Difference between revisions of "008A Sample Final A, Question 1"
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(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...") |
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Foundations | + | !Foundations: |
|- | |- | ||
|1) How would you find the inverse for a simpler function like <math>f(x) = 3x + 5</math>? | |1) How would you find the inverse for a simpler function like <math>f(x) = 3x + 5</math>? | ||
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− | |2) How | + | |2) How do you remove the <math>\log_3</math> in the following equation: <math>\log_3(x) = y?</math> |
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|Answers: | |Answers: | ||
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|1) you would replace f(x) by y, switch x and y, and finally solve for y. | |1) you would replace f(x) by y, switch x and y, and finally solve for y. | ||
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− | |2) By | + | |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> |
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{| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step 1: | + | !Step 1: |
|- | |- | ||
|We start by replacing f(x) with y. | |We start by replacing f(x) with y. | ||
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{|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step 2: | + | !Step 2: |
|- | |- | ||
|Now we swap x and y to get <math>x = \log_3(y + 3) - 1</math> | |Now we swap x and y to get <math>x = \log_3(y + 3) - 1</math> | ||
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{|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step 3: | + | !Step 3: |
|- | |- | ||
− | | | + | |From <math>x = \log_3(y + 3) - 1</math>, we add 1 to both sides to get |
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|<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>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>. | + | |<math>y + 3 = 3^{x+1}</math>. |
|} | |} | ||
{|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Step 4: | + | ! Step 4: |
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|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 | |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 | ||
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{|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
− | ! Final Answer: | + | !Final Answer: |
|- | |- | ||
|<math>f^{-1}(x) = 3^{x+1} - 3</math> | |<math>f^{-1}(x) = 3^{x+1} - 3</math> |
Latest revision as of 22:56, 25 May 2015
Question: Find for
Foundations: |
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1) How would you find the inverse for a simpler function like ? |
2) How do you remove the in the following equation: |
Answers: |
1) you would replace f(x) by y, switch x and y, and finally solve for y. |
2) By the definition of when we write the equation we mean y is the number such that |
Solution:
Step 1: |
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We start by replacing f(x) with y. |
This leaves us with |
Step 2: |
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Now we swap x and y to get |
In the next step we will solve for y. |
Step 3: |
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From , we add 1 to both sides to get |
Now we will use the relation in Foundations 2) to swap the log for an exponential to get |
. |
Step 4: |
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After subtracting 3 from both sides we get . Replacing y with we arrive at the final answer that |
Final Answer: |
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