Is my solution correct in terms of proof? I can't prove directly that $f(x)=4$.

Question

Given $f:\mathbb{R}\rightarrow \mathbb{R}^+$ and the following conditions:

1. $f(x+2)\cdot f(x+3)=16$,
2. $f(x)+f(-x)=8$,

we have to find integral $\int_{-8}^8 f(x-2019)dx$. So, if we plug in $0$, then $f(0)+f(0)=8$. Therefore, $f(0)=4$. Then plug in $-2$ in first equation, we get $f(0)\cdot f(1)=16$, therefore $f(1)=4$, analogically plug $x=-1$ and I think that $f(x)=4$, so answer will be $64$ (value after integrating). Is this correct? How can i prove it in a better manner?

Answer 1

From 1 we have $f(x+1)f(x)=16$, so $\displaystyle f(x+2) = \frac{16}{f(x+1)} = \frac{16}{16/f(x)} = f(x)$. Since $f$ has period $2$ then $$\int_{-8}^8 f(x-2019)\,dx = \int_{-8}^8 f(x-1)\,dx = 8 \int_0^2f(x-1)\,dx$$ $$ = 8 \int_{-1}^1 f(x)\,dx = 8\int_{-1}^0f(x)\,dx+8\int_{0}^1 f(x)\,dx$$ $$=8 \int_{0}^1f(-x)\,dx + 8\int_{0}^1 f(x)\,dx = 8 \int_0^1f(x)+f(-x)\,dx = 8 \int_0^1 8\,dx = 64$$

Answer 2

Here's a more direct solution:

From 1 we have $\displaystyle f(1+x)=\frac{16}{f(x)}$ and from 2 we have $f(x) = 8-f(-x)$.

Then $$f(x) = f(1+x-1) = \frac{16}{f(x-1)} = \frac{16}{8-f(1-x)} = \cfrac{16}{8-\cfrac{16}{f(-x)}} = \cfrac{16}{8-\cfrac{16}{8-f(x)}}$$

Solving for $f(x)$ we get $f(x)=4$.