Paul 's copy of Max and Min 4

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Paul Howes

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Paul Howes 8 years, 9 months ago

Created this as a copy of Max and Min 4.

Name	Status	Author	Last Modified
Max and Min 4	Ready to use	Bill Foster	01/06/2016 09:47
Paul 's copy of Max and Min 4	Ready to use	Paul Howes	11/06/2016 19:09
Paul 's copy of Max and Min 4	draft	Xiaodan Leng	11/07/2019 05:56

There are 8 other versions that do you not have access to.

Question statement

Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be the function given by:

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

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	-2
b	integer	8
valmin	number	-0.125
valmax	number	0.125
lmi	integer	8
s	integer	-1
tol	number	0.001
lma	integer	-8

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.

Can an exam override the value of this variable?

Generated value: `integer`

-2

← Depends on:

→ Used by:

lma
lmi
valmin

Statement
Advice
"Part b)" → "Unnamed gap" - Correct answer
"Part e)" → "Gap 0." - Correct answer
"Part f)" → "Gap 1." - Choice text

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
1. Gap Gap 0. Mathematical expression
  Add
2. Gap Gap 1. Mathematical expression
  Add
Part e) Gap-fill
1. Gap Gap 0. Mathematical expression
  Add
2. Gap Gap 1. Mathematical expression
  Add
3. Gap Gap 2. Number entry
  Add
4. Gap Gap 3. Mathematical expression
  Add
5. Gap Gap 4. Number entry
  Add
Part f) Gap-fill

Choose one from a list

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

Is $g(x)$ continuous at all points of $\mathbb{R}$ ?

Choose Yes or No.

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Advanced settings

Selection type

Number of display columns

(not displayed in columns)

Custom marking matrix?

Yes

Marks: Distractor Message:
No

Marks: Distractor Message:

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 $g'(x)=0$ .

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

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 Limits.

If we divide $g(x)$ top and bottom by $x^2$ (OK as $x \neq 0$ at any time) we obtain:

$g(x)=\simplify[std]{({a}/x)/(1+{b^2}/x^2)}$

Then using the fact that $\displaystyle \frac{1}{x}$ and $\displaystyle \frac{1}{x^2}$ both tend to $0$ as $x \rightarrow \pm\infty$ we see that

$\displaystyle \lim_{x \to \infty}g(x)=\frac{0}{1}=0$ and similarly

$\lim_{x \to -\infty}g(x)=0$

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 $\var{valmin}$ and the local maximum $\var{valmax}$

Then:

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

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

So we have shown

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

Give a worked solution to the whole question.

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Part	Test	Passed?
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Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Paul Howes 8 years, 9 months ago

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

Choose one from a list

Advanced settings

Stationary Points.

The Limits.

Global Maximum and Minimum

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Paul 's copy of Max and Min 4

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Paul Howes 8 years, 9 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Choose one from a list

Advanced settings

Stationary Points.

The Limits.

Global Maximum and Minimum

Generated value: `integer`