Given that $f(x)$ is differentiable function of $x$ and that $f(x)\cdot f(y) = f(x) + f(y) + f(xy) - 2$ and that $f(2) = 5$. Then find value of $f(3)$.
By putting $x=2$ and $y=1$ , I got $f(1)=2$ but when I put $x=1$ and $y=3$, $f(3)$ gets cancelled. Could someone help me with this?
$f(x)f(y) = f(x) + f(y) + f(xy) - 2$ holding y contant and differentiating by x
$f'(x)f(y) = f'(x) + yf'(xy)$
Now set $x = 1$ let $f'(1) = C$
$Cf(y) = C + yf'(y)$
Now we have a differential equation.
$\frac {f'(y)}{f(y)-1} = \frac C{y}\\ \ln (f(y) - 1) = C\ln y\\ f(y) = y^c + 1$
Fit the initial condition:
$f(2) = 2^c + 1 = 5\\ c = 2\\ f(y) = y^2 + 1$
check it...
$(x^2 + 1)(y^2+1) = (x^2+1) + (y^2 + 1) + (x^2y^2+1) - 2\\ x^2y^2 + x^2 + y^2 + 1 = x^2y^2 + x^2 + y^2 + 1$
Seems to work.
Note that continuity is sufficient.
If you set $g(x)=f(x)-1$ then
$\require{cancel} f(x)f(y)=(g(x)+1)(g(y)+1)=\cancel{g(x)}+\cancel{g(y)}+g(x)g(y)+\cancel{1}=\cancel{g(x)}+\cancel{1}+\cancel{g(y)}+\cancel{1}+g(xy)+\cancel{1}-\cancel{2}$
So $g$ verifies the functionnal equation $$g(xy)=g(x)g(y)$$
By continuity $g$ is entirely determined : $g(x)=x^t$
[See for instance this thread : If $f(xy)=f(x)f(y)$ then show that $f(x) = x^t$ for some t ]
Pluging $f(2)=5\iff g(2)=4$ gives $t=2$.
And finally $\ \bbox[5px,border:1px solid]{f(x)=x^2+1}$
