Max and Min 1 and 2

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Xiaodan Leng 5 years, 8 months ago

Created this as a copy of Paul 's copy of Max and Min 1 and 2.

Name	Status	Author	Last Modified
Max and Min 1 and 2	Ready to use	Bill Foster	01/06/2016 09:47
CF Maths In class test two mock paper question 4	Ready to use	Jinhua Mathias	28/08/2024 12:15
Max and Min (5 parts) 24/11	Ready to use	Katie Dexter	01/06/2016 09:47
CF Maths January test mock paper finding stationary points	Ready to use	Jinhua Mathias	28/08/2024 12:15
Paul 's copy of Max and Min 1 and 2	Ready to use	Paul Howes	11/06/2016 19:08
Paul 's copy of Max and Min 1 and 2	Ready to use	Paul Howes	11/06/2016 19:08
Problema 1 máximo y mínimo Global	draft	Luis Hernandez	02/12/2020 15:42
Max and Min v2	draft	Daniel Mansfield	22/01/2021 01:16
Max and Min 1 and 2	draft	Xiaodan Leng	11/07/2019 05:58

There are 13 other versions that do you not have access to.

Question statement

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

$g(x) = \simplify{{c}/3*x^3+ {-c*(a+b)}/2*x^2+{c*a*b}*x+{d}}$

Give any introductory information the student needs.

			Name	Type	Generated Value

temp2	number	17.00
temp1	number	-47.00
gmi	number	-1.5
valmin	integer	-47
gma	number	7.5
s	integer	1
m1	number	7
valbegin	number	-89.25
a	integer	1
valmax	integer	17
valgmax	number	59.25
valgmin	number	-89.25
rawvalend	number	59.25
rawvalmax	number	17
b	integer	5
rawvalbegin	number	-89.25
rawvalmin	number	-47
rtemp2	number	17
c	integer	6
rtemp1	number	-47
d	integer	3
m	number	7.5
l	number	-1.5
valend	number	59.25
l1	number	-1

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
precround(rtemp2,2)

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: `number`

17.00

← Depends on:

rtemp2

→ Used by:

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
Part c) Gap-fill
Part d) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part e) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Number entry
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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 linear factors in the form $g'(x)=c(x-a)(x-b)$ for suitable integers $a$ , $b$ and $c$ :

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

Advice

Differentiating, we have:

$g'(x)=\simplify{{c}*x^2+{-c*(a+b)}*x+{c*a*b}={c}*(x+{-a})(x-{b})}$

Note that we have already factorised the derivative.

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

So the least stationary point is $x=\var{a}$ and the greatest is $x=\var{b}$ .

Since $\var{a} > \var{l}$ and $\var{b} \lt \var{m}$ we have that both stationary points are in $I$ .

The second derivative is given by

$g''(x)=\simplify{{2*c}*x-{c*(a+b)}}$

Local Maximum

At the stationary point $x=\var{a}$ we have $g''(\var{a})=\var{c*a-c*b} \lt 0$ .

Hence at this value of $x$ we have a local maximum.

The value of the function $g$ at this local maximum is $g(\var{a})= \var{valmax}$ .

Local Minimum

At the stationary point $x=\var{b}$ we have $g''(\var{b})=\var{c*b-c*a} \gt 0$ .

Hence this point is a local minimum.

The value of the function $g$ at this local minimum is $g(\var{b})= \var{valmin}$ .

Global Maximum

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

$g(\var{l})=\var{valbegin}$ to 3 decimal places.

$g(\var{m})=\var{valend}$ to 3 decimal places.

To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$ .

So we proceed by comparing the values of the function at the endpoints with the local maximum.

a) If the value at the local maximum is greater than either of the values at the endpoints then this is the global maximum on the interval.

b) Otherwise if the greatest value of the function at the endpoints is greater than the local maximum then this is the global maximum.

$\begin{array}{c|c|c|c} x & \mbox{Local Maximum}=\var{a} & \var{l} \in I & \var{m} \in I \\ \hline\\ g(x)& \var{valmax} & \var{valbegin} & \var{valend} \\ \end{array}$

So for our example we see that the global maximum occurs at $x=\var{gma}$ and $g(\var{gma})=\var{valgmax}$ .

Global Minimum

We proceed as for the global maximum by comparing the values of the function at the endpoints with the local minimum.

a) If the value at the local minimum is less than either of the values at the endpoints then this is the global minimum on the interval.

b) Otherwise if the least value of the function at the endpoints is less than the local minimum then this is the global minimum.

$\begin{array}{c|c|c|c} x & \mbox{Local Minimum}=\var{b} & \var{l} \in I & \var{m} \in I \\ \hline\\ g(x)& \var{valmin} & \var{valbegin} & \var{valend} \\ \end{array}$

In our example we see that the global minimum occurs at $x=\var{gmi}$ and $g(\var{gmi})=\var{valgmin}$ .

Give a worked solution to the whole question.

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Taxonomy: mathcentre

Taxonomy: Kind of activity

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Contributors

History

Comment

Checkpoint description

Xiaodan Leng 5 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: `number`

Parts

Gap-fill

Local Maximum

Local Minimum

Global Maximum

Global Minimum

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Error

Max and Min 1 and 2

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Xiaodan Leng 5 years, 8 months ago

Error in variable testing condition

Error

Warning

Generated value: number

Parts

Gap-fill

Local Maximum

Local Minimum

Global Maximum

Global Minimum

Generated value: `number`