Max and Min 5

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Christian Lawson-Perfect

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Ready to use

10 years, 2 months ago

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Christian Lawson-Perfect 10 years, 2 months ago

Gave some feedback: Ready to use

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Bill Foster 55 years, 4 months ago

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Name	Status	Author	Last Modified
Max and Min 5	Ready to use	Bill Foster	01/06/2016 09:47
Paul 's copy of Max and Min 5	Ready to use	Paul Howes	17/09/2019 15:07
Paul 's copy of Max and Min 5	draft	Xiaodan Leng	11/07/2019 05:55

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Question statement

Let $I=[\var{c},\var{d}]$ be an interval and let $g: I \rightarrow I$ be the function given by:

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

Answer the following questions. There are seven parts and you may need to scroll down to complete all parts.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	-1
c	integer	-6
b	integer	3
d	integer	5
valmax	number	0.167
s	integer	-1
lmi	integer	3
valend	number	-0.147
tol	number	0.001
valbegin	number	0.133
lma	integer	-3
valmin	number	-0.167

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
s*random(1..9)

JME function reference

Description

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

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Generated value: `integer`

-1

← Depends on:

→ Used by:

lma
lmi
valbegin
valend
valmin

This variable doesn't seem to be used anywhere.

Parts

Part a) Choose one from a list
Part b) Gap-fill
1. Gap Mathematical expression
  Add
Part c) Choose one from a list
Part d) Gap-fill
Part e) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Mathematical expression
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3. Gap Gap 2. Number entry
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4. Gap Gap 3. Mathematical expression
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5. Gap Gap 4. Number entry
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Part f) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part g) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Ask the student a question, and give any hints about how they should answer this part.

Is $g(x)$ continuous at all points of $I$ ?

Choose Yes or No.

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Advice

The function $g(x)$ is continuous and differentiable at all points in $\mathbb{R}$ .

Using the quotient rule for differentiation we see that

$\begin{eqnarray*}g'(x)&=&\simplify{({a}*(x^2+{b^2})-{2*a}*x^2)/(x^2+{b^2})^2}\\ &=&\simplify{({-a}*(x-{b})(x+{b}))/(x^2+{b^2})^2} \end{eqnarray*}$

Stationary Points.

The stationary points are given by solving $g'(x)=0$ .

$g'(x)=0 \Rightarrow \simplify{{-a}*(x-{b})(x+{b})=0} \Rightarrow x=\var{b} \mbox{ or } x=\var{-b}$

We see that both stationary points are in the inerval $I$ .

The second derivative can be found by applying the quotient rule to the derivative of $g(x)$ and we obtain:

Using the quotient rule for differentiation we see that

$\begin{eqnarray*}g''(x)&=&\simplify[std]{({-2*a}*x*(x^2+{b^2})^2+{4*a}*x*(x^2-{b^2})(x^2+{b^2}))/(x^2+{b^2})^4}\\ &=&\simplify[std]{({2*a}*x*(x^2-{3*b^2}))/(x^2+{b^2})^3} \end{eqnarray*}$

The nature of the stationary points are determined by evaluating $g''(x)$ at the stationary points.

For $x= \var{lma}$ we have:

$g''(\var{lma})= \simplify[std]{-{abs(a)}/{2*b^3}} \lt 0$

Hence is a local maximum.

Evaluating the function at $x=\var{lma}$ gives $g(\var{lma})=\var{valmax}$ to 3 decimal places.

For $x= \var{lmi}$ we have:

$g''(\var{lmi})= \simplify[std]{{abs(a)}/{2*b^3}} \gt 0$

Hence is a local minimum.

Evaluating the function at $x=\var{lmi}$ gives $g(\var{lmi})=\var{valmin}$ to 3 decimal places.

The values of $g$ at the endpoints are:

$g(\var{c})=\var{valbegin}$ and $g(\var{d})=\var{valend}$ to 3 decimal places.

Global Maximum and Minimum

Since $g$ has a finite limit of $0$ as $x \rightarrow \pm\infty$ and we have that $0$ lies between the local minimum value $\var{valmin}$ and the local maximum value $\var{valmax}$ (and these occur at values in $I$ ).

then:

Global Maximum: The local maximum of $g$ we have found at $x=\var{lma} \in I$ must be a global maximum and similarly,

Global Minimum: The local minimum of $g$ we have found at $x=\var{lmi} \in I$ must be a global minimum.

So we have shown

$\forall x \in \mathbb{R},\;\;\var{valmin} \le g(x) \le \var{valmax}$

(all to 3 decimal places).

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Taxonomy: mathcentre

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Christian Lawson-Perfect 10 years, 2 months ago

Bill Foster 55 years, 4 months ago

Error in variable testing condition

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Generated value: `integer`

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Stationary Points.

Global Maximum and Minimum

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Max and Min 5

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 10 years, 2 months ago

Bill Foster 55 years, 4 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Choose one from a list

Advanced settings

Stationary Points.

Global Maximum and Minimum

Generated value: `integer`