When x is 2014

I notice that people here seem to have been plugging in values to find the period of the function. I decided to do it differently.

$f(x+2) = \frac{f(x)-5}{f(x)-3}\\\\ f(x) = \frac{f(x-2)-5}{f(x-2)-3}$

From the first equation, we can get the following,

$f(x+2)f(x) - 3f(x+2) = f(x) - 5\\\\ f(x+2)f(x) -f(x)=3f(x+2)-5\\\\f(x) = \frac{3f(x+2)-5}{f(x+2)-1}$

Equating these definitions of f(x), we can cross multiply.

$f(x-2)f(x+2) - f(x-2) - 5f(x+2) +5 = 3f(x+2)f(x-2)-5f(x-2)-9f(x+2)+15$

$4f(x-2) + 4(x+2)= 2f(x+2)f(x-2) + 15$

This can be used to solve for f(x+2) and f(x-2)

$4f(x-2) - 15 = 2f(x+2)f(x-2) - 4f(x+2)$

$f(x+2) = \frac{4f(x-2) - 15}{2f(x-2) - 4}$

$f(x) = \frac{4f(x-4)-15}{2f(x-4)-4}$

Similarly,

$4f(x+2) - 15 = 2f(x+2)f(x-2) - 4f(x-2)$

$f(x-2) = \frac{4f(x+2) - 15}{2f(x+2)-4}$

$f(x) = \frac{4f(x+4)-15}{2f(x+4)-4}$

Setting these two new formulae for f(x) equal to each other and cross multiplying,

$\frac{4f(x+4)-15}{2f(x+4)-4} = \frac{4f(x-4)-15}{2f(x-4)-4}$

$8f(x-4)f(x+4) - 16f(x+4) - 30f(x-4) + 60 = 8f(x-4)f(x+4) - 30f(x+4) - 16f(x-4) + 60$

$14f(x+4) = 14f(x-4)$

$f(x+4) = f(x-4)\$

$f(x) = f(x-8)$ )

From this, it is seen that the function f(x) has a period of 8, and therefore f(2014) = f(2006)

assign any value for a function, lets take it as f(0). then keep on getting f(2....8) then @ f(8) the value of f(0) repeats. as this satisfies for any value assigned to the given function, by inspection we can say that the function repeats for every intervals of 8. hence the solution is 2006

Let $a,b,c,d$ such that $ad-bc \neq 0$ . For each $g(x) = \dfrac{ax +b}{cx+d}$ we can associate the matrix $A =\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ . We will write $g \sim A$ . Its easy to prove that if $g_1 \sim A_1$ and $g_2 \sim A_2$ then $(g_1 \circ g_2) \sim A_1A_2$

Now consider $g \sim A = \begin{pmatrix} 1 &-5 \\ 1 & - 3\end{pmatrix}$ . Notice that $A^4 = \lambda I$ . Therefore, $g^4(x) = g(g(g(g(x)))) = x$ . As $f(x+2) = g(f(x))$ , then $f(x+8) = g^4(f(x)) = f(x)$ . So, $f(2006) = f(2008)$

Can you show that this function is periodic with period $8$ ??

In fact if we replace $2$ with any real $P$ then the function is still periodic with period $4P$ .

take f(0) =5 and calculate f2,f4,f6,f8 ..... u will find f0=f8 thus function is periodic with period 8 this is completely a objective question i didnt found any subjective approach to it