find the equation of tangent to the curve f(x)=x^3-3x^2+x-1 at x=3
solution,Here, f(x) = y = x^3-3x^2+x-1
Slope(m) = first derivative of y = 3x^2 - 6x + 1
when x =3 then slope(m) = 10
Again x = 3 then y = 2
we have equation of tangent to the given curve at (3, 2) is
or, y -y1 = m( x- x1 )
or, y - 2 = 10 ( x - 3 )
y = 10x - 28 is required equation of tangent
