We know that, by the definition of a periodic function, if #f# is periodic with period #rho#, then

#f(x)=f(x+n rho)# for any integer #n#

Now, in order to find all the solutions of #f(x) = 1# for our function, let us apply the same property:

#f(x+4n) = 1#, #forall n in ZZ#.

Let #n = 0#.

#f(x) = 1#

We only have a formula for #f(x)# within the interval #[0,2]#. Does there exist a value of #x# in this interval such that #f(x) = 1#?

Yes. We see that #x=1# is a solution.

Thus, we must have

#f(1+4n) = 1#, #forall n in ZZ#.

In order to find all the solutions of this equation in the #[-10, 20]#, we must find all the values of #n# such that

#-10<= 1+4n <= 20#

#:.-11 <= 4n <= 19#

#:.-11/4 <= n <= 19/4 =>-2.75<=n<=4.75#

Since #n# is an integer, we have the following:

#n in {-2, -1, 0, 1, 2, 3, 4}#

And so, **some** solutions to the equation #f(x) = 1# in #[-10,20]# are

#x in {-7,-3,1,5,9,13,17}#

**The rest of the solutions**

Note that the function #f(x)# is defined to be #2^x-1# in the interval #[0,2]# - which is not one complete period. It is defined in the interval #[-2,0)# by the fact that it is an even function. Since this means that #f(-x) = f(x)#, we have

#f(-1) = f(1) = 1#

as well. Thus #x=-1# is a solution.

Using the periodicity of the function we see that in addition to those found above, the following are solutions to the equation in the interval #[-10,20]# too :

# {-9,-5,-1,3,7,11,15,19} #

So, the complete set of solutions to #f(x) = 1# in the interval #[10,20]# are

# {-9,-7,-5,-3,-1,1,3,5,7,9,11,13,15,17,19}#