If f(x) monotonically increases on (a, b), the derivative f' (x) of f(x) >: =0 holds on (a, b).
(1)
But (a) for any x, f' (x) >; 0, so it is wrong.
A counterexample: f (x) = x 3 is a simple increase, but when x=0, f'(x)=0.
(2)
According to the meaning of the question, when x1>; X2, both have f(x 1)>f(x2).
For any x, if f'(-x)≤0 and x∈(-∞, +∞), then -x∈(-∞, +∞).
Obviously, for any x, f'(-x)≤0 does not hold.
(3)
When x 1 >; At x2, if f(x 1)>F(x2), there is
When -x 1
(4)
According to the meaning of the question, when x1>; X2, both have f(x 1)>f(x2).
∴f(x) single increase
∴f'(x)≥0
∫-x 1 ≤- x2
∴f(-x 1)≤f(-x2)
That is, -f(x 1)≥-f(x2)
∴-f(-x) single increase
Obviously, you should choose D.
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