设函数f(x)=x^3-3ax+b(a不等于0) (1)若曲线y=f(x)在点(2,f(2))处与直线y=3x+8相切,求a,b的值?解法是：f'(x)=3x^2-3a f'(2)=12-3a =0 所以a=4为什么?f'(x)是怎么求的?

设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=-1/3x^3+2ax^2-3a^2x+b,0 设函数f(x)=x^3+ax^2-9x-1(a 设函数f(x)=x^3+ax^2-9x-1(a 设函数f(x)=x^3+ax^2-9x-1(a 设函数f(x)=x^3+ax^2-9x-1(a 设函数　f(x)=x的3次方－3ax+b(a不等于0）求函数f(x)的单调区间与极值点 设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b? 设函数f(x)=ax+b,其中a,b为实数,f1(x)=f(x),fn+1(x)=f(fn(x)),n=1,2,3…若f7(x)=128x+381,求a+b? 设函数f(x)=-1/3x^3+2ax^2+1/3a(0 设函数f(x)=-x^2+4ax-3a^2,若0 设函数f(x)=-x^2+4ax-3a^2.若0 设函数f(x)=-x^2+4ax-3a^2,若0 已知函数f(x)=x^3-3ax+b(a,b∈R) .（2）设b=0,且g(x)=|f(x)|,（|x|≤1）,求函数g(x)的最大值h(a)