write the equation of line tangent to f(x)=(1/3x^3)-x^2+x-1 when x=1

Answer 1

I meant to write the equation like this: f(x)=(1/3)x^3-x^2+x-1 when x=1

Answer 2

(1/3x^3)-x^2+x-1 when x=1
F'(x) = x^2 -2x + 1
F'(1) = 1-2+ 1 = 0
f(x) = 1/3-1+ 1 -1
f(x) =( 1-3+3-3)/3
f(x) = -2/3
y - (-2/3 ) = 0(x -1)
y + 2/3 = 0
y = -2/3