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Directrix
 4 years ago
There are an infinite number of polynomials P for which P(x+5)  P(x) = 2 for all x. What is the least possible value of P(4)  P(2)?
Directrix
 4 years ago
There are an infinite number of polynomials P for which P(x+5)  P(x) = 2 for all x. What is the least possible value of P(4)  P(2)?

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dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0Let \[P(x) = ax^{n}, .... P(x+5) = a(x+5)^{n}\] Then \[a(x+5)^{n} ax^{n} = 2\] \[\rightarrow a = \frac{2}{(x+5)^{n} x^{n}}\] since it applies for all x, let x=0 \[a = \frac{2}{5^{n}}\] Evaluate P(4)P(2) \[\rightarrow \frac{2}{5^{n}}4^{n}  \frac{2}{5^{n}}2^{n} = \frac{2(4^{n}2^{n})}{5^{n}}\] Take limit as n> infinity \[\lim_{n \rightarrow \infty}\frac{2(4^{n}2^{n})}{5^{n}} = 0\] smallest possible value goes to 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@dumbcow: why the initial assumption of \( P(x) = ax^{n} \)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[P(x+5)P(x)=2\]\[P(x+5)P(x)2=0\] if this is the function, given two value of x, 4 and 2, to be subtracted. . \[[P(4+5)P(4)2] [P(2+5)P(2)2]=0\]\[[P(9)P(4)2][P(7)P(2)2]=0\]\[P(5)2 P(5)2=0\]\[p(0)4=0\]\[P(0)=4\] kindly check my solution. i got 0 as the least value of x, and 4 is the result. .

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0@FoolFormath because it was the simplest polynomial to work with and since it doesn't matter what polynomial it is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@dumbcow, how about my solution? is it acceptable. .

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0in one of your steps(3rd line) what happened to P(9) and P(7)

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0oh i got it but how can you assume P(9) P(4) = P(5) ?

Directrix
 4 years ago
Best ResponseYou've already chosen the best response.3No correct answer yet. Hint: Consider the value(s) of n a polynomial P of degree n would have in order for P(x+5)  P(x) to be constant.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It seems that I have understood the solution (with some help) the polynomial should be for the form \( P(x)=\frac{2}{5}x+c \) so \( P(4)P(2)=\frac{4}{5} \)

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0i see what i did wrong now. isn't the question a little misleading though, there really is only 1 possible value for P(4)P(2)
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