Solveeit Logo
Login

Question

Question: Find the maximum slope of the curve \(y = - {x^3} + 3{x^2} + 2x - 27\)....

Find the maximum slope of the curve y=−x3+3x2+2x−27y = - {x^3} + 3{x^2} + 2x - 27.

Explanation

Solution

Slope of the curve is given by m=dydxm = \dfrac{{dy}}{{dx}}. So, in this problem first we will find dydx\dfrac{{dy}}{{dx}}. To find the maximum value of slope, we will use a second derivative test. Let us say dydx=f(x)\dfrac{{dy}}{{dx}} = f\left( x \right). Now we will find the first derivative f′(x)f'\left( x \right). Then, we will equate f′(x)f'\left( x \right) to zero for finding critical points. Now we will find the second derivative f′′(x)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=−x3+3x2+2x−27.y = - {x^3} + 3{x^2} + 2x - 27. Slope of the curve is m=dydxm = \dfrac{{dy}}{{dx}}. Therefore,
dydx=−3x2+6x+2\dfrac{{dy}}{{dx}} = - 3{x^2} + 6x + 2. Note that here we used the differentiation formula ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}.
Let us say dydx=f(x)\dfrac{{dy}}{{dx}} = f\left( x \right). Therefore, f(x)=−3x2+6x+2f\left( x \right) = - 3{x^2} + 6x + 2.
Now we will find the first derivative f′(x)f'\left( x \right). Therefore, f′(x)=−6x+6f'\left( x \right) = - 6x + 6.
Now we will equate f′(x)f'\left( x \right) to zero for finding critical points. Therefore,
f′(x)=0 ⇒−6x+6=0 ⇒−6(x−1)=0 ⇒x−1=0 ⇒x=1  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=1x = 1.
Now we are going to find the second derivative off(x)f\left( x \right) and its value at a critical point. Therefore,
f′′(x)=−6f''\left( x \right) = - 6.
Forx=1x = 1, we get f′′(x)=−6<0f''\left( x \right) = - 6 < 0.
Note that here f′′(x)f''\left( x \right) is negative for all xx because f′′(x)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=1x = 1 in f(x)f\left( x \right). Therefore, the maximum slope is
(−3)(1)2+6(1)+2=5.\left( { - 3} \right){\left( 1 \right)^2} + 6\left( 1 \right) + 2 = 5.
Hence, the maximum slope of the given curve y=−x3+3x2+2x−27y = - {x^3} + 3{x^2} + 2x - 27 is 55.

Note: To find the maxima of slope of the curve, put x=1x = 1 in the given equation of curve. We get, y=−(1)3+3(1)2+2(1)−27=−23y = - {\left( 1 \right)^3} + 3{\left( 1 \right)^2} + 2\left( 1 \right) - 27 = - 23. Therefore, the slope is maximum at point (1,−23)\left( {1, - 23} \right). The point(1,−23)\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.