And f(-0)=-f(0), so f(0)=0.

F(x) monotonically decreases at (0, +∞) because it monotonically decreases at (-∞, 0).

① The solution of f (x) ≤ 0 is

②x & lt； 0，f (x)

X>0, f(x)>0, so the image is in the first quadrant. 0

To sum up, -2