(1) First, take the derivative of f(x) and substitute it into a=4/3, so that f ′ (x) = 0. After the solution, the signs of the derivative function on both sides of the root are judged.

(2) Since A > 0, f(x) is increasing function on R, and f ′ (x) ≥ 0 is constant on R, it is transformed into a quadratic function constant problem as long as △≤0.

Answer:

Solution: Derive f(x) to get f ′ (x) = [(1+ax 2? 2ax)/( 1+ax^2)^2]×e^x

(1) when a=4/3, if f ′ (x) = 0, then 4x 2-8x+3 = 0, and the solution is X 1 = 3/2, X2 = 1/2.

Combination 1, can you know?

Therefore, x 1 = 3/2 is the minimum point and x 1 = 1/2 is the maximum point.

(2) If f(x) is a monotone function on R, then f ′ (x) is invariant on R. Combining ① with condition A > 0, it is known that AX 2-2ax+1≥ 0 is invariant on R, so △ = 4a 2-4a = 4a (a-/kloc-0.