Revision as of 17:13, 16 March 2015

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 =Precalculus=
-Presented below is the template for one of the sample questions Parker presented during 302.
-. 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'''
-. 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=

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Revision as of 17:13, 16 March 2015

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