Given the function f (x) = a ^ X-2 root sign (4-A ^ x) - 1 (a > 0 and a is not equal to 1) (1), find the definition domain and value range of the function (2) When the definition domain is (2, + ∞), f (x) ≥ 0 is always true

Given the function f (x) = a ^ X-2 root sign (4-A ^ x) - 1 (a > 0 and a is not equal to 1) (1), find the definition domain and value range of the function (2) When the definition domain is (2, + ∞), f (x) ≥ 0 is always true

(1) 4 - A ^ x ≥ 0A ^ x ≤ 4 does not hold when 000 ≤ 4 - A ^ x < 40 ≤ T < 2F (x) = 4 - t? - 2T - 1 = - (T + 1) A2 + 4 ∈ (- 5,3] (2) the domain of definition is (2, + ∞)  0 - (2 + 1) 2 + 4 = - 5F (x) ≥ 0

The minimum value of x ^ 4 + 4 under the function y = x ^ 2 + 5 / radical The wrong number is x ^ 2 + 4 under the root sign

Let the root sign (x ^ 2 + 4) = t > = 2, then y = (T ^ 2 + 1) / T = t + 1 / T = t / 4 + 1 / T + 3t / 4 > = 2 radical (T / 4 * 1 / T) + 3t / 4 > = 1 + 1.5 = 2.5
When t / 4 = 1 / T and T = 2, the minimum value is 2.5

Find the minimum value of the sum of the function y = root sign (x ^ 2 + 1) and root sign (x ^ 2-4x + 8)

y=√[(x-0)^2+(0+1)^2]+√[(x-2)^2+(0-2)^2]
So y is the sum of the distances from a point P (x, 0) on the X axis to a (0, - 1) and B (2,2)
Obviously, when APB is in a straight line and P is between AB, there is a minimum
AB is on both sides of the x-axis
The minimum is the length of ab
=√[(0-2)^2+(-1-2)^2]=√13

Find the minimum value of the radical sign x + 1 of the function y = x + 2

-1
Substitution method
Let √ x + 1 = t ≥ 0, then x = T ^ 2-1
The original function becomes y = T ^ 2 + 2t-1 (t ≥ 0)
y=(t+1)^2-2≥1-2=-1
The minimum value of the function is - 1
Monotonicity method
Because Y1 = x (x ∈ R) is an increasing function, y2 = 2 √ x + 1 (x ≥ - 1) is also an increasing function
So y = x + 2 √ x + 1 (x ≥ - 1) is also an increasing function
So the minimum value is y = - 1 when x = - 1

Let the maximum value of the function y = acosx + B (A and B are constants) be 1 and the minimum value be - 7, then the maximum value of acosx + bsinx is () A. 1 B. 4 C. 5 D. 7

∵ the maximum value of the function y = acosx + B (A and B are constants) is 1 and the minimum value is - 7,

It is known that the maximum value of the function y = acosx + B, (a > 0) is 1 and the minimum value is - 3. Try to determine the monotone increasing interval of F (x) = bsin (AX + π / 3)

A> When cosx = 1, the function value is the largest, that is, a + B = 1. When cosx = - 1, the function value is the smallest, that is - A + B = - 3 solution, a = 2, B = - 1, then f (x) = bsin (AX + π / 3) = - sin (2x + π / 3)

Find the maximum and minimum of the function y = acosx + B

Because the value range of cosx is [- 1,1]
When a < 0, the range of acosx is [a, - A] y minimum a + B max - A + B
When a > 0, the range of acosx is [- A, a] y minimum - A + B Max a + B

If the function y = (acosx + bsinx) cosx has a maximum value of 2 and a minimum value of - 1, then the value of real number (AB) 2 is______ .

y=acos2x+bsinx•cosx
=a•1+cos2x
2+1
2b•sin2x
=1
2•
a2+b2cos（2x-φ）+a
2（φ=arctanb
A confirm)
∵1
Two
a2+b2+a
2=2，-1
Two
a2+b2+a
2=-1，
A = 1, B = ± 2
2．
∴（ab）2=8．
So the answer is: 8

It is known that the maximum value of the function y = acosx + B is 1 and the minimum value is - 3. Find the value range of the function y = bcos2x + cosx + a

The value range of cosx is [- 1,1] function y = acosx + B is [(a + b), (- A + b)], so a + B = 1, - A + B = - 3 or a + B = - 3, - A + B = 1. The solution of a = 2, B = - 1 or a = - 2, B = - 1, cos2x = 2 (cosx) 2 - 1, and the range of [0,1] function y = bcos2x + cosx + a becomes y = - [2 (cosx) 2 - 1

Find the sum of the maximum and minimum values of the function f (x) = (x ^ 2 + cosx SiNx + 1) / (x ^ 2 + cosx + 1) (x is a real number)

If f (x) = (x ^ 2 + cosx SiNx + 1) / (x ^ 2 + cosx + 1) (x is a real number) f (x) = 1-sinx / (x ^ 2 + cosx + 1) Let G (x) = - SiNx / (x ^ 2 + cosx + 1), then f (x) = 1 + G (x)