Pages For External Wiki - Revision history

Matt Lee at 18:36, 30 May 2015

2015-05-30T18:36:17Z

Matt Lee at 17:40, 12 May 2015

2015-05-12T17:40:23Z

Kayla Murray at 23:07, 28 April 2015

2015-04-28T23:07:18Z

Grad at 01:20, 21 March 2015

2015-03-21T01:20:29Z

Grad at 19:50, 20 March 2015

2015-03-20T19:50:20Z

Grad: Replaced content with "=Precalculus= =Calculus="

2015-03-17T00:13:17Z

Replaced content with "=Precalculus= =Calculus="

Grad: Created page with "=Precalculus= Presented below is the template for one of the sample questions Parker presented during 302. 2. Question Statement {| class="mw-collapsible mw-collapsed" ! Fo..."

2015-03-16T20:53:48Z

Created page with "=Precalculus= Presented below is the template for one of the sample questions Parker presented during 302. 2. Question Statement {| class="mw-collapsible mw-collapsed" ! Fo..."

New page

=Precalculus=

Presented below is the template for one of the sample questions Parker presented during 302.

2. Question Statement

{| class="mw-collapsible mw-collapsed"
! Foundations
|-
|'''The foundations:'''
|-
| Provide an short explanation about the prerequisite material required to complete this problem.
|}

Solution:

{| class = "mw-collapsible mw-collapsed"
! Step 1:
|-
|Provide as many steps as necessary to complete the problem.
|-
|The steps should split the solution based on the foundation topics
|}

{| class = "mw-collapsible mw-collapsed"
! Step 2:
|-
|Additional step provided to make the template longer
|}

'''Example'''

2. Find the domain of the following function. Your answer should use interval notation.
f(x) = <math style="vertical-align:-17%;">\displaystyle{\frac{1}{\sqrt{x^2-x-2}}}</math>

{| class="mw-collapsible mw-collapsed"
! Foundations
|-
|'''The foundations:'''
|-
| What is the domain of g(x) = <math style="vertical-align:-17%;">\frac{1}{x}</math>?
|-
|The function is undefined if the denominator is zero, so x <math>\neq </math>0.
|-
|Rewriting"x <math>\neq </math>0" in interval notation( <math>-\infty</math>, 0) <math>\cup</math>(0, <math>\infty</math>)
|-
|What is the domain of h(x) = <math>\sqrt{x}</math>?
|-
|The function is undefined if wwe have a negative number inside the square root, so x <math>\ge</math> 0
|}

Solution:

{| class = "mw-collapsible mw-collapsed"
! Step 1:
|-
|Factor <math style="vertical-align:-17%;">x^2 - x - 2</math>
|-
|<math style="vertical-align:-17%;">x^2 - x - 2 = (x + 1) (x - 2)</math>
|-
| So we can rewrite f(x) as f(x) = <math style="vertical-align:-17%;">\displaystyle{\frac{1}{\sqrt{(x+1)(x-2)}}}</math>
|}

{|class = "mw-collapsible mw-collapsed"
! Step 2:
|-
|When does the denominator of f(x) = 0?
|-
|<math>sqrt{(x + 1)(x - 2)} = 0</math>
|-
|(x + 1)(x - 2) = 0
|-
|(x + 1) = 0 or (x - 2) = 0
|-
|x = -1 or x = 2
|-
|So, since the function is undefiend when the denominator is zero, x <math>\neq</math> -1 and x <math>\neq</math> 2
|}

{|class = "mw-collapsible mw-collapsed"
! Step 3:
|-
|What is the domain of h(x) = <math style="vertical-align:-17%;">\sqrt{(x + 1)(x - 2)}</math>
|-
|critical points: x = -1, x = 2
|-
|Test points:
|-
|x = -2: (-2 + 1)(-2 - 2): (-1)(-4) = 4 > 0
|-
|x = 0: (0 + 1)(0 - 2) = -2 < 0
|-
|x = 3: (3 + 1)(3 - 2): 4*1 = 4 > 0
|-
|So the domain of h(x) is (<math>-\infty</math>, -1] <math>\cup</math> [2, <math>\infty</math>)
|}

{|class = "mw-collapsible mw-collapsed"
! Step 4:
|-
|Take the intersection (i.3. common points) of Steps 2 and 3. ( <math>- \infty</math>, -1) <math>\cup</math> (2, <math>\infty</math>)
|}

=Calculus=

← Older revision		Revision as of 18:36, 30 May 2015
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	=Calculus=		=Calculus=
		+	[[022 Sample Final A]]

@@ Line 6: / Line 6: @@
 Self-explanatory <br>
 [[Matt Lee's pages]] <br>
 [[Kayla's pages]]
 ===Completed===

@@ Line 5: / Line 5: @@
 ===In Progress===
 Self-explanatory <br>
-[[Matt Lee's pages]]
+[[Matt Lee's pages]] <br>
 ===Completed===
 Once you think your page is ready to be moved to the external wiki move your link here and contact one of the Pre-calculus project members. We will let you know if it needs to be changed. Once it is approved we will let you know. Make sure you follow up to make sure you get credit for your work

@@ Line 4: / Line 4: @@
 ===In Progress===
-Self-explanatory
+Self-explanatory <br>
 [[Matt Lee's pages]]
 ===Completed===

← Older revision		Revision as of 19:50, 20 March 2015
Line 1:		Line 1:
	=Precalculus=		=Precalculus=
		+	===Good Ideas/Help===
		+	This is a section for things you want to see as part of the external wiki, Precalculus section. Provide a piped link to your page where you describe what you want to build/need help with
		+
		+	===In Progress===
		+	Self-explanatory
		+	[[Matt Lee's pages]]
		+	===Completed===
		+	Once you think your page is ready to be moved to the external wiki move your link here and contact one of the Pre-calculus project members. We will let you know if it needs to be changed. Once it is approved we will let you know. Make sure you follow up to make sure you get credit for your work


	=Calculus=		=Calculus=

← Older revision		Revision as of 00:13, 17 March 2015
Line 1:		Line 1:
	=Precalculus=		=Precalculus=
−
−	~~Presented below is the template for one of the sample questions Parker presented during 302.~~
−
−	~~2. Question Statement~~
−
−	~~{\| class="mw-collapsible mw-collapsed"~~
−	~~! Foundations~~
−	\|-
−	~~\|'''The foundations:'''~~
−	\|-
−	~~\| Provide an short explanation about the prerequisite material required to complete this problem.~~
−	\|}
−
−
−	~~Solution:~~
−
−	~~{\| class = "mw-collapsible mw-collapsed"~~
−	~~! Step 1:~~
−	\|-
−	~~\|Provide as many steps as necessary to complete the problem.~~
−	\|-
−	~~\|The steps should split the solution based on the foundation topics~~
−	\|}
−
−	~~{\| class = "mw-collapsible mw-collapsed"~~
−	~~! Step 2:~~
−	\|-
−	~~\|Additional step provided to make the template longer~~
−	\|}
−
−
−
−	~~'''Example'''~~
−
−	~~2. Find the domain of the following function. Your answer should use interval notation.~~
−	~~f(x) = <math style="vertical-align:-17%;">\displaystyle{\frac{1}{\sqrt{x^2-x-2}}}</math>~~
−
−	~~{\| class="mw-collapsible mw-collapsed"~~
−	~~! Foundations~~
−	\|-
−	~~\|'''The foundations:'''~~
−	\|-
−	~~\| What is the domain of g(x) = <math style="vertical-align:-17%;">\frac{1}{x}</math>?~~
−	\|-
−	~~\|The function is undefined if the denominator is zero, so x <math>\neq </math>0.~~
−	\|-
−	~~\|Rewriting"x <math>\neq </math>0" in interval notation( <math>-\infty</math>, 0) <math>\cup</math>(0, <math>\infty</math>)~~
−	\|-
−	~~\|What is the domain of h(x) = <math>\sqrt{x}</math>?~~
−	\|-
−	~~\|The function is undefined if wwe have a negative number inside the square root, so x <math>\ge</math> 0~~
−	\|}
−
−
−	~~Solution:~~
−
−	~~{\| class = "mw-collapsible mw-collapsed"~~
−	~~! Step 1:~~
−	\|-
−	~~\|Factor <math style="vertical-align:-17%;">x^2 - x - 2</math>~~
−	\|-
−	~~\|<math style="vertical-align:-17%;">x^2 - x - 2 = (x + 1) (x - 2)</math>~~
−	\|-
−	~~\| So we can rewrite f(x) as f(x) = <math style="vertical-align:-17%;">\displaystyle{\frac{1}{\sqrt{(x+1)(x-2)}}}</math>~~
−	\|}
−
−	~~{\|class = "mw-collapsible mw-collapsed"~~
−	~~! Step 2:~~
−	\|-
−	~~\|When does the denominator of f(x) = 0?~~
−	\|-
−	~~\|<math>sqrt{(x + 1)(x - 2)} = 0</math>~~
−	\|-
−	~~\|(x + 1)(x - 2) = 0~~
−	\|-
−	~~\|(x + 1) = 0 or (x - 2) = 0~~
−	\|-
−	~~\|x = -1 or x = 2~~
−	\|-
−	~~\|So, since the function is undefiend when the denominator is zero, x <math>\neq</math> -1 and x <math>\neq</math> 2~~
−	\|}
−
−	~~{\|class = "mw-collapsible mw-collapsed"~~
−	~~! Step 3:~~
−	\|-
−	~~\|What is the domain of h(x) = <math style="vertical-align:-17%;">\sqrt{(x + 1)(x - 2)}</math>~~
−	\|-
−	~~\|critical points: x = -1, x = 2~~
−	\|-
−	~~\|Test points:~~
−	\|-
−	~~\|x = -2: (-2 + 1)(-2 - 2): (-1)(-4) = 4 > 0~~
−	\|-
−	~~\|x = 0: (0 + 1)(0 - 2) = -2 < 0~~
−	\|-
−	~~\|x = 3: (3 + 1)(3 - 2): 4*1 = 4 > 0~~
−	\|-
−	~~\|So the domain of h(x) is (<math>-\infty</math>, -1] <math>\cup</math> [2, <math>\infty</math>)~~
−	\|}
−
−	~~{\|class = "mw-collapsible mw-collapsed"~~
−	~~! Step 4:~~
−	\|-
−	~~\|Take the intersection (i.3. common points) of Steps 2 and 3. ( <math>- \infty</math>, -1) <math>\cup</math> (2, <math>\infty</math>)~~
−	\|}


	=Calculus=		=Calculus=