Let f (x) = x ^ 2-2lnx, H (x) = x ^ 2-x + A. let K (x) = f (x) - H (x), if K (x) has exactly two zeros on [1.3] Finding the value range of real number a

Let f (x) = x ^ 2-2lnx, H (x) = x ^ 2-x + A. let K (x) = f (x) - H (x), if K (x) has exactly two zeros on [1.3] Finding the value range of real number a

k(x)=f(x)-h(x)＝(x^2-2lnx)-(x^2-x+a)=x-2lnx-a；
The derivative of function K (x) is = 1-2 / X
So: when x = 2, the derivative is equal to 0;
When x > 2, derivative > 0, the original function is increasing;
When 0

The zeros of the function f (x) = x3-2x2-x + 2 are______ ．

From the function f (x) = x3-2x2-x + 2 = X2 (X-2) - (X-2) = (x + 1) (x-1) (X-2), Let f (x) = 0, the solution is x = - 1 or 1 or 2. The zeros of function f (x) are - 1, 1, 2. So the answer is - 1, 1, 2

Let f (x) = x2-4x-4 (t ≤ x ≤ T + 1), find the expression of the minimum value g (x) of F (x)
Let f (x) = x2-4x-4 (t ≤ x ≤ T + 1), find the expression of the minimum value g (x) of F (x)
Question 2: given the function f (x) = (x2 + 2x + a) / x, X belongs to [1, positive infinity]
(1) When a = 1 / 2, find the minimum value of function f (x)
(2) If f (x) > 0 is constant for any x belonging to [1, positive infinity], find the value range of real number a

When t is less than or equal to 1, G (x) = t2-2t-7
When t is greater than 1 and less than 2, G (x) = - 8
When t is greater than or equal to 2, G (x) = t2-4t-4
T2 is the square of T
The second question 1: take the minimum when x = 1, 3.5
2: A greater than - 2 root sign 2 less than 2 root sign 2

Let f (x) = x2-4x-4, X belong to [T, t + 1] (t belongs to R) and find the analytic expression of the minimum value g (T) of function f (x)

Solution: F (x) when x ∈ R, the opening of F (x) image is upward with respect to x = 2 symmetric image
(-, 2) monotonically decreasing [2, +] increasing
therefore
When t ≥ 2, GT = f (x) min = f (T)
When t

Given that f (x) = x2-4x-3, X ∈ R, the function g (T) represents the minimum value of F (x) on [T, t + 2], and the expression of G (x) is obtained

Axis of symmetry: x = 2
① When t + 2

The minimum value of function f (x) = x & # 178; - 4x + 1 in [T, 4] is g (T). Find the analytic expression of G (T) and the minimum value of G (T)

f(x)=(x-2)^2-3
The opening is upward and the axis of symmetry is x = 2
When t

The minimum value of the function f (x) = x & # 178; - 4x-4 in the closed interval [T, t + 1] (t ∈ R) is denoted as G (T)
How to know 1

f(x)=(x-2)²-8
The opening is upward and the axis of symmetry is x = 2
When 1

The minimum value of the function f (x) = x ^ 2-2x + 2 in the closed interval [T, t + 1] (t belongs to all real numbers) is denoted as G (T). (1) try to write out the G (T) expression of the function

f(x)=x^2-2x+2
=(x-1)^2+1
When t + 1

The minimum value of the function f (x) = x ^ 2-4x-4 in the closed interval [T, t + 1] (t ∈ R) is denoted as the functional expression of G (T) for G (x)

（1）
F (x) = x ^ 2 - 4x - 4 = (x - 2) ^ 2 - 8, so the vertex coordinates are (2, - 8)
When t + 1 = 2, G (T) = f (T);
When t > 2-1 and t

The minimum value of function f (x) = xx-4x-4 in the interval [T, t + 1] is g (T) --- write the function expression of G (T)

If t belongs to [1,2], G (T) = f (2) = - 8 2. If t belongs to (negative infinite, 1), G (T) = f (T + 1) = (T + 1) ^ 2-4 (T + 1) - 4 = T ^ 2-2t-7 3