If the two real roots X1 and X2 of the equation x ^ 2-2ax + a = 0 satisfy that X1 belongs to (- 1,0) and X2 belongs to (0,1), then the value range of real number a is

If the two real roots X1 and X2 of the equation x ^ 2-2ax + a = 0 satisfy that X1 belongs to (- 1,0) and X2 belongs to (0,1), then the value range of real number a is

Y = f (x) = x ^ 2-2ax + a opening upward
X1 belongs to (- 1,0), X2 belongs to (0,1)
Draw an image
f(-1)>0,f(0)0
So 1 + 2A + a > 0
a0
That is a > - 1 / 3
a

Find the extremum of function f (x) = x / 1 + X & # 178 in the interval (- 3 / 2,1 / 2)

f(x)=x/（1+x²）,
f'(x)=(1+x^2-2x^2)/(1+x^2)^2=-(x+1)(x-1)/(1+x^2)^2,
-1

A ∈ R, f (x) = (X & # 178; - 4) (x-a), the function f (x) has extremum at x = - 1
If there is a common point between the image of the line y = m and the function f (x), then the value range of M is the same
If f (x) - C ≤ 0 is constant when x ∈ [- 2,4], then the range of C is constant

f'(x)=2x(x-a)+x²-4=3x²-2ax-4
Because f (x) has an extreme value at x = - 1
So f '(- 1) = 0, that is, 3 + 2a-4 = 0, a = 1 / 2
f(x)=(x²-4)(x-1/2)
f'(x)=3x²-x-4=（3x-4）（x+1）
Let f '(x) = 0, then x = 4 / 3, - 1
When x ＜ - 1, f '(x) > 0, is an increasing function
When - 1 ＜ x ＜ 4 / 3, f '(x) ＜ 0 is a decreasing function
When x > 4 / 3, f '(x) > 0, is an increasing function
So the maximum f (- 1) = 9 / 2
Minimum f (4 / 3) = - 50 / 27
If y = m has a common point with the image of function f (x), then M < f (4 / 3) = - 50 / 27, or M > F (- 1) = 9 / 2
When x ∈ [- 2,4], f (x) - C ≤ 0 is constant
When x ∈ [- 2,4], the maximum value is f (- 1) = 9 / 2, so only C ≥ f (- 1) = 9 / 2 is needed
So m ＜ - 50 / 27 or m ＞ 9 / 2
c≥9/2

Let f (x) = x (x-1) & #, x > 0 (1) find the extremum of F (x) (2) let 0
The solution is urgent
Let f (x) = x (x-1) &# 178;, x > 0
(1) Finding the extreme value of F (x)
(2) Let 0 ＜ a ≤ 1, Let f (x) be the maximum value on (0, a)] be f (a), and find the minimum value of function g (a) = f (a) / A
(3) Let g (x) = ㏑ x-2x & # 178; + 4x + T (t is a constant), if G (x) ≤ x + m ≤ f (x) has and only has one real number m which is constant on x > 0, find the values of real numbers m and t

(1) F & # 39; (x) = (x - 1) & # 178; + X * 2 (x - 1) = (x - 1) (3x - 1) F & # 39; (x) = 0, x = 1, x = 1 / 30 & lt; X & lt; 1 / 3: X - 1 & lt; 0,3x - 1 & lt; 0, F & # 39; (x) & gt; 0, increasing function 1 / 3 & lt; X & lt; 1: X - 1 & lt; 0,3x - 1