Question

求出曲线y=(1/3)x^3上与直线x-4y-5=0平行的切线方程.

Answer 1

y=(1/3)x^3

y'=x^2

与直线x-4y-5=0平行,即斜率k=1/4

由y'=x^2=1/4,解得：x=1/2,或x=-1/2

又y(1/2)=1/24,y(-1/2)=-1/24

所以这样的切线有2条：

y=1/4*(x-1/2)+1/24=x/4-1/12

y=1/4*(x+1/2)-1/24=x/4+1/12