Problem: If f(2x)=22+xf(2 x)=\dfrac{2}{2+x}f(2x)=2+x2 for all x>0x>0x>0, then 2f(x)=2 f(x)=2f(x)=
Answer Choices:
A. 21+x\dfrac{2}{1+x}1+x2
B. 22+x\dfrac{2}{2+x}2+x2
C. 41+x\dfrac{4}{1+x}1+x4
D. 42+x\dfrac{4}{2+x}2+x4
E. 84+x\dfrac{8}{4+x}4+x8
Solution:
2f(x)=2f(2⋅x2)=2(22+x2)=2(44+x)=84+x.2 f(x)=2 f\left(2 \cdot \dfrac{x}{2}\right)=2\left(\dfrac{2}{2+\dfrac{x}{2}}\right)=2\left(\dfrac{4}{4+x}\right)=\dfrac{8}{4+x}. 2f(x)=2f(2⋅2x)=2⎝⎛2+2x2⎠⎞=2(4+x4)=4+x8.