SOLUTION: The function f:x->px+q, xER, is such that f^-1 (6)=3 and f^-1 (29)=-2. Find f^-1 (27). I have spent a long time trying to figure this out and I can't get it. The answer giv

Algebra ->  Functions -> SOLUTION: The function f:x->px+q, xER, is such that f^-1 (6)=3 and f^-1 (29)=-2. Find f^-1 (27). I have spent a long time trying to figure this out and I can't get it. The answer giv      Log On


   

Question 941055: The function f:x->px+q, xER, is such that f^-1 (6)=3 and f^-1 (29)=-2.
Find f^-1 (27).
I have spent a long time trying to figure this out and I can't get it.
The answer given in the textbook is 6.
I have found the inverse function to be f^(-1): 1/p (x-q) so I am guessing that means I subtract or add something to x and then divide it.
Any help on this would be greatly appreciated.
Thanks
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
f%5E%28-1%29%286%29=3 so f%283%29=6
f%5E%28-1%29%2829%29=-2 so f%28-2%29=29
f%28x%29=px%2Bq
1.f%283%29=3p%2Bq=6
2.f%28-2%29=-2p%2Bq=29
Subtract eq. 1 from eq. 2,
-2p%2Bq-3p-q=29-6
-5p=23
p=-23%2F5
Then use either equation to solve for q
3%28-23%2F5%29%2Bq=6
-69%2F5%2Bq=30%2F5
q=30%2F5%2B69%2F5
q=99%2F5
So then,
f%28x%29=%28-23%2F5%29x%2B99%2F5
-%2823%2F5%29x%2B99%2F5=27
-%2823%2F5%29x=135%2F5-99%2F5
-%2823%2F5%29x=36%2F5
x=-36%2F23
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The answer of 6 doesn't make sense.
Since the solution is linear and 27 is between 6 and 29, the solution needs to be between 3 and -2, closer to -2 since 27 is closer to 29. My solution is about -1.5, close compared to -2. 6 is not even in the possible range of solutions.