It is known that the real number a is greater than zero. The function f (x) = ax (X-2) ^ 2. X belongs to the real number, and the maximum value is 32. Find the value of A

It is known that the real number a is greater than zero. The function f (x) = ax (X-2) ^ 2. X belongs to the real number, and the maximum value is 32. Find the value of A

(1) ∵ f (x) = ax (X-2) 2 = ax3-4ax2 + 4ax, ∵ f ′ (x) = 3ax2-8ax + 4a. From F ′ (x) = 0, we get 3ax2-8ax + 4A = 0. ∵ a ≠ 0, ∵ 3x2-8x + 4 = 0
Given the real number a > 0, the function f (x) = ax (X-2) 2 (x ∈ R) has a maximum of 32. (1) find the value of real number a; (2) find the monotone interval of function f (x)
(1) ∵ f (x) = ax (X-2) 2 = ax3-4ax2 + 4ax, ∵ f ′ (x) = 3ax2-8ax + 4a. From F ′ (x) = 0, we get 3ax2-8ax + 4A = 0. ∵ a ≠ 0, ∵ 3x2-8x + 4 = 0. The solution is that when x = 2 or x = 23. ∵ a > 0, ∵ x < 23 or x > 2, f ′ (x) > 0; when 23 < x < 2, f ′ (x) < 0. ∵ when x = 23, f (x) has a maximum of 32, that is 827a-169a = 32, ∵ a = 27. (2) ∵ when x < 23 or x > 2, f (x) ? when x < 2, f (x) ? when x ? when x ∵ When ′ (x) > 0, the function f (x) increases monotonically. When 23 < x < 2, f ′ (x) < 0, the function f (x) decreases monotonically. F (x) is an increasing function on (- ∞, 23) and (2, + ∞), and a decreasing function on (23, 2)
Given the function f (x) = x ^ 2 + ax + B. (1) if f (x + 1) = f (1-x) holds for any real number x, find the value of real number a; (2) if f (x) is an even function, find the value of real number a; (3) if f (x) increases on [1, + ∞), find the value range of real number a
(1) We can know that the function image is symmetric with respect to 1, so a / (- 2) = 1 gives a = - 2
(2) Because it is an even function, the axis of symmetry is y-axis, so a = 0
(3) Because f (x) is increasing on [1, + ∞), the axis of symmetry is less than or equal to 1
That is a / (- 2) = - 2
The function f (x) whose domain is r satisfies: for any real number x, y, f (x + y) = f (x) + F (y) holds, and f (x) < 0 holds when x > 0. (1) judge the parity of function f (x) and prove your conclusion; (2) prove that f (x) is a decreasing function; if f (x) always holds f (x) ≤ 6 on [- 3, 3], try to determine the conditions that f (1) should satisfy; (3) prove that f (x) is a decreasing function ）The solution of inequality 1NF (AX2) − f (x) > 1NF (a2x) − f (a), (n is a given natural number, a < 0)
(1) Let f (x + y) = f (x) + F (y) be constant for any x ∈ R, y ∈ R, let x = y = 0, then f (0 + 0) = f (0) + F (0), Let f (0) = 0, let x = - y, then f (x-x) = f (x) + F (- x) = 0. For any x, f (- x) = - f (x) is an odd function
If the maximum value of function f (x) = loga (x) in the interval [3,5] is greater than the minimum value by 1, then a=___
The maximum value of F (x) = logax in the interval [3,5] is 1 larger than the minimum value
When 0 ＜ a ＜ 1, f (3) - f (5) = 1, loga3 - loga5 = 1, the solution is a = 3 / 5
When a > 1, f (5) - f (3) = 1, loga5 - loga3 = 1, the solution is a = 5 / 3
If the function f (x) = MX / (4x-3) (x ≠ 3 / 4) has f [f (x)] = x in the domain of definition, then M is a real number=____
F[f(x)]=m[(mx）/（4x-3）]÷[4(mx）/（4x-3）-3]
=m^2x/(4mx-12x+9)=x
m^2/(4mx-12x+9)=1
(4m-12)*x+(9-m^2)=0
It holds for any X
If M ≠ 3, then the left is a function of degree about X, which cannot be equal to 0
So 4mx-12x = 0, and 9-m ^ 2 = 0
Then M = 3
Given that the function f (x) = loga (x + 1) (a > 0, a ≠ 1) in the interval [2,8], the maximum value is 1 / 2 larger than the minimum value, find the value of A
a> 1:00
F (x) = loga (x + 1) increases monotonically in the interval [2,8]
Max min = f (8) - f (2)
=loga9-loga3
=loga3
=1/2
A=9
A
The first floor is right
Let the domain of definition of function f (x) be r. there are three propositions as follows: 1) if there is a constant m such that f (x) ≤ m for any x ∈ R, then M is the maximum of function f (x); 2) if there is x0 ∈ r such that f (x) < f (x0) for any x ∈ R and X ≠ x0, then f (x0) is the maximum of function f (x); 3) if there is x0 ∈ r such that f (x) ≤ f (x0) for any x ∈ R ）In these propositions, the number of true propositions is ()
A. 0B. 1C. 2D. 3
① Wrong. Reason: m is not necessarily a function value, maybe "=" cannot be obtained. Because the definition of the maximum value of a function is that if there is a function value greater than all other function values, then this function value is the maximum value of the function. Therefore, select C
If the function y = loga (X & # 178; - ax + 1) has a minimum value, the value range of a is obtained
x^2-ax+1=(x-a/2)^2+1-(a/2)^2
X = A / 2 function y = loga (x ^ 2-ax + 1) has minimum value
y=loga[1-(a/2)^2}
1-(a/2)^2>0 ,4-a^2>0 ,a>2
If the constant T is known to be a negative real number, then the domain of the function f (x) = 12t2 − TX − X2 is______ .
From the meaning of the question, we get that 12t2-tx-x2 ≥ 0, that is - x2 TX + 12t2 ≥ 0, that is - (x-3t) (x + 4T) ≥ 0. Because T < 0, the solution is 3T ≤ x ≤ - 4T. So the definition domain of function f (x) = 12t2 − TX − X2 is [3T, - 4T]. So the answer is: [3T, - 4T]