Computing Value of a Functional Equation
Posted: Sun Jun 27, 2021 8:02 pm
A Function $f(x)$ is such that for any integer $x$, $f(x) + x f(2-x) = 6$ . Compute $-2019 f(2020)$
$f(x)+xf(2-x)=6\cdots\cdots(1)$sakib17442 wrote: ↑Sun Jun 27, 2021 8:02 pmA Function $f(x)$ is such that for any integer $x$, $f(x) + x f(2-x) = 6$ . Compute $-2019 f(2020)$
Replacing $x\mapsto2-x$ gives,
$f(2-x)+(2-x)f(x)=6\cdots\cdots(2)$
Multiplying $x$ with $(2)$ and subtracting that from $(1)$ gives us,
$f(x)-(2x-x^2)f(x)=6-6x$
$\Rightarrow (1-x)^2f(x)=6(1-x)\cdots\cdots(3)$
Substituting $x=2020$ in $(3)$ gives us
$(-2019)^2f(2020)=6\cdot(-2019)$
$\Rightarrow\boxed{-2019f(2020)=6}$