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Let us find slope of the curve $y = - {x^3} + 3{x^2} + 2x - 27.$ Slope of the curve is $m = \dfrac{{dy}}{{dx}}$. Therefore,

$\dfrac{{dy}}{{dx}} = - 3{x^2} + 6x + 2$. Note that here we used the differentiation formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$.

Let us say $\dfrac{{dy}}{{dx}} = f\left( x \right)$. Therefore, $f\left( x \right) = - 3{x^2} + 6x + 2$.

Now we will find the first derivative $f'\left( x \right)$. Therefore, $f'\left( x \right) = - 6x + 6$.

Now we will equate $f'\left( x \right)$ to zero for finding critical points. Therefore,

$

f'\left( x \right) = 0 \\

\Rightarrow - 6x + 6 = 0 \\

\Rightarrow - 6\left( {x - 1} \right) = 0 \\

\Rightarrow x - 1 = 0 \\

\Rightarrow x = 1 \\

$

Therefore, the critical point is $x = 1$.

Now we are going to find the second derivative of$f\left( x \right)$ and its value at a critical point. Therefore,

$f''\left( x \right) = - 6$.

For$x = 1$, we get $f''\left( x \right) = - 6 < 0$.

Note that here $f''\left( x \right)$ is negative for all $x$ because $f''\left( x \right)$ is constant.

Here the value of the second derivative is negative at a critical point. So, we can say that the slope is maximum.

To find the maximum value of slope, we will put $x = 1$ in $f\left( x \right)$. Therefore, the maximum slope is

$\left( { - 3} \right){\left( 1 \right)^2} + 6\left( 1 \right) + 2 = 5.$

Hence, the maximum slope of the given curve $y = - {x^3} + 3{x^2} + 2x - 27$ is $5$.