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