Exam Templates - Revision history

Grad: Replaced content with "We should put generic templates here, nothing class specific We should probably create a course directory that will house class specific resources"

2015-03-16T20:43:49Z

Replaced content with "We should put generic templates here, nothing class specific We should probably create a course directory that will house class specific resources"

Grad at 16:53, 16 March 2015

2015-03-16T16:53:37Z

WikiSysop: Created page with "We should put generic templates here, nothing class specific We should probably create a course directory that will house class specific resources"

2015-03-10T00:13:55Z

Created page with "We should put generic templates here, nothing class specific We should probably create a course directory that will house class specific resources"

New page

We should put generic templates here, nothing class specific

We should probably create a course directory that will house class specific resources

← Older revision		Revision as of 20:43, 16 March 2015
Line 2:		Line 2:

	We should probably create a course directory that will house class specific resources		We should probably create a course directory that will house class specific resources
−
−	~~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>)~~
−	\|}

← Older revision		Revision as of 16:53, 16 March 2015
Line 2:		Line 2:

	We should probably create a course directory that will house class specific resources		We should probably create a course directory that will house class specific resources
		+
		+	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>)
		+	\|}