已知函数f(x)=2x^3+3/2tx^2-3t^2xf(x)=2x^3+3/2tx^2-3t^2x+(t-1)/2,x∈R,其中t∈R(1)当t=1时,求曲线y=f(x)在点(0,f(0))处的切线方程
问题描述:
已知函数f(x)=2x^3+3/2tx^2-3t^2x
f(x)=2x^3+3/2tx^2-3t^2x+(t-1)/2,x∈R,其中t∈R
(1)当t=1时,求曲线y=f(x)在点(0,f(0))处的切线方程
答
1)t=1时,
f(x)=2x^3+3/2x^2-3x
f'(x)=6x^2+3x-3
f(0)=0
f'(0)=-3
切线方程为:y=-3(x-0)+0=-3x