CB5 - Finding turning points - Numbas at mathcentre.ac.uk

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Will Morgan	said	Ready to use	1 year ago
Andrew Neate	said	Needs to be tested	1 year ago

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Will Morgan 11 months, 2 weeks ago

Published this.

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Andrew Neate 1 year ago

Created this as a copy of Differentiation: Stationary Points and their Nature 1.

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
Differentiation: Using a Table of Derivatives 4	Ready to use	Ben McGovern	26/09/2022 12:11
Differentiation: Using a Table of Derivatives 5a	Ready to use	Ben McGovern	26/09/2022 12:11
Differentiation: Using a Table of Derivatives 5b	Ready to use	Ben McGovern	26/09/2022 12:12
Differentiation: Using a Table of Derivatives 5c	Ready to use	Ben McGovern	26/09/2022 12:12
Differentiation (Harder powers - negative and fractions) 2	Ready to use	Ben McGovern	04/05/2023 05:42
Differentiation: Stationary Points and their Nature 1	Ready to use	Ben McGovern	26/09/2022 11:56
Differentiation: Stationary Points and their Nature 2	Ready to use	Ben McGovern	26/09/2022 11:56
Integration: Using a Table of Integrals 2	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Using a Table of Integrals 3a	Ready to use	Ben McGovern	13/02/2023 10:37
Integration: Using a Table of Integrals 3b	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Using a Table of Integrals 4	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Using a Table of Integrals 5a	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Using a Table of Integrals 5b	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Using a Table of Integrals 5c	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Using a Table of Integrals 6	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 1a i	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 1b	Ready to use	Ben McGovern	23/02/2023 12:45
Integration: By Substitution 1c	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 2a	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 2b	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 2c	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 3a i	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 3c	Ready to use	Ben McGovern	27/09/2022 09:01
Integration (Negative and Fractional Powers) 2	Has some problems	Ben McGovern	27/09/2022 09:01
Integration (Negative and Fractional Powers) 3	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Definite Integrals 2	Ready to use	Ben McGovern	26/09/2022 12:18
Integration: By Substitution 1a ii (Negative and Fractional Powers)	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: By Substitution 3a ii (Negative and Fractional Powers)	Ready to use	Ben McGovern	27/09/2022 09:01
Integration: Definite Integrals 3	Ready to use	Ben McGovern	26/09/2022 12:18
Integration: Definite Integrals 4	Ready to use	Ben McGovern	26/09/2022 12:19
CB2 Differentiating polynomials 2	Ready to use	Mash Sheffield	09/04/2024 15:44
CD3 Integration - Substitution	Ready to use	Andrew Neate	09/04/2024 15:56
CB4 Definite integration	Ready to use	Andrew Neate	09/04/2024 15:45
CB6 Differentiating Trig 1	Ready to use	Andrew Neate	09/04/2024 15:46
CB7 Differentiating Trig 2	Ready to use	Andrew Neate	09/04/2024 15:46
CB8 Differentiating with Trig 3	Ready to use	Andrew Neate	09/04/2024 15:52
CB9 Differentiating with Exponentials	Ready to use	Andrew Neate	09/04/2024 15:52
CB10 Differentiation with logs	Ready to use	Andrew Neate	09/04/2024 15:43
CB3 - Indefinite integration - polynomials	Ready to use	Andrew Neate	09/04/2024 15:45
CB5 - Finding turning points	Ready to use	Andrew Neate	09/04/2024 15:46

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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	3
b	integer	-4
c	integer	0
d	integer	-3
solx1	number	0.8888888889
soly1	number	-4.0534979424
solx2	number	0
soly2	number	-3
check	number	8.00
check2	number	-8.00
solx1dp	number	0.89
solx2dp	number	0.00
soly1dp	number	-4.05
soly2dp	number	-3.00
x1	string	(2 d.p.)
x2	string
y1	string	(2 d.p.)
y2	string

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
random(1..3)

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

→ Used by:

check
check2
solx1
solx2
soly1
soly2

This variable doesn't seem to be used anywhere.

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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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CB5 - Finding turning points

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Will Morgan 11 months, 2 weeks ago

Will Morgan 1 year ago

Andrew Neate 1 year ago

Andrew Neate 1 year ago

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CB5 - Finding turning points

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

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Comment

Checkpoint description

Will Morgan 11 months, 2 weeks ago

Will Morgan 1 year ago

Andrew Neate 1 year ago

Andrew Neate 1 year ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`