Question

F(x) = 4^x and g(x) = 4^1/2x

Answer 1

There are a few different ways to do this one. We can just note

$g(x) = 4^{\frac 1 2 x} = (4^\frac 1 2)^x = 2^x$

so

$A=g(-2) = 2^{-2} = \frac 1 4,\quad \quad B=g(0)=2^0 = 1, \quad\quad C=g(2)=2^2=4$

That's one way. We might also note

$g(x) = 4^{\frac 1 2 x} = (4^x)^{\frac 1 2 } = \sqrt{4^x} = \sqrt{f(x)}$

so we can just take the square root of the corresponding value of f,

$A=\sqrt{\frac 1 {16}} = \frac 1 4, \quad \quad B=\sqrt{1}=1, \quad\quad C=\sqrt{16} =4$

Finally nothing prevents us from substituting directly into the given g(x):

$A = g(-2) = 4^{\frac 1 2(-2)} = 4^{-1} = \frac 1 4$

$B = g(0) = 4^{\frac 1 2 (0)} = 4^0 = 1$

$C = g(2) = 4^{\frac 1 2 (2)} = 4^1 = 4$

That's three different ways and they all give the same answer, which they have to, otherwise math doesn't work.

Answer: A=1/4, B=1, C=4

Answer 2

Answer: A=1/4 B=1 C=4

i hope this helps :)