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

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.