Single Absolute Value

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Johnny Yi 6 years, 3 months ago

Created this as a copy of Caitríona's copy of Inequalities that involve a single absolute value.

Name	Status	Author	Last Modified
Inequalities involving a rational function/hyperbola	Ready to use	Ben Brawn	08/08/2017 08:15
Inequalities that involve a single absolute value	Ready to use	Ben Brawn	28/08/2017 12:44
Caitríona's copy of Inequalities that involve a single absolute value	draft	Caitríona Ní Shé	02/12/2020 15:42
Caitríona's copy of Caitríona's copy of Inequalities that involve a single absolute value	draft	Caitríona Ní Shé	02/12/2020 15:42
Inequalities que involucra valor absoluto	draft	Luis Hernandez	17/01/2019 18:46
Single Absolute Value	draft	Johnny Yi	24/01/2019 02:04
Inequalities that involve a single absolute value	draft	Xiaodan Leng	11/07/2019 01:35

There are 6 other versions that do you not have access to.

Question statement

You are given the inequality

$|\simplify{{a}x+{b}}| \var{latex(sym)} \var{c}.$

Give any introductory information the student needs.

			Name	Type	Generated Value

choice	integer	3
sym	string	<
a	integer	-4
c	integer	2
d	integer	4
b	integer	10
leadCoeff	number	-1
prod	number	-3
one	boolean	true
rootList	list	[ 2, 3 ]
r1	rational	2
r2	rational	3
strict	boolean	true
backChoice	integer	1
backsym	string	>
extremeY	rational	1/4
backprod	number	-1
dr	rational	1
hole	rational	-5/2
nonhole	rational	-24/7

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Name

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Value

xxxxxxxxxx
 
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Description

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Generated value: `integer`

→ Used by:

backChoice
one
prod
strict
sym

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Parts

Gap-fill
1. Gap Gap 0. Choose one from a list
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Choose one from a list
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4. Gap Gap 3. Choose one from a list
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5. Gap Gap 4. Number entry
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6. Gap Gap 5. Choose one from a list
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7. Gap Gap 6. Number entry
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8. Gap Gap 7. Choose one from a list
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9. Gap Gap 8. Number entry
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Gap-fill

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

The solution to the above inequality

is made up of Gap 0.

Given by the condition Gap 1. Gap 2. $x$ Gap 3. Gap 4..

Given by the condition, $x$ Gap 5. Gap 6. or $x$ Gap 7. Gap 8..

Advice

Short-cut Method

For questions of this type (where the absolute value is 'less than' or 'less than and equal to' something positive) we can set out our working in a shorter way:

\begin{alignat}{2} &&|\simplify{{a}x+{b}}| &\var{latex(sym)} \var{c}\\ \var{-c} &\var{latex(sym)}&\simplify{{a}x+{b}}&\var{latex(sym)}\var{c}\\ \var{-c-b} &\var{latex(sym)} &\simplify{{a}x}\quad&\var{latex(sym)}\var{c-b}\\ \var[fractionNumbers]{r1}&\var{latex(sym)}&x\quad&\var{latex(sym)}\var[fractionNumbers]{r2}.\end{alignat}

\begin{alignat}{2} &&|\simplify{{a}x+{b}}| &\var{latex(sym)} \var{c}\\ \var{-c} &\var{latex(sym)}&\simplify{{a}x+{b}}&\var{latex(sym)}\var{c}\\ \var{-c-b} &\var{latex(sym)} &\simplify{{a}x}\quad&\var{latex(sym)}\var{c-b}\\ \var[fractionNumbers]{r2}&\var{latex(backsym)}&x\quad&\var{latex(backsym)}\var[fractionNumbers]{r1}\\\var[fractionNumbers]{r1}&\var{latex(sym)}&x\quad&\var{latex(sym)}\var[fractionNumbers]{r2}.\end{alignat}

Cases Method

Recall that the absolute value is defined as the piecewise function \[|x|=\begin{cases}x, &\text{ for } x\ge 0\\ -x, &\text{ for } x<0. \end{cases}\]

This means that $|\simplify{{a}x+{b}}|$ is actually $\simplify{{a}x+{b}}$ when $\simplify{{a}x+{b}}\ge 0$ but it is $-(\simplify{{a}x+{b}})$ when $\simplify{{a}x+{b}}<0$. We have two cases:

Case 1: $\simplify{{a}x+{b}}\ge 0$

Rearranging $\simplify{{a}x+{b}}\ge 0$ for $x$ gives $x \ge \simplify[fractionNumbers]{{-b/a}}$ $x \le \simplify[fractionNumbers]{{-b/a}}$ and so case 1 is only relevant for $x \ge \simplify[fractionNumbers]{{-b/a}}$ $x \le \simplify[fractionNumbers]{{-b/a}}$.

In case 1, our inequality is simply $\simplify{{a}x+{b}} \var{latex(sym)} \var{c}$, and so rearranging it for $x$ gives $x \var{latex(sym)} \simplify[fractionNumbers,simplifyFractions,!unitDenominator]{{(c-b)/a}}$ $x \var{latex(backsym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$.

So in conclusion for case 1, we require that $x \ge \simplify[fractionNumbers]{{-b/a}}$ $x \le \simplify[fractionNumbers]{{-b/a}}$ and $x \var{latex(sym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $x \var{latex(backsym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ which is the same as $\simplify[fractionNumbers]{-{b/a}}\le x\var{latex(sym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}\var{latex(sym)} x\le \simplify[fractionNumbers]{{-b/a}}$ $x\var{latex(sym)}\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$ $x\var{latex(backsym)}\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(c-b)/a}}$.

Case 2: $\simplify{{a}x+{b}}< 0$

Rearranging $\simplify{{a}x+{b}}< 0$ for $x$ gives $x < \simplify[fractionNumbers]{{-b/a}}$ $x > \simplify[fractionNumbers]{{-b/a}}$ and so case 2 is only relevant for $x < \simplify[fractionNumbers]{{-b/a}}$ $x > \simplify[fractionNumbers]{{-b/a}}$.

In case 2, our inequality is actually $-(\simplify{{a}x+{b}}) \var{latex(sym)} \var{c}$, and so rearranging it for $x$ gives $x \var{latex(backsym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x \var{latex(sym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$.

So in conclusion for case 2, we require that $x < \simplify[fractionNumbers]{-{b/a}}$ $x > \simplify[fractionNumbers]{-{b/a}}$ and $x \var{latex(backsym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x \var{latex(sym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ which is the same as $\simplify[fractionNumbers]{{-(b+c)/a}}\var{latex(sym)} x< \simplify[fractionNumbers,simplifyFractions,unitDenominator]{{-b}/{a}}$ $\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{-b/a}}< x \var{latex(sym)} \simplify[fractionNumbers,simplifyFractions,unitDenominator]{-{(b+c)/a}}$ $x\var{latex(backsym)}\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$ $x\var{latex(sym)}\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{(-c-b)/a}}$.

Therefore the solution to our original inequality $|\simplify{{a}x+{b}}| \var{latex(sym)} \var{c}$ is

\[\var[fractionNumbers]{r1}<x<\var[fractionNumbers]{r2}.\]

\[\var[fractionNumbers]{r1}\le x \le\var[fractionNumbers]{r2}.\]

\[x< \var[fractionNumbers]{r1}, \, x> \var[fractionNumbers]{r2}. \]

\[x\le \var[fractionNumbers]{r1}, \, x\ge\var[fractionNumbers]{r2}. \]

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Taxonomy: mathcentre

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Johnny Yi 6 years, 3 months ago

Error in variable testing condition

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Generated value: `integer`

Parts

Gap-fill

Short-cut Method

Cases Method

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Single Absolute Value

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

History

Comment

Checkpoint description

Johnny Yi 6 years, 3 months ago

Error in variable testing condition

Error

Warning

Generated value: integer

Parts

Gap-fill

Short-cut Method

Cases Method

Generated value: `integer`