Let the function f (x) = (x-a) ^ 2lnx and a belong to R (1). If x = e is the extreme point of y = f (x), find a (2) and find the value range of the real number a, so that for any x belonging to (0,3e), it is true that f (x) is less than or equal to 4E ^ 2 I can't understand the answer. The second question is, how do you get a 0

Let the function f (x) = (x-a) ^ 2lnx and a belong to R (1). If x = e is the extreme point of y = f (x), find a (2) and find the value range of the real number a, so that for any x belonging to (0,3e), it is true that f (x) is less than or equal to 4E ^ 2 I can't understand the answer. The second question is, how do you get a 0

(1) The derivative is f '(x) = 2 (x-a) LNX + = (x-a) (2lnx + 1 -),
Because x = e is the extreme point of F (x),
So f '(E) = 0
The solution is a = e or a = 3E
After inspection, it meets the meaning of the title,
So a = e, or a = 3E
(2) ① when 0 < 3a ≤ 1, f (x) ≤ 0 < 4e2 holds for any real number x ∈ (0,3a), that is, 0 < a ≤ 1 / 3
② When 3A > 1, that is, when a > 1 / 3, it is known from ① that when x ∈ (0,1], the inequality is constant, so the maximum value of the function on (1,3a] is studied,
Firstly, there is f (3a) = (3a-a) 2ln3a = 4a2ln3a. This value increases with the increase of a, so it should be
4a2ln3a ≤ 4e2, that is, a2ln3a ≤ E2,
Therefore, the value range of parameters is 0 < a ≤ 1 / 3 or a > 1 / 3 and a2ln3a ≤ E2,

Given f (x) = (x power of 2 - 1) divided by (x power of 2 + 1) (x belongs to the real number set), find the value range of the function

f(x)=(2^x +1 -2)/(2^x +1)
=1-2/(2^x +1)
Let t = 2 ^ x + 1 ∈ (1, + ∞)
y=1-2/t
2/t∈(0,2)
y∈(-1,1)
The value range of F (x) is (- 1,1)
Answer: the value range of F (x) is (- 1,1)

Given that the function f (x) = a + 1 / (x-1 of 2) + is an odd function, find the value of real number a and the value range of F (x)

F (x) is an odd function, f (- x) = - f (x)
f(x)+f(-x)
=a+1/(2^x-1)+a+1/[2^(-x)-1]
=2a+1/(2^x-1)+1/[2^(-x)-1]
={2^x-1+2^(-x)-1+2a(2^x-1)[2^(-x)-1]}/{(2^x-1)*[2^(-x)-1]}
={[2^x+2^(-x)]*(1-2a)+4a-2}/{(2^x-1)*[2^(-x)-1]}
={[2^x+2^(-x)]*(1-2a)-2(1-2a)}/{(2^x-1)*[2^(-x)-1]}
=0
Obviously, 1-2a = 0, a = 1 / 2
f(x)=1/2+1/(2^x-1)
2^x>0
2^x-1>-1
1/(2^x-1)∈(-∞,-1)∪(0,+∞)
Value range: (- ∞, - 1 / 2) ∪ (1 / 2, + ∞)

If the function f (x) = a + 1 / 4 to the power of X + 1 is an odd function, find the value of the real number a

f(x)=(a+1/4)x ² Is an odd function
So f (- x) = - f (x)
f(-x)=(a+1/4)(-x) ²=)= (a+1/4)x ²
-f(x)=-(a+1/4)x ²
(a+1/4)x ²=- (a+1/4)x ²
(a+1/4)x ²= 0
a=-1/4

The quadratic function f (x) = AX2 + BX + C is known, if f (x) + F (x + 1) = 2x2-2x + 13 (1) Find the analytical formula of function f (x); (2) Draw an image of the function; (3) When x ∈ [T, 5], find the maximum value of function f (x)

(1) F (x) + F (x + 1) = AX2 + BX + C + a (x + 1) 2 + B (x + 1) + C = 2ax2 + (2a + 2b) x + A + B + 2C ∵ f (x) + F (x + 1) = 2x2-2x + 13 ∵ 2A = 22a + 2B = - 2A + B + 2C = 13 ∵ a = 1b = - 2C = 7 ∵ f (x) = x2-2x + 7 (2) the function is that the axis of symmetry is x = 1, the vertices are (1, 6), and there is no relation with the X axis

It is known that the quadratic function f (x) = ax ^ 2 + BX (a, B are constants, and a is not equal to 0) satisfies the following conditions: F (x-1) = f (3-x), and the equation f (x) = 2x has equal roots, (1) find the analytical formula of F (x) (2) whether there are real numbers m, n (m)

(1)
Because f (x-1) = f (3-x),
So the axis of symmetry is x = (x-1 + 3-x) / 2 = 1,
So - B / 2A = 1,
Equation f (x) = 2x has equal roots,
So ax ^ 2 + BX = 2x,
ax^2+bx-2x=0,
(b-2) ^ 2-4 * a * 0 = 0 and a is not equal to 0,
Solution of equations - B / 2A = 1;
(b-2)^2-4*a*0=0
B = 2, a = - 1,
So f (x) = - x ^ 2 + 2x
(2)
Derivation of F (x): x1, f (x) decreasing
Case 1, M