Question

已知函数f(x)=2x3-3x2+3求曲线Y=f(x)在点x=2处切线方程

Answer 1

f(x)=2x3-3x2+3f ' (x)=6x^2-6xf '(2)=6*4-6*2=12f (2)=2*8-3*4+3=7曲线Y=f(x)在点x=2处切线方程:y-7=f'(2)(x-2) y-7=12(x-2) 12x-y-17=0