Max and Min (6 parts) 24/11

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Katie Dexter

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Christian Lawson-Perfect

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Ready to use

8 years, 9 months ago

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Christian Lawson-Perfect 8 years, 9 months ago

Gave some feedback: Ready to use

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Katie Dexter 9 years, 6 months ago

Created this as a copy of Max and Min 3.

Name	Status	Author	Last Modified
Max and Min 3	Ready to use	Bill Foster	01/06/2016 09:47
Max and Min (6 parts) 24/11	Ready to use	Katie Dexter	01/06/2016 09:47
Paul 's copy of Max and Min 3	Ready to use	Paul Howes	11/06/2016 19:08
Max and Min 3	draft	Xiaodan Leng	11/07/2019 05:57

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Question statement

Let $I=[\var{l},\var{m}]$ be an interval in $x$ and let $g$ be a function defined on this interval given by :

$g(x) = \simplify{(x-{a})*(x-{b})^3}$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	8
c	integer	6
b	integer	32
d	integer	7
gma	number	729
k	integer	14
m	integer	35
xma	integer	35
l	integer	11
valend	number	729
gmi	number	-34992
valbegin	number	-27783
third2	integer	144
third1	integer	-288
s1	integer	1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
s1*random(1..9)

JME function reference

Description

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

Can an exam override the value of this variable?

Generated value: `integer`

← Depends on:

→ Used by:

b
gma
gmi
k
valbegin
valend

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Mathematical expression
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Part b) Gap-fill
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1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Number entry
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1. Gap Gap 0. Number entry
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Part f) Gap-fill
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Gap-fill

Ask the student a question, and give any hints about how they should answer this part.

Input the first derivative of $g$ here, factorised into a product of two factors in the form $g'(x)=c(x-a)(x-b)^2$ for suitable integers $a$ , $b$ and $c$ (these can be zero):

$g'(x)=\;\;$ Unnamed gap

Advice

First derivative

Differentiating we have:

$\begin{eqnarray*} g'(x)&=&\simplify{(x-{b})^3+3*(x-{a})*(x-{b})^2}\\ &=&\simplify{(x-{b})^2(3*(x-{a})+x-{b})}\\ &=&\simplify{4*(x-{k})*(x-{b})^2} \end{eqnarray*}$ and we have factorised the expression.

Stationary points

These are given by solving $g'(x)=0 \Rightarrow x=\var{k},\;\;\mbox{or }x=\var{b}$

Therefore the least stationary point is $x=\var{k}$ and the greatest is $x=\var{b}$ and we see that both stationary points are in $I$ .

Second derivative.

The second derivative is given by:

$\begin{eqnarray*} g''(x)&=&\simplify{4*(x-{b})^2+8*(x-{k})(x-{b})}\\ &=&\simplify{4*(x-{b})(3*x-{2*k+b})} \end{eqnarray*}$

Local Minimum

At the stationary point $x=\var{k}$ we have $g''(\var{k})=\var{4*(k-b)^2} \gt 0$ .

Hence $x=\var{k}$ is a local minimum.

The other stationary point

The value at $x=\var{b}$ is $g(\var{b})= 0$ .

Hence this test fails at this point and we proceed to use the third derivative to see in more information can be gained.

Third derivative

We see that $g'''(x)=\simplify{8*(3*x-{k+2*b})}$ .

Testing the stationary point using the third derivative gives:

$g'''(\var{b})=\var{8*(b-k)} \neq 0$ .

Therefore there cannot be an extremum point at $x=\var{b}$ .

Finding the global maximum and minimum on $I$

First we find the values at the endpoints of the interval $I=[\var{l},\var{m}]$ are:

$g(\var{l})=\var{valbegin}$ .

$g(\var{m})=\var{valend}$ .

Global Maximum

To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$ and since $g$ does not have a local maximum on $I$ it must take its maximum value at one of the end points of $I$ .

We see from the values at the end points obtained above that the global maximum value on $I$ is at $x=\var{xma}$ .

We have $g(\var{xma})=\var{gma}$ .

Global Minimum.

$g$ has only one local minimum on $I$ at $x=\var{k}$ and so this must be the global minimum on $I$ .

We have $g(\var{k})=\var{(k-a)*(k-b)^3}$ .

More information on maxima and minima is chapter 3 of your Mathematical Physics book, sections 3.2 and 3.3, pages 51-53.

Give a worked solution to the whole question.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

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Checkpoint description

Christian Lawson-Perfect 8 years, 9 months ago

Katie Dexter 9 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: `integer`

Parts

Gap-fill

First derivative

Stationary points

Second derivative.

Local Minimum

The other stationary point

Third derivative

Finding the global maximum and minimum on $I$

Global Maximum

Global Minimum.

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Error

Max and Min (6 parts) 24/11

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 8 years, 9 months ago

Katie Dexter 9 years, 6 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

First derivative

Stationary points

Second derivative.

Local Minimum

The other stationary point

Third derivative

Finding the global maximum and minimum on II

Global Maximum

Global Minimum.

Generated value: `integer`

Finding the global maximum and minimum on $I$