Answer:
1/2,1,2,4,8 in order
Step-by-step explanation:
Given:
[tex]\displaystyle \large{f(x)=2(2)^x}[/tex]
The function can be simplified to:
[tex]\displaystyle \large{f(x)=2^{x+1}}[/tex] via law of exponent.
Recall:
[tex]\displaystyle \large{a^m\cdot a^n=a^{m+n}}[/tex]
Next, substitute given x-values in:
[tex]\displaystyle \large{f(-2)=2^{-2+1}=2^{-1}=\dfrac{1}{2}}\\\displaystyle \large{f(-1)=2^{-1+1}=2^0=1}\\\displaystyle \large{f(0)=2^{0+1}=2^1=2}\\\displaystyle \large{f(1)=2^{1+1}=2^2=4}\\\displaystyle \lage{f(2)=2^{2+1}=2^3=8}[/tex]
Therefore, the solution is 1/2,1,2,4,8