Function f is symmetric to the origin and periodic with period 8. If f(2)=3, what is the value of f(4)+f(6)?

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Answer 1 · 2018-07-26T19:20:15

$- 3$ $-3$

Let $f (x)$ $f(x)$ be the function symmetric to the origin i.e. $f (x)$ $f(x)$ is odd hence

$f (- x) = - f (x)$ $f(-x)=-f(x)$

Since function $f (x)$ $f(x)$ is periodic with period $8$ $8$ hence we have

$f (x + 8) = f (x)$ $f(x+8)=f(x)$

setting $x = - 4$ $x=-4$ in above equation we get

$f (- 4 + 8) = f (- 4)$ $f(-4+8)=f(-4)$

$f (4) = f (- 4)$ $f(4)=f(-4)$

$f(4)=-f(4)\quad (\because f(-x)=-f(x))$

$2f(4)=0$

$f(4)=0$

Again setting $x=-2$ in above function we get

$f(-2+8)=f(-2)$

$f(6)=-f(2) \quad (\because f(-x)=-f(x))$

$f(6)=-3 \quad (\because f(2)=3)$

$\therefore f(4)+f(6)$

$=0-3$

$=-3$