The slope of the tangent to the curve $y = –x^3 + 3x^2 + 9x – 27$ is maximum when x equals

1 3 1/2 -1/2

1

If m be the slope of the tangent to the given curve, then $m = = – 3x^2 + 6x + 9$ $= – 6x + 6, = – 6$ Now $= 0 – 6x + 6 = 0 x = 1$ $= – 6 < 0 $ so at x = 1 the slope m will be maximum