If \(f(x) \) is a polynomial in \(x\) satisfying \(f(x) f(y) + 2 = f(x) + f(y) + f(xy) \) and \(f(0) = 0, f'(1) = 2 \), find \(f(x) \).

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TopNewestIs the question all right?? Put y=0 in the given functional equation to get: \[0+2=f(x)+0+0~~~~\small{\text{(since f(0)=0)}}\] \[\Large f(x)=2\] which does not satisfy the condition f'(1)=2 !!

(It might work if indeed it was given f(1)=2 and f'(0)=0) – Rishabh Cool · 11 months, 2 weeks ago

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The question is wrong . I will prove it . Put \(x,y=0\) \[f(0)^{2}-3f(0)+2=0\] This implies \[f(0)=1,2\] now put \(x,y=1\) this gives \(f(1)=1,2\) Differentiating the function partially with respect to \(x\) we get \[f'(x)f(y)=f'(x)+yf'(xy).......(1)\] Now put \(x,y=1\) in \((1)\) We get \(f(1)=2\) Now put \(x,y=0\) in \((1)\) We get\(f(0)=1..or..f'(0)=0\) Now put \(x=1\) in (1) we get \(\frac{2f(y)-2}{y}=f'(y)...(2)\) Now let \(f(0)=1\) , we have \(f(1)=2\)then we observe that \[f(x)=x+1\] also in satisfies main functional equation and (2),(1) but \[f'(x)=1\] hence question is wrong . If only main functional equation is given without any condition then \[f(x)=x+1\]. – Shivam Jadhav · 11 months, 1 week ago

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Hey, if you see the other post, which is the same question, here,

you'll see we get,

\( f(x) - 1 = x^t \)

For no value of t, will both the given conditions be satisfied- there is no such function, as a result. Try it out! I think you should check your question again. – Ameya Daigavane · 11 months, 1 week ago

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f(1)=2 and f'(0)=0 should be there for the question to be correct – Parv Mor · 11 months, 2 weeks ago

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– Rishabh Cool · 11 months, 2 weeks ago

Ya.. I also reached the same conclusion...Log in to reply

– Akhilesh Prasad · 11 months, 2 weeks ago

But your suggestion makes the functional equation pretty much useless, as for \(x=1\), and \(y=c\), \(c\in dom(f)\) we do not get any relation to solve for \(f(x)\)Log in to reply

Please if you can post a solution@parv mor – Akhilesh Prasad · 11 months, 2 weeks ago

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I also came up with the same conclusion as you, i.e. \(f(x)=2\), so i thought of using the \(f'(1)=2\). So as we know from the definition of derivatives

\(f'(1)=\frac { f(1+h)-f(1) }{ h }\\ \)

\(\Longrightarrow f'(1)=\frac { f(1(1+h))-f(1) }{ h } \)

From the functional equation we have,\( \\ \)

\(f(1)f(1+h))+2=f(1)+f(1+h)+f(1(1+h))\)

\(\Longrightarrow f(1+h)=f(1)f(1+h)-f(1)-f(1+h)+2\)

\(\therefore \quad f^{ 1 }\left( x \right) =\frac { f(1)f(1+h)-f(1)-f(1+h)+2-f(1) }{ h } =\frac { f(1)(f(1+h)-2)-(f(1+h)-2) }{ h }\)

\(\Longrightarrow f^{ 1 }\left( x \right) =\frac { (f(1)-1)(f(1+h)-2) }{ h } \)

\(\Longrightarrow 2h=(f(1)-1)(f(1+h)-2)\)

From here on I am not able to conclude anything useful, so if you would help me lead in the right direction or tell me something that i am doing wrong. – Akhilesh Prasad · 11 months, 2 weeks ago

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– Ameya Daigavane · 11 months, 1 week ago

You have to take the limit for the derivative, as h goes to 0.Log in to reply