Number of solutions of the function

Question

Let $'f'$ be an even periodic function with period 4 such that $f(x) = 2^x -1$, $0\le x \le2$. The number of solutions of the equation $f(x) = 1$ in $[-10,20]$ are?

Since the given function is even,

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

since the given function is periodic,

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

So,

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

Well, according to the function defined for $x \in [0,2]$ we get one solution for $f(x) = 1$ at $x=1$

I don't know how do I proceed from here.

Any help would be appreciated.

Answer 1

$f $ is even thus if $-2\le x\le 0, $ $f (x)=2^{-x}-1$.

we have $f (1)=f (-1)=2-1=1$

and

$$[-10,20]= $$ $$\cup_{k=0}^6 [-10+4k,-10+4k+4] \cup [18,20] .$$

in each intervall of the form $[-10+4k,-6+4k ] $, there are two roots and there one root at $[18,20] $ thus

$$Total =2×7+1=15 \; roots$$ which are ; $$S=\{-9+2n, n=0,14\}$$ $$=\{-9,-7,-5,...,17,19\} $$

Answer 2

First, we observe that as $f$ is even, the solutions $f(x) = 1$ on $[-2,2]$ is 1 and -1. Then, use the periodicity, $f(x+4) = f(x)$ to extend this function to find all the roots in your interval. For instance, we have $f(1) = f(5) = f(9)$ ...