(2014卓越11)
已知$f(x)$为$R$上的可导函数,且对$forall x_0in R$ 有$0<f^{'}(x+x_0)-f^{'}(x_0)<4x(x>0)$.
(1)对$forall x_0in R$,证明:$f^{'}(x_0)<dfrac{f(x+x_0)-f(x_0)}{x} (x>0)$
(2)若$|f(x)|le1,xin R$,证明:$|f^{'}(x)|le 4$.
分析:(1)构造$g(x)=f(x+x_0)-f(x_0)-xf^{'}(x_0) (x>0)$. 易证$g(x)>g(0)=0$.
(2)构造$h(x)=f(x+x_0)-f(x_0)-xf^{'}(x_0)-2x^2 (x>0)$易证$h(x)<h(0)=0(x>0)$
故$f^{'}(x_0)>dfrac{f(x+x_0)-f(x_0)}{x}-2x(x>0)$
结合(1)知$|f^{'}(x_0)|ledfrac{|f(x+x_0)-f(x_0)|}{x}+2xledfrac{2}{x}+2x$ 对任意$x>0$恒成立.
故$|f^{'}(x_0)|le 4$由$x_0$的任意性知$|f^{'}(x)|le4$
注:这里主要是构造函数的技巧.类似的技巧在数学分析中证明拉格朗日中值定理和柯西中值定理时也有体现
