008A Sample Final A, Question 1 - Revision history

MathAdmin at 06:56, 26 May 2015

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MathAdmin at 06:49, 26 May 2015

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MathAdmin at 06:46, 26 May 2015

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MathAdmin at 05:55, 23 May 2015

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MathAdmin at 05:52, 23 May 2015

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MathAdmin: Created page with "'''Question:''' Find $f^{-1}(x)$ for $f(x) = \log_3(x+3)-1$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Foundations |- |1) How..."

2015-04-28T21:03:59Z

Created page with "'''Question:''' Find <math>f^{-1}(x)</math> for <math>f(x) = \log_3(x+3)-1</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" ! Foundations |- |1) How..."

New page

'''Question:''' Find <math>f^{-1}(x)</math> for <math>f(x) = \log_3(x+3)-1</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations
|-
|1) How would you find the inverse for a simpler function like <math>f(x) = 3x + 5</math>?
|-
|2) How are <math>log_3(x)</math> and <math>3^x</math> related?
|-
|Answers:
|-
|1) you would replace f(x) by y, switch x and y, and finally solve for y.
|-
|2) By stating <math>y = \log_3(x)</math> we also get the following relation <math>x = 3^y</math>
|}

Solution:

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|We start by replacing f(x) with y.
|-
|This leaves us with <math>y = \log_3(x + 3) - 1</math>
|}

{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|Now we swap x and y to get <math>x = \log_3(y + 3) - 1</math>
|-
|In the next step we will solve for y.
|}

{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|Starting with <math>x = \log_3(y + 3) - 1</math>, we start by adding 1 to both sides to get
|-
|<math>x + 1 = \log_3(y + 3).</math> Now we will use the relation in Foundations 2) to swap the log for an exponential to get
|-
|<math>y + 3 = 3^{x+1}</math>. All we have to do is subtract 3 from both sides to yield the final answer
|}

{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
|-
|After subtracting 3 from both sides we get <math>y = 3^{x+1}-3</math>. Replacing y with <math>f^{-1}(x)</math> we arrive at the final answer that
|-
|<math>f^{-1}(x) = 3^{x+1} - 3</math>
|}

{|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|<math>f^{-1}(x) = 3^{x+1} - 3</math>
|}

[[008A Sample Final A|<u>'''Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Foundations: &nbsp;
+!Foundations: &nbsp;
 |-
 |1) How would you find the inverse for a simpler function like <math>f(x) = 3x + 5</math>?
@@ Line 20: / Line 20: @@
 {| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 1: &nbsp;
+!Step 1: &nbsp;
 |-
 |We start by replacing f(x) with y.
@@ Line 28: / Line 28: @@
 {|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 2: &nbsp;
+!Step 2: &nbsp;
 |-
 |Now we swap x and y to get <math>x = \log_3(y + 3) - 1</math>
@@ Line 36: / Line 36: @@
 {|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 3: &nbsp;
+!Step 3: &nbsp;
 |-
 |From <math>x = \log_3(y + 3) - 1</math>, we add 1 to both sides to get
@@ Line 54: / Line 54: @@
 {|class = "mw-collapsible mw-collapsed" style = "text-align:left;"
-! Final Answer: &nbsp;
+!Final Answer: &nbsp;
 |-
 |<math>f^{-1}(x) = 3^{x+1} - 3</math>

@@ Line 38: / Line 38: @@
 ! Step 3:
 |-
-|Starting with <math>x = \log_3(y + 3) - 1</math>, we start by adding 1 to both sides to get
+|From <math>x = \log_3(y + 3) - 1</math>, we add 1 to both sides to get
 |-
 |<math>x + 1 = \log_3(y + 3).</math> Now we will use the relation in Foundations 2) to swap the log for an exponential to get
 |-
-|<math>y + 3 = 3^{x+1}</math>. All we have to do is subtract 3 from both sides to yield the final answer
+|<math>y + 3 = 3^{x+1}</math>.
 |}

@@ Line 7: / Line 7: @@
 |1) How would you find the inverse for a simpler function like <math>f(x) = 3x + 5</math>?
 |-
-|2) How are <math>log_3(x)</math> and <math>3^x</math> related?
+|2) How do you remove the <math>\log_3</math> in the following equation: <math>\log_3(x) = y?</math>
 |-
 |Answers:
@@ Line 13: / Line 13: @@
 |1) you would replace f(x) by y, switch x and y, and finally solve for y.
 |-
-|2) By stating <math>y = \log_3(x)</math> we also get the following relation <math>x = 3^y</math>
+|2) By the definition of <math>\log_3</math> when we write the equation <math>y = \log_3(x)</math> we mean y is the number such that <math>3^y = x</math>
 |}