函数(x)=dfrac (1)(3)(x)^3+dfrac (1)(2)(x)^2的单调递增区间是( )A.(x)=dfrac (1)(3)(x)^
dfrac (1)(2)x+dfrac (1)(3)=dfrac (1)(4)x-dfrac (1)(5)
[题目]设函数f (x)在闭区间(0,1)上连续,在开区间-|||-(0,1)内可导,且 (0)=0, (1)=dfrac (1)(3),-|||-证明:存在
已知(x-dfrac (1)(x))=dfrac ({x)^3-x}(1+{x)^4}-|||-__,求(x-dfrac (1)(x))=dfrac ({x)^
[题目]函数 (x)=dfrac (|x|sin (x-1))(x(x-1)(x-2)) 在下列区间内有界-|||-的是 ()-|||-A.(0,1)-|||-
[题目]设函数f (x)在闭区间[0,1]上连续,在开区间-|||-0,1)内大于零,并满足 (x)=f(x)+dfrac (3a)(2)(x)^2 为常数),
3、设 (x,y)=arctan dfrac (x)(y), 则 (1,1)=-|||-(A)1; (B)0; (C) dfrac {1)(2),dfrac
[单选题]函数f(x)=(x+1)/x在[1,2]上符合拉格朗日定理条件的ζ值为:()A . B . -C . 1/D . -1/
(1) =dfrac (x+1)(x(x-1)) 在区间 (-1,0) (0,1)、(1,2)、(2,3)上的有界性;
(B) dfrac (1)(2)(X)_(1)+dfrac (1)(2)(X)_(2)-|||-(C) dfrac (1)(2)(X)_(1)+dfrac (1