Difference between revisions of "004 Sample Final A, Problem 1"
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Kayla Murray (talk | contribs) (Created page with "Find <math>f^{-1}(x)</math> for <math style = "vertical-align: -17%;>f(x) = \frac{3x-1}{4x+2}</math>") |
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− | Find <math>f^{-1}(x)</math> for <math style = " | + | <span class="exam"> Find <math>f^{-1}(x)</math> for <math>f(x) = \frac{3x-1}{4x+2}</math> |
+ | |||
+ | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Foundations | ||
+ | |- | ||
+ | | How would you find the inverse for a simpler function like <math>f(x)=2x+4?</math> | ||
+ | |- | ||
+ | |Answer: | ||
+ | |- | ||
+ | |You would replace <math>f(x)</math> with <math>y</math>. Then, switch <math>x</math> and <math>y</math>. Finally, we would solve for <math>y</math>. | ||
+ | |} | ||
+ | |||
+ | |||
+ | Solution: | ||
+ | |||
+ | {| class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 1: | ||
+ | |- | ||
+ | |We start by replacing <math>f(x)</math> with <math>y</math>. | ||
+ | |- | ||
+ | |This leaves us with <math>y=\frac{3x-1}{4x+2}</math> | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 2: | ||
+ | |- | ||
+ | |Now, we swap <math>x</math> and <math>y</math> to get <math>x=\frac{3y-1}{4y+2} </math>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 3: | ||
+ | |- | ||
+ | |Starting with <math>x=\frac{3y-1}{4y+2} </math>, we multiply both sides by <math>4y+2</math> to get | ||
+ | |- | ||
+ | |<math>x(4y+2)=3y-1</math>. | ||
+ | |- | ||
+ | |Now, we need to get all the <math>y</math> terms on one side. So, adding <math>1</math> and <math>-4xy</math> to both sides we get | ||
+ | |- | ||
+ | |<math> 2x+1=3y-4xy</math>. | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Step 4: | ||
+ | |- | ||
+ | |Factoring out <math>y</math>, we get <math> 2x+1=y(3-4x) </math>. Now, dividing by <math>(3-4x)</math>, we get | ||
+ | |- | ||
+ | |<math>\frac{2x+1}{3-4x}=y</math>. Replacing <math>y</math> with <math>f^{-1}(x)</math>, we arrive at the final answer | ||
+ | |- | ||
+ | |<math>f^{-1}(x)=\frac{2x+1}{3-4x}</math> | ||
+ | |} | ||
+ | |||
+ | {|class = "mw-collapsible mw-collapsed" style = "text-align:left;" | ||
+ | ! Final Answer: | ||
+ | |- | ||
+ | |<math>f^{-1}(x)=\frac{2x+1}{3-4x}</math> | ||
+ | |} | ||
+ | |||
+ | [[004 Sample Final A|<u>'''Return to Sample Exam</u>''']] |
Latest revision as of 21:08, 28 April 2015
Find for
Foundations |
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How would you find the inverse for a simpler function like |
Answer: |
You would replace with . Then, switch and . Finally, we would solve for . |
Solution:
Step 1: |
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We start by replacing with . |
This leaves us with |
Step 2: |
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Now, we swap and to get . |
Step 3: |
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Starting with , we multiply both sides by to get |
. |
Now, we need to get all the terms on one side. So, adding and to both sides we get |
. |
Step 4: |
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Factoring out , we get . Now, dividing by , we get |
. Replacing with , we arrive at the final answer |
Final Answer: |
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