The monotone decreasing interval of function f (x) = 2 ^ x ^ 2-1 is

The monotone decreasing interval of function f (x) = 2 ^ x ^ 2-1 is

2>1
SO 2 ^ x is an increasing function
So the decreasing interval of F (x) is the decreasing interval of index
The symmetry axis of X & sup2; - 1 is Y-axis with the opening upward
So x

F (x) = 4x cube-108x, find monotone decreasing interval, maximum and minimum

df(x)/dx=12x²-108
Let DF / DX = 0
df/dx=12x²-108=x²-9=0
(x+3)(x-3)=0
When X3, DF / DX > 0, f (x) increases monotonically in these two intervals
When - 3 ≤ x ≤ 3, DF / DX ≤ 0, in this interval, f (x) decreases monotonically
When x = - 3, f (x) is the maximum, f (x) = 216
When x = 3, f (x) is the minimum, f (x) = - 216

Let f (x) = (2-x) (x + 2) ^ 2 (1) find the maximum and minimum of f999 (x) and (2) find the monotone interval of F (x)
There is f (x) in the stem of the question. There is no 999 in front of it

Solution:
f(x)=(2-x)(x+2)²=-x³-2x²+4x+8
f'(x)=-3x²-4x+4
Let f '(x) = 0, the solution is x = - 2 or x = 2 / 3
Let f '(x) > 0, we get - 2