Difference between revisions of "1967 AHSME Problems/Problem 14"
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Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us | Since we know that <math>y=f(x)</math>, we can solve for <math>y</math> in terms of <math>x</math>. This gives us | ||
| − | <math>y=\frac{ | + | <math>y=\frac{x}{1-x}</math> |
| − | <math>\Rightarrow y(1-x)= | + | <math>\Rightarrow y(1-x)=x</math> |
| − | <math>\Rightarrow y-yx= | + | <math>\Rightarrow y-yx=x</math> |
| − | <math>\Rightarrow yx | + | <math>\Rightarrow y=yx+x</math> |
| − | <math>\Rightarrow x | + | <math>\Rightarrow y=x(y+1)</math> |
| − | Therefore, we want to find the function with <math>y</math> that outputs <math>\frac{y | + | <math>\Rightarrow x=\frac{y}{y+1}</math> |
| − | Listing out the possible outputs from each of the given functions we get | + | |
| + | Therefore, we want to find the function with <math>y</math> that outputs <math>\frac{y}{y+1}</math> Listing out the possible outputs from each of the given functions we get | ||
<math>f\left(\frac{1}{y}\right)=\frac{1}{y-1}</math> | <math>f\left(\frac{1}{y}\right)=\frac{1}{y-1}</math> | ||
Revision as of 21:02, 29 May 2022
Problem
Let 
, 
. If 
, then 
can be expressed as
Solution
Since we know that , we can solve for 
in terms of . This gives us
Therefore, we want to find the function with that outputs 
Listing out the possible outputs from each of the given functions we get
Since 
the answer must be 
.
See also
| 1967 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
