一道微积分函数题证明：当x＞0时,x-x^2/6＜sinx＜x

x-x³/6＜sinx＜x1。令 f(x) = x - sinx ,f(0)=0,f '(x) = 1 - cosx ≥ 0 => f(x) 单增,当x>0时,f(x) > f(0) = 0=> f(x) > 0,即 sinx < x2。令 g(x) = sinx - (x-x³/6),g(0)=0,g '(x) = cosx - 1 + x²/2,g '(0)=0当x>0时,g ''(x) = x - sinx > 0 => g '(x) 单增 => 当x>0时,g '(x) > g '(0) = 0 => g (x) 单增 => 当x>0时,g(x) > g(0) = 0即 当x>0时,x-x³/6＜sinx