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Find the maximum slope of the curve $y = - {x^3} + 3{x^2} + 2x - 27$.

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Hint: Slope of the curve is given by $m = \dfrac{{dy}}{{dx}}$. So, in this problem first we will find $\dfrac{{dy}}{{dx}}$. To find the maximum value of slope, we will use a second derivative test. Let us say $\dfrac{{dy}}{{dx}} = f\left( x \right)$. Now we will find the first derivative $f'\left( x \right)$. Then, we will equate $f'\left( x \right)$ to zero for finding critical points. Now we will find the second derivative $f''\left( x \right)$. Then, we will find the value of the second derivative at the critical points. If this value is negative then we can say that the slope is maximum.

Complete step-by-step answer:
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 $5$.

Note: To find the maxima of slope of the curve, put $x = 1$ in the given equation of curve. We get, $y = - {\left( 1 \right)^3} + 3{\left( 1 \right)^2} + 2\left( 1 \right) - 27 = - 23$. Therefore, the slope is maximum at point $\left( {1, - 23} \right)$. The point$\left( {1, - 23} \right)$ is called the maxima of the slope. In the second derivative test, if the value of the second derivative is positive at a critical point then we will get the minimum value of a function.