1.设 (x)=dfrac (1)(1-{x)^2}, 求 (-x),f[ f(x)] ,f[ dfrac (1)(f(x))]
设 函数 f ( x ) 在 x = 1 处可导且lim _(xarrow 0)dfrac (f(1)-f(1-x))(2x)=1则 lim _(xarrow
24 题 设 (x)=dfrac (1)(ln x)-dfrac (1)(x-1), in (0,1), 则补充定义 f(1)= ()-|||-可使f(x)在
[题目]-|||-设f(x)为连续函数,且 (x)=(int )_(dfrac {1)(x)}^ln xf(t)dt, 则F(x)等于 ()-|||-(A) d
1.已知 (x)=dfrac (1)(x(1+2ln x)) 且 f(1)=1, 则f(x)等于_ __-|||-
8.设函数f(x)在点 x=1 处连续,且 lim _(xarrow 1)dfrac (f(x)-2)(x-1) 存在,则 f(1)= __
设f(x)=((1-{x^2)})/(({x^2)+1)},则f(((1)/(x)))等于( )A. f(x)B. -f(x)C. $\frac{1}{f(x)
设f(x)=dfrac (2)(3)(x)^3,xleqslant 1 (x)^2,xgt 1,则f(x)在x=1处的( )设f(x)=,则f(x)在x=1处
设f(x)=dfrac (2)(3)(x)^3,xleqslant 1 (x)^2,xgt 1,则f(x)在x=1处的( )设f(x)=,则f(x)在x=1处
设lim _(xarrow 0)dfrac (ln (1+x+dfrac {f(x))(x))}(x)=3,则lim _(xarrow 0)dfrac (ln