# Help: Functional Equation

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 $$f(x)$$.

2 years, 10 months ago

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Is 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)

- 2 years, 10 months ago

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 $f(x)=x+1$.

- 2 years, 9 months ago

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.

- 2 years, 9 months ago

f(1)=2 and f'(0)=0 should be there for the question to be correct

- 2 years, 10 months ago

Ya.. I also reached the same conclusion...

- 2 years, 10 months 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 $$f(x)$$

- 2 years, 10 months ago

Please if you can post a solution@parv mor

- 2 years, 10 months ago

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.

- 2 years, 10 months ago

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

- 2 years, 9 months ago