Let the image of the function f (x) = ax & sup2; + BX + A-3 be symmetric about the y-axis, and its domain of definition is [A-4, a] (a, B ∈ R), and find the range of F (x)

Let the image of the function f (x) = ax & sup2; + BX + A-3 be symmetric about the y-axis, and its domain of definition is [A-4, a] (a, B ∈ R), and find the range of F (x)

F (- x) = f (x) ax & sup2; - BX + A-3 = ax & sup2; + BX + A-3, so B = 0. The domain of definition of even function is symmetric about the origin, so A-4 + a = 0, a = 2
It is known that x belongs to [- 3 / pie, 3 / 2 pie] to find the range of function y = cosx
It is [- π / 3,2 π / 3] generally, when inputting fractions, they are input in the order of numerator, fractional line and denominator
When x = 2 π / 3, the function has the minimum value cos (2 π / 3) = - 1 / 2,
When x = 0, the function has the maximum cos0 = 1,
So the function range is [- 1 / 2,0]
The range of function y = = √ (9-x ^ 2) + cosx is
9-x ^ 2 ≥ 0, - 3 ≥ x ≤ 3
The function y = = √ (9-x ^ 2) + cosx is an increasing function on [- 3,0] and a decreasing function on [0,3]
When x = 0, the maximum value of the function is 4;
When x = + - 3, the function takes the minimum value cos3
The function range is [cos3,4]
Cos3 to 4
Let f (x) = ln (1 + x) - 2x / (x + 2), we prove that when x is greater than 0, f (x) is greater than 0;
Derivation
Given f (x) = Cos2 θ + 2msin θ - 2m-2 (θ∈ R), the maximum value g (m) of F (θ) can be obtained for any m ∈ R
Expand the original formula to f (θ) = - 2 [sin θ - (M / 2)] ^ 2 + (m ^ 2) / 2-2m-1
Because sin θ∈ [- 1,1]
Therefore, classification discussion:
M/2
It is proved that the function f (x) = ln (x + 1) - 2x / x + 1 is an increasing function on (1, positive infinity)
I don't know if you've learned derivative. It's very simple to use derivative. First, we get f '(x) = 1 / (x + 1) - 2 / [(x + 1) ^ 2] = [1 / (x + 1)] × [1-2 / (x + 1)]. Then we prove that f' (x) is constant greater than zero on (1, positive infinity) when x > 1,2 / (x + 1) 0; 1 / (x + 1) > 0, so f '(x) is constant on (1, positive infinity)
I don't know math!!!!!!!!!!!
It is known that the domain of definition of the function f (x) = sin (x + θ) + cos (x + θ) is R. (1) when θ = 0, find the monotone increasing interval of F (x). (2) if θ∈ (0, π), and SiNx is not = 0, f (x) is an even function when θ is a value
(1)
F (x) = radical 2Sin (x + π / 4)
So the monotone increasing interval of F (x) is (- 3 / 4 π, π / 4)
(2)
If f (x) is even function, then f (0) = sin θ + cos θ, take the maximum root sign 2 / 2
θ=π/4
I don't know if it's right
I can only confirm the first question
Knowing the function f (x) = ln (1 + x) / x, it is proved that if x is greater than or equal to 1, then f (x) is less than or equal to LN2
Problem 2 if for any x greater than or equal to 0, f (x) greater than 1 + PX is constant, find P max
f'(x)=[x/(x+1)-ln(x+1)]/x^2=[x-(x+1)ln(x+1)]/(x+1)x^2
Because x ≥ 1, so denominator (x + 1) x ^ 2 > 0, we only need to judge the sign of the numerator;
Let g (x) = x - (x + 1) ln (x + 1), then G '(x) = 1-LN (x + 1) - 1 = - ln (x + 1),
Because x ≥ 1, then ln (x + 1) > 0, so G '(x)
Uncle, help you
x>=1
Zero
It is known that the domain of definition of the function f (x) = sin (x + θ) + cos (x + θ) is r, (1) when θ = 0, find the monotone interval of F (x); (2) if θ ∈ (0, π), and SiNx ≠ 0, when the value of θ is, f (x) is an even function
(1) When θ = 0, f (x) = SiNx + cosx = 2ssin (x + π 4) and 2K π - π 2 ≤ 2K π (2k π - π 2 (2k π - π 2 ≤ 2K π + π 4 ≤ 2K π + π 2 ≤ 2K π (2k (x) = SiNx + cosx = 2ssin (x + π 4) and 2K π - π 2 (2k π - π 2 ≤ 2K (2k - π ~ 2 ≤ 2K - π - π 2, 2K π - π - 2, 2K π-3 π (2k π π-3 π [2K π π π π π\\\\\\\\\\\\\\4], K ∈ Z; (2) f (x) = sin (x)+ θ) If f (x) is an even function, then θ + π 4 = π 2 + K π, that is, there is θ = π 4 + K π, K ∈ Z, if θ ∈ (0, π), and SiNx ≠ 0, then θ = π 4 when k = 0
The function f (x) = ln (1 + x) x is known. (I) it is proved that if x ≥ 1, then & nbsp; f (x) ≤ LN2; (II) if f (x) > 1 + PX holds for any x > 0, the maximum value of P is obtained
(I) the derivative of function f (x) = ln (1 + x) x is f / (x) = X1 + X − ln (1 + x) x2. In [0, + ∞), consider the function g (x) = X1 + X − ln (1 + x). From g / (x) = 1 (1 + x) 2 − 11 + X ≤ 0, we can know that G (x) decreases monotonically. Combined with G (0) = 0, when x > 0, G (x) < 0, so f '(x) < 0