Help: Functional Equation 3

f(x)f(y)=f(x)+f(y)+f(xy)2f\left( x \right) f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2

If f(x)f\left( x \right) is a polynomial satisfying the equation above for all xx and yy and f(2)=5f\left( 2 \right)=5 , then compute limx2f(x)\displaystyle \lim _{ x\rightarrow 2 }{ f^{ ' }\left( x \right) } .

If anyone solves this one, please do tell me what is the error in this one

Note by Akhilesh Prasad
5 years, 2 months ago

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1 vote

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Hey there!
We have,
f(x)f(y)f(x)f(y)+1=f(xy)1(f(x)1)(f(y)1)=f(xy)1g(x)=f(x)1g(x)g(y)=g(xy) f(x)f(y)-f(x)-f(y)+1\quad =\quad f(xy)-1\\ (f(x)-1)(f(y)-1)\quad =\quad f(xy)-1\quad \\ g(x)=f(x)-1 \\ \Rightarrow g(x)g(y) = g(xy) \\

This is famous, see here.
We then have that,
g(x)=xt,t=2.g(x)=x2f(x)=x2+1f(x)=2xlimx2f(x)=4 g(x) = x ^ t, t = 2. \\ g(x) = x^2 \\ f(x) = x^2 + 1 \\ f'(x) = 2x \\ \Rightarrow \lim _{ x\rightarrow 2 }{ f^{ ' }(x) } = 4

Ameya Daigavane - 5 years, 2 months ago

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Put x and y as 2 which will give you the value of f(4) as 17

Put x as 4 and y as 2 and you get f(8) as 65. Similarly you can check with other values and this leads to the conclusion that f(x) is one greater than the square of x.

Hence the limit computation will be 4

Please excuse me if you were looking for a formal proof or evaluation of the same.

Jitin Nair - 5 years, 2 months ago

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@parv mor

Akhilesh Prasad - 5 years, 2 months ago

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@Rishabh Cool, this question again.

Akhilesh Prasad - 5 years, 2 months ago

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