The monotone increasing interval of function y = - x ^ 2 + 4x-7, X ∈ (- 1,3) is

The monotone increasing interval of function y = - x ^ 2 + 4x-7, X ∈ (- 1,3) is

From x = - B / 2a, the symmetry axis of the function image can be calculated as x = 2. Drawing the function image, it is obvious that in the (- 1,3) interval, the monotone increasing interval of the function is (- 1,2)

Monotone increasing interval of function f (x) = (1 / 2) ^ (- x ^ 2 + 4x)

The value of F (x) = (1 / 2) ^ (- x ^ 2 + 4x) will decrease with the increase of (- x ^ 2 + 4x),
So its monotone increasing interval is the monotone decreasing interval of y = (- x ^ 2 + 4x)
Y=-x^2+4x=-（x-2)^2+4
Monotone decreasing interval of y = (- x ^ 2 + 4x) when x is greater than or equal to 2
Therefore, the monotone increasing interval of function f (x) = (1 / 2) ^ (- x ^ 2 + 4x) is [2, positive infinity]

What is the monotone increasing interval of function f (x) = 4x ^ 2 + 1 / x?

Derivation
f(x)'=8x-1/x^2
Let f (x) '> 0
Then x ^ 3-1 / 8 > 0
(x-1/2)(x^2+x/2+1/4)>0
(x-1/2)[(x+1/4)^2+3/16]>0
The solution is x > 1 / 2
So the answer is (1 / 2, + ∞)

Let f (x) = x3-4.5x2 + 6x-a
1. For any real number x, if f (x) is greater than or equal to m, the maximum value of M is obtained
2. If the equation f (x) = 0 has only one real root, the value range of a is obtained

f'(x)=3x²-9x+6=3(x²-3x+2)=3(x²-3x+9/4)-3/4=3(x-3/2)²-3/4≥-3/4
So the maximum value of M is - 3 / 4
f'(x)=3x²-9x+6=3(x²-3x+2)=3(x-1)(x-2)
F (x) takes the maximum at x = 1 and the minimum at x = 2
∵ f (x) = 0 has and has only one real root
∴f(1)0
If f (1) 0, that is 8-18 + 12-A > 0, a2.5 or a