#### Question

Find the maximum slope of the curve y = \[- x^3 + 3 x^2 + 2x - 27 .\]

#### Solution

\[\text { Given: } \hspace{0.167em} y = - x^3 + 3 x^2 + 2x - 27 ............\left( 1 \right)\]

\[\text { Slope } = \frac{dy}{dx} = - 3 x^2 + 6x + 2\]

\[\text { Now,} \]

\[M = - 3 x^2 + 6x + 2\]

\[ \Rightarrow \frac{dM}{dx} = - 6x + 6\]

\[\text { For maximum or minimum values of M, we must have }\]

\[\frac{dM}{dx} = 0\]

\[ \Rightarrow - 6x + 6 = 0\]

\[ \Rightarrow 6x = 6\]

\[ \Rightarrow x = 1\]

\[\text { Substituing the value of x in eq. } \left( 1 \right),\text { we get }\]

\[y = - 1^3 + 3 \times 1^2 + 2 \times 1 - 27 = - 23\]

\[\frac{d^2 M}{d x^2} = - 6 < 0\]

\[\text { So, the slope is maximum when x = 1 and y } = - 23 . \]

\[ \therefore At \left( 1, - 23 \right): \]

\[\text { Maximum slope } = - 3 \left( 1 \right)^2 + 6\left( 1 \right) + 2 = - 3 + 6 + 2 = 5\]