone interval of Y is opposite to the monotone interval of exponent
So monotonically increasing interval (- ∞, - 1)
Monotone decreasing interval (- 1, + ∞)

Find the value range of the function y = x square + X negative square + X + X negative first power

Let t = x + 1 / x, then t > = 2 or T = 0

Find the value range of (x-1 / 2) - power-3.2's X-5 on the interval [0,2]

The value range of y = 4 ^ (x-1 / 2) - 3 * 2 ^ x + 5 on interval [0,2]
[explanation]
Let a = 2 ^ X
Zero

Let the value range of function f (x) = radical sign (the x power of 16-4) be a, and the solution set of inequality LG (x-1) < 1 be B, find the set a and B

Because 4 ^ x > 0, 0 < = 16-4 ^ x < 16, then 0 < = √ (16-4 ^ x) < 4,
So a = [0,4]
From LG (x-1) < 1, 0 is obtained, so B = (1,11)

What is the value range of the function y = the third power of root x plus 3

y=√(x^3+3)
Because the root is greater than or equal to 0
The value range of x ^ 3 + 3 is r
therefore
The range of y = √ (x ^ 3 + 3) is [0, + ∞)

Find the value range of x power of function y = 2 + x power of 2 / 1 - 1

Y = 2 to the x power + 2 / 1 to the x power - 1
=1 - [2 / (x power of 2 + 1)]
Because the x power of 2 is > 0
So: the maximum value of Y tends to be 1-0 = 1
The minimum value of Y tends to be 1-2 = - 1
The value range of the function is: (- 1,1)

Find the range of (1-x) power of function y = (1 / 3)

This problem is very simple. 1 / 3 (1-x) can be written as 1 / 3 - (x-1) can be written as X-1 of 3
X-1 of F (x) = 3 is the right translation of X of F (x) = 3 by one unit length. Therefore, the range is 0 to positive infinity