Differentiation: Stationary Points and their Nature 1

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Created this as a copy of Differentiation: Using a Table of Derivatives 2.

Name	Status	Author	Last Modified
Differentiation: Using a Table of Derivatives 2	Ready to use	Ben McGovern	26/09/2022 12:10
Differentiation: Using a Table of Derivatives 3	Ready to use	Ben McGovern	26/09/2022 12:11
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Integration: Using a Table of Integrals 5b	Ready to use	Ben McGovern	27/09/2022 09:01
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Question statement

Given the function

$\simplify{y={a}x^3+{b}x^2+{c}x+{d}} ,$ find its stationary points and determine their nature.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	2
b	integer	-2
c	integer	-5
d	integer	-5
solx1	number	1.3051586491
soly1	number	-10.4861548968
solx2	number	-0.6384919825
soly2	number	-3.1434747329
check	number	11.66
check2	number	-11.66
solx1dp	number	1.31
solx2dp	number	-0.64
soly1dp	number	-10.49
soly2dp	number	-3.14
x1	string	(2 d.p.)
x2	string	(2 d.p.)
y1	string	(2 d.p.)
y2	string	(2 d.p.)

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There is a minimum point at (Gap 0., Gap 1.) and a maximum point at (Gap 2. , Gap 3.).

(Give the coordinates of the stationary points to 2 decimal places where necessary.)

Advice

To find the stationary points of the function, we must solve $\tfrac{dy}{dx}=0$ for $x$ . For the function $\simplify{y={a}x^3+{b}x^2+{c}x+{d}}$ ,

$\frac{dy}{dx} = \simplify{{3a}x^2+{2b}x+{c}}.$

Setting $\frac{dy}{dx}=0$ and solving for $x$ :

$\simplify{{3a}x^2+{2b}x+{c}} =0 \\ \\ \implies x=\var{solx1dp} \var{x1} \text{ and } x=\var{solx2dp} \var{x2}.$

Hence, the function has two stationary points at $x=\var{solx1dp}$ and $x=\var{solx2dp}$ . To find the corresponding $y$ -coordinates, we want to plug these values back into the initial equation.

When $x=\var{solx1dp}$ ,

$\begin{split} y &\,= \simplify[unitFactor,!cancelTerms]{{a}*({solx1dp})^3+{b}*({solx1dp})^2+{c}*({solx1dp})+{d}} \\ &\,=\simplify{{soly1dp}} \var{y1}. \end{split}$

When $x=\var{solx2dp}$ ,

$\begin{split} y &\,= \simplify[unitFactor,!cancelTerms]{{a}*({solx2dp})^3+{b}*({solx2dp})^2+{c}*({solx2dp})+{d}} \\ &\,=\simplify{{soly2dp}} \var{y2}. \end{split}$

Therefore, the stationary points of $y=\simplify{{a}x^3+{b}x^2+{c}x+{d}}$ are

$(\simplify{{solx1dp}},\, \simplify{{soly1dp}}) \, , \,(\simplify{{solx2dp}},\, \simplify{{soly2dp}}).$

Finally, we need to determine the nature of the stationary points. To do this we want to calculate the second derivative of the initial function and then evaluate it for each $x$ -value of the stationary points.

Recall:

If $\tfrac{d^2y}{dx^2}<0$ the stationary point is a maximum;
If $\tfrac{d^2y}{dx^2}>0$ the stationary point is a minimum;
If $\tfrac{d^2y}{dx^2}<0$ the stationary point is a point of inflection.

To calculate $\tfrac{d^2y}{dx^2}$ , we want to differentiate $\tfrac{dy}{dx}$ again with respect to $x$ :

$\begin{split} &\frac{dy}{dx} = \simplify{{3a}x^2+{2b}x+{c}}, \\ \\\implies &\frac{d^2y}{dx^2} = \simplify{{6a}x+{2b}}. \end{split}$

For $(\simplify{{solx1dp}},\, \simplify{{soly1dp}})$ , $\frac{d^2y}{dx^2} = \simplify{{check}}$ , so it is a minimum.

For $(\simplify{{solx2dp}},\, \simplify{{soly2dp}})$ , $\frac{d^2y}{dx^2} = \simplify{{check2}}$ , so it is a maximum.

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Differentiation: Stationary Points and their Nature 1

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