If a non-zero polynomial f ( x ) satisfies the condition f ( 2 x ) = f ′ ( x ) f ′ ′ ( x ) , find the value of f ( 3 ) .
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Use \implies for ⟹ symbol.
Note that the case of "n=0" has to be dealt with separately, since the degree of the 0 polynomial is not -1.
This is why your solution missed the crucial requirement that the polynomial is "non-zero".
Is f ( x ) = 0 a polynomial too? If so, the answer could be 1 2 or 0
Read question one more time..it's written that if f(x) be a non zero polynomial.
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I didn't see it when I answer it......
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@Calvin Lin – If I have problems with Latex, where I do report it? (Just kidding). I need the codes of Latex because the ones I know uses the $$$ sign. Where can I find helpful hints with latex codes. Thanks.
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@Hana Wehbi – We use \ ( \ )instead of $ $ to surround Latex. Otherwise, the "codes" are all the same.
You can hover your mouse over the Latex image to see the underlying code.
E.g.
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@Calvin Lin – Thanks, that was really helpful. I will give it a try.
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Let degree of function = n
therefore, D E G f ( 2 x ) = n
D E G f ′ ( x ) = ( n − 1 )
D E G f ′ ′ ( x ) = ( n − 2 )
so,
n = ( n − 1 ) + ( n − 2 )
=> n = 3
let f ( x ) = a x 3 + b x 2 + c x + d
=> f ( 2 x ) = 8 a x 3 + 4 b x 2 + 2 c x + d
also, f ′ ( x ) = 3 a x 2 + 2 b x + c
and f ′ ′ ( x ) = 6 a x + 2 b
as f ( 2 x ) = f ′ ( x ) . f ′ ′ ( x )
=> 8 a x 3 + 4 b x 2 + 2 c x + d = ( 3 a x 2 + 2 b x + c ) ( 6 a x + 2 b ) )
by comparing the coeffs.
8 a = 1 8 a 2
=> a = 9 4
also, b = c = d = 0
therefore,
f ( x ) = 9 4 x 3
therefore,
f ( 3 ) = 9 4 * 3 3
=> f ( 3 ) = 1 2 .