Graphs and derivatives

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Paul Verheyen

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Created this as a copy of no graph Graphs and derivatives.

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Question statement

Consider the function $f: \mathbb{R} \to \mathbb{R} : x \mapsto \ln{ \frac{\var{c} \cdot x-\var{a}}{x-\var{b}} }\, .$

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	17
b	integer	5
c	integer	3
d	integer	2
nulp1	integer	5
nulp2	rational	17/3
teken1	string	\|
teken2	string	0
ordeafg1	string	>
stijgafg1	string	increases
opI1tekenafg2	string	+
opI2tekenafg2	string	-
tekenafg1	string	+

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
b*c+d

JME function reference

Description

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

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Generated value: `integer`

← Depends on:

This variable doesn't seem to be used anywhere.

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Mathematical expression
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Part c) Gap-fill
1. Gap Mathematical expression
  Add
Part d) Choose one from a list
Part e) Choose one from a list

Gap-fill

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

The domain of $ f $ is given by:

$ \left ] - \infty , \right. $ Gap 0. $ \left. \right[ \quad \cup \quad \left] \right. $ Gap 1. $ \left. \, , +\infty \right[ $

Advice

a) In order to obtain the domain of $ f $ , we need to solve the inequality \[ \frac{\var{c}\cdot x-\var{a}}{x-\var{b}} > 0 \]

Since the sign of $ \frac{\var{c}\cdot x-\var{a}}{x-\var{b}} $ is given by

\[
\begin{array}{ c | c c c c c}
x   &            & \simplify[all,fractionNumbers]{{nulp1}} & & \simplify[all,fractionNumbers]{{nulp2}} &    \\
    \hline
\frac{\var{c}\cdot x-\var{a}}{x-\var{b}} & + & \var{teken1} & - & \var{teken2} & +    \\
\end{array}
\]
we get
\[ dom f = \left ] - \infty , \simplify[all,fractionNumbers]{{nulp1}} \right[ \quad \cup \quad \left] \simplify[all,fractionNumbers]{{nulp2}} , +\infty \right[ .\]

b) Since $ \mathrm{D} \ln x = \frac{1}{x} $ , the chain rule leads to
\[ \mathrm{D} \ln{\left( \frac{\var{c}\cdot x-\var{a}}{x-\var{b}}\right) } = \frac {x-\var{b}} {\var{c}\cdot x-\var{a}} \cdot \mathrm{D} \frac{\var{c}\cdot x-\var{a}}{x-\var{b}} , \]
i.e.
\begin{align} f'(x) &= \frac {x-\var{b}} {\var{c}\cdot x-\var{a}} \cdot \frac{(x-\var{b})\cdot \var{c} - (\var{c}\cdot x-\var{a}) \cdot 1}{(x-\var{b})^2} \\
&= \frac{\var{c}\cdot (x-\var{b}) - (\var{c}\cdot x-\var{a}) }{(\var{c}\cdot x-\var{a}) \cdot (x-\var{b})} \\
&= \frac{\var{a-b*c} }{(\var{c}\cdot x-\var{a}) \cdot (x-\var{b})}
\end{align}

Remark: you can get the same result (making abstraction of the domain) by using
\[ \ln{\left( \frac{\var{c}\cdot x-\var{a}}{x-\var{b}}\right) } = \ln{\left( {\var{c}\cdot x-\var{a}}\right) } - \ln{\left( {x-\var{b}}\right) } , \]
such that
\begin{align}
\mathrm{D} \ln{\left( \frac{\var{c}\cdot x-\var{a}}{x-\var{b}}\right) } &= \mathrm{D} \left( \ln { \left( {\var{c}\cdot x-\var{a}} \right)} \right) - \mathrm{D} \left( \ln { \left( {x-\var{b}} \right)} \right) \\
&= \frac{1}{\var{c}\cdot x-\var{a}} \cdot \var{c}- \frac{1}{x-\var{b}}\\
&= \frac{ \var{c} \cdot ({x-\var{b}}) - (\var{c} \cdot x-\var{a}) }{(\var{c} \cdot x-\var{a}) \cdot (x-\var{b})} \\
&= \frac{\var{a-b*c} }{(\var{c} \cdot x-\var{a}) \cdot (x-\var{b})}
\end{align}

c) The second order derivative can be calculated as follows:
\[ \begin{align}
\mathrm{D} \frac{\var{a-b*c} }{(\var{c} \cdot x-\var{a}) \cdot (x-\var{b})} &= {\var{a-b*c}} \cdot \mathrm{D} {\left( (\var{c} \cdot x-\var{a}) \cdot (x-\var{b}) \right) ^{-1}}\\
&= {\var{-a+b*c}} \cdot {\left( (\var{c} \cdot x-\var{a}) \cdot (x-\var{b}) \right) ^{-2}} \cdot \mathrm{D} {\left( (\var{c} \cdot x-\var{a}) \cdot (x-\var{b}) \right) }\\
&= {\var{-a+b*c}} \cdot {\frac {1}{\left( (\var{c} \cdot x-\var{a}) \cdot (x+\var{b}) \right) ^{2}}} \cdot {\left( \var{c} \cdot (x-\var{b}) + (\var{c} \cdot x-\var{a}) \right) }\\
&= {\var{-a+b*c}} \cdot {\frac {1}{\left( (\var{c} \cdot x-\var{a}) \cdot (x+\var{b}) \right) ^{2}}} \cdot {\left( \var{2*c} \cdot x - \var{a+b*c} \right) }\\
&= \frac{{\var{-a+b*c}} \cdot {\left( \var{2*c} \cdot x - \var{a+b*c} \right) }}{ (\var{c} \cdot x-\var{a})^2 \cdot (x+\var{b})^2}
\end{align}
\]

d) On each subinterval$ f'(x) $ {ordeafg1} $ 0 $, which implies $ f $ {stijgafg1} on both $ I_- $ and $ I_+ $.

e) Since\[   \simplify[all,fractionNumbers]{{nulp1}} < \simplify[all,fractionNumbers]{{a+b*c}/{2*c}} < \simplify[all,fractionNumbers]{{nulp2}} \]
we find
\begin{array}{ c | c c c c c c c}
x   &            & \simplify[all,fractionNumbers]{{nulp1}} & & \simplify[all,fractionNumbers]{{a+b*c}/{2*c}} & & \simplify[all,fractionNumbers]{{nulp2}} &    \\
    \hline
f''(x) & \var{{opI1tekenafg2}} & | & // & // & // & | & \var{{opI2tekenafg2}}
\end{array}
Consequently the curvature of $ f$ can be determined from
\begin{array}{ c | c c c c c c c}
x   &            & \simplify[all,fractionNumbers]{{nulp1}} & & \simplify[all,fractionNumbers]{{a+b*c}/{2*c}} & & \simplify[all,fractionNumbers]{{nulp2}} &    \\
    \hline
f''(x) & \var{{opI1tekenafg2}} & | & // & // & // & | & \var{{opI2tekenafg2}}    \\
    \hline
f'(x) & \var{{tekenafg1}} & | & // & // & // & | & \var{{tekenafg1}}
\end{array}

{GetekendeGrafiek()}

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Paul Verheyen 4 years, 7 months ago

Paul Verheyen 4 years, 7 months ago

Paul Verheyen 4 years, 7 months ago

Consider the function $f: \mathbb{R} \to \mathbb{R} : x \mapsto \ln{ \frac{\var{c} \cdot x-\var{a}}{x-\var{b}} }\, .$

Error in variable testing condition

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Warning

Generated value: `integer`

Parts

Gap-fill

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Graphs and derivatives

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Paul Verheyen 4 years, 7 months ago

Paul Verheyen 4 years, 7 months ago

Paul Verheyen 4 years, 7 months ago

Consider the function f:R→R:x↦ln{c}⋅x−{a}x−{b}. f: \mathbb{R} \to \mathbb{R} : x \mapsto \ln{ \frac{\var{c} \cdot x-\var{a}}{x-\var{b}} }\, .

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Consider the function $f: \mathbb{R} \to \mathbb{R} : x \mapsto \ln{ \frac{\var{c} \cdot x-\var{a}}{x-\var{b}} }\, .$

Generated value: `integer`