Question

Maximum slope of the curve $y=–x^3+3x^2+9x–27$ is

• A. 0.0
• B. 12
• C. 16
• D. 32

Answer 1

The correct option is B 12

Given that, $y=–x^3+3x^2+9x–27\therefore \mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{curve}=\frac{\mathrm{dy}}{\mathrm{dx}}=-3x^2+6x+9=-3\left(x^2-2x-3\right)\frac{d^{2}y}{{\mathrm{dx}}^2}=-6x+6=-6\left(x-1\right)\frac{d^{2}y}{{\mathrm{dx}}^2}=0\Rightarrow x=1\frac{d^{3}y}{{\mathrm{dx}}^3}=-6<0\therefore x=1\mathrm{is}a\mathrm{point}\mathrm{of}\mathrm{maxima}\mathrm{of}\mathrm{the}\mathrm{slope}\frac{\mathrm{dy}}{\mathrm{dx}}.\mathrm{Maximum}\mathrm{value}\mathrm{of}\mathrm{slope}={\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x=1}=-3\left(1-2-3\right)=12.$