y' = 3x^2+2x-1 = (3x-1)(x+1)
y" = 6x+2 = 3(2x+1)
(a)
(i) y'=0
(ii) max if y" < 0, min if y" > 0
(b)
(i) y(1) = 0, y'(1) = 4; use point-slope form
(ii) -1/y' = -1/4
The equation of a curve is y = x^3 + x^2 - x -1
(a) Determine:
(i) the stationary points of the curve;
(ii) the nature of the stationary points in (a)(i) above.
(b) Determine:
(i) the equation of the tangent to the curve at x = 1;
(ii) the equation of the normal to the curve at x = 1
1 answer
1 answer