f ( 1 ) f ( 2 ) f ( 3 ) f ( 4 ) = = = = 1 4 9 1 6
I am thinking of some polynomial f ( x ) satisfying the equations above. What is f ( 5 ) ?
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Oh wow! I understand now. Thanks. I have learnt something.
The easiest explanation to the question is that, It is not given that the polynomial is a monic polynomial so, it cannot be determined what the value of f(5) will be. It could be anything, tbh
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Eh? What if I say that this is a monic quintic (fifth degree) polynomial? There will still be infinitely many possible values of f ( 5 )
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Agreed , being monic doesn't matters for a quintic polynomial.
If it is mentioned that is a quartic polynomial, then f(5) has a unique value. But true, if it is mentioned that the polynomial is a quintic monic polynomial, f(5) can still assume infinite values.
I was just commenting with regard to this question, No need to get mad :P
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@Mehul Arora – I was just commenting based on your comments, I'm not getting mad at all. Just wanna avoid confusing others :)
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@Hung Woei Neoh – LOL Did I say something wrong? I think that the question can easily be solved by observation of the fact that f(x) is not monic.
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@Mehul Arora – Your comment might lead some to think this:
"We cannot determine f ( 5 ) because it is not given that the polynomial is monic. Therefore, if the question states that it is a monic polynomial, we can determine a unique f ( 5 ) "
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@Hung Woei Neoh – Lol ,You guys done arguing ? Hung woei neoh wins ! Okay he is correct , Mehul arora , it might confuse the readers ! :-)
Nice! Kudos to you for finding all the functions which match the answer options!
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Thanks! It wasn't hard with the help of an online equation system calculator
Very nice explanation(+1)!
Data is insufficient ; the degree of the polynomial must be stated , because without it we will not be able to frame the polynomial.
Hence ,
" C a n n o t b e D e t e r m i n e d "
There seems to be sufficient data to determine f(x). Can someone explain why f(5) is not 25?
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f ( x ) can either be x 2 or k ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) + x 2 where k is the leading coefficient. In fact there are infinitely many polynomials of f ( x ) which satisfy the conditions.
All 7 options can be f ( 5 ) . Check out my solution
Note that besides what Svatejas mentioned, we can also have infinitely many polynomials of degree 5 or higher that satisfy the conditions above. Therefore, unless the degree is 2 , even with the degree mentioned, we probably still will not be able to determine the value unless more information is provided
The answer varies with degree. If degree is 5, then infinite possibilites will be there, because we are given 4 equations and degree 5 needs 5 equation. So coefficients cannot be determined and hence infinite values of f(5) are there. Here I gave example and not only degree 5 but many infinite degrees have infininite answers so answer is obviously cannot be determined
Mod: Agree that there seems to be sufficient data to determine f(x), at least answer should be 25.
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Nah, the problem is there are so many equations satisfying the conditions.. You can make f(5) to be any of the given options in fact not only 25.
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There are infinitely many possible values of f ( 5 ) . In fact, all the given options could be f ( 5 ) :
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ f ( x ) = 6 2 5 x 4 − 3 1 2 5 x 3 + 6 8 8 1 x 2 − 3 6 2 5 x + 1 0 0 f ( x ) = 2 5 x 4 − 2 5 0 x 3 + 8 7 6 x 2 − 1 2 5 0 x + 6 0 0 f ( x ) = − 8 3 x 4 + 4 1 5 x 3 − 8 9 7 x 2 + 4 7 5 x − 9 f ( x ) = x 2 f ( x ) = − 2 4 2 5 x 4 + 1 2 1 2 5 x 3 − 2 4 8 5 1 x 2 + 1 2 6 2 5 x − 2 5 f ( x ) = − x 4 + 1 0 x 3 − 3 4 x 2 + 5 0 x − 2 4 f ( x ) = 2 4 1 1 x 4 − 1 2 5 5 x 3 + 2 4 4 0 9 x 2 − 1 2 2 7 5 x + 1 1 f ( 5 ) = 1 2 5 f ( 5 ) = 6 2 5 f ( 5 ) = 1 6 f ( 5 ) = 2 5 f ( 5 ) = 0 f ( 5 ) = 1 f ( 5 ) = 3 6
I can list other values for f ( 5 ) , or higher degree polynomials for the multiple choice options, but you get the picture. We do not have sufficient information, therefore f ( 5 ) Cannot be Determined