Question

what is the tangent line of function f(x)=x^3-6x^2+11x-6. the derivative of f(x)is f'(x)=3x^2-12x+11
f(4)=6 so the tangent line in question has equation y-6=11(x-4), or y=11x-38. Which one is it?

Answer 1

set f'(x) to 0 so that slope is 0
0=\[3x ^{2}-12x+11\]
x=\[\sqrt{3}/3+2=x\]
then substute it to the original solving for the y
now you have (x1,y1) and (x2,y2)
you apply it to y-yA=0
you need the ordinate only
x1= sq3/3 +2=-sqr3/3 +2

im not sure >.<

Answer 2

x1=sq3/3+2 ,x2=-sq3/3 +2 >.<