Max and Min 6

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Christian Lawson-Perfect

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Christian Lawson-Perfect 9 years, 11 months ago

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Bill Foster 55 years, 2 months ago

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Name	Status	Author	Last Modified
Max and Min 6	Ready to use	Bill Foster	01/06/2016 09:47
Katie's copy of Max and Min 6	Ready to use	Katie Dexter	01/06/2016 09:47
Paul 's copy of Max and Min 6	Ready to use	Paul Howes	11/06/2016 19:09
Paul 's copy of Max and Min 6	draft	Xiaodan Leng	11/07/2019 05:56

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Question statement

Let $I=[\var{l},\var{m}]$ be an interval and let $g: I \rightarrow \mathbb{R}$ be the function given by:
$g(x)=\simplify{x^2/(x-{c})^({a}/{b})}$

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	9
c	decimal	dec("11")
b	integer	7
valmin	decimal	dec("20.416")
m	integer	34
l	integer	14
xma	integer	14
lmi	integer	-7
s	integer	-1
u	integer	88
valend	decimal	dec("20.519")
tol	number	0.001
l1	integer	14
valbegin	decimal	dec("47.733")
lma	integer	7
statpoint	number	30.8
xmi	integer	-7
valsd	decimal	dec("0.013")

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
if(b=3,random(1,2,4,5),if(b=5,random(1,2,3,4,6,7,8,9),random(2,3,4,5,6,8,9)))

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Description

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Generated value: `integer`

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Parts

Part a) Choose one from a list
Part b) Gap-fill
1. Gap Mathematical expression
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Part d) Gap-fill
Part e) Gap-fill
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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. Mathematical expression
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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 $\mathbb{R}$ as $\var{c} \notin I$ .

On differentiating we see that
$\begin{eqnarray*}g'(x)&=&\simplify{2*x/(x-{c})^({a}/{b}) - {a}*x^2/({b}(x-{c})^({a+b}/{b}))}\\ &=&\simplify{({2*b}x*(x-{c})-{a}*x^2)/({b}(x-{c})^({a+b}/{b}))}\\ &=&\simplify{(x*({2*b-a}x-{2*b*c}))/({b}(x-{c})^({a+b}/{b}))} \end{eqnarray*}$

Stationary Points.

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

$\displaystyle g'(x)=0 \Rightarrow \simplify{x*({2*b-a}x-{2*b*c})=0} \Rightarrow x=0 \mbox{ or } x=\simplify[std]{{2*b*c}/{2*b-a}}$

We see that $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}}$ is the only stationary point in $I$ .

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

Hence $r(x)=\simplify[std]{({a^2-3*a*b+2*b^2}*x^2+{4*b*c*(a-b)}*x+{2*c^2*b^2})}$

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

There is only one stationary point $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}}$ in $I$ and at that point we have:
$g''\left(\simplify[std]{{2*b*c}/{2*b-a}}\right)=\var{valsd} \gt 0$

Hence this point is a local minimum.

Evaluating at end points of the interval.

The values of $g$ at the endpoints are:

$g(\var{l})=\var{valbegin}$ and $g(\var{m})=\var{valend}$ , both to 3 decimal places.

Global Maximum and Minimum

Global Maximum: Since $g$ does not have a local maximum in the interval $I$ , it must take a global maximum value at one of the end points of the interval.

From the values found for $g(\var{l})$ and $g(\var{m})$ found above, we see that $x=\var{xma}$ is the global maximum for $g$ in the interval $I$ .

Global Minimum: The local minimum of $g$ given by $\displaystyle x=\simplify[std]{{2*b*c}/{2*b-a}} \in I$ is the only local minimum and must be a global minimum in $I$ .

Note that the global minimum value for $g$ on $I$ is:

$g\left(\simplify[std]{{2*b*c}/{2*b-a}}\right)=\var{valmin}$

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Taxonomy: mathcentre

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Christian Lawson-Perfect 9 years, 11 months ago

Bill Foster 55 years, 2 months ago

Error in variable testing condition

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Generated value: `integer`

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Stationary Points.

Evaluating at end points of the interval.

Global Maximum and Minimum

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Max and Min 6

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 9 years, 11 months ago

Bill Foster 55 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Choose one from a list

Advanced settings

Stationary Points.

Evaluating at end points of the interval.

Global Maximum and Minimum

Generated value: `integer`