y=−x3+3x2+2x−27 On differentiating w.r.t. x, we get dxdy=−3x2+6x+2 Slope of the curve =−3x2+6x+2 Let m=−3x2+6x+2 ∴ dxdm=−6x+6 Put dxdm=0 to find the critical points ⇒ −6x+6=0 ⇒ x=1(dx2d2m)x=1=−6<0 Slope is maximum at x=1 The maximum slope, mmax=−3(1)2+6(1)+2=−3+6+2=5