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

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

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

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TopNewestHey there!

We have,

\(

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,

\( g(x) = x ^ t, t = 2. \\ g(x) = x^2 \\ f(x) = x^2 + 1 \\ f'(x) = 2x \\ \Rightarrow \lim _{ x\rightarrow 2 }{ f^{ ' }(x) } = 4 \) – Ameya Daigavane · 9 months, 2 weeks 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 · 9 months, 3 weeks ago

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@parv mor – Akhilesh Prasad · 9 months, 3 weeks ago

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@Rishabh Cool, this question again. – Akhilesh Prasad · 9 months, 3 weeks ago

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