Answer
$g(0)=0$
$g(3)=\frac{3}{2}$
$5g(a)=\frac{5a}{\sqrt{a+1}}$
$\frac{1}{2}g(4a)=\frac{2a}{\sqrt{4a+1}}$
$g(a^2)=\frac{a^2}{\sqrt{a^2+1}}$
$[g(a)]^2=\frac{a^2}{a+1}$
$g(a+h)=\frac{a+h}{\sqrt{a+h+1}}$
$g(x-a)=\frac{x-a}{\sqrt{x-a+1}}$
Work Step by Step
$g(x)=\frac{x}{\sqrt{x+1}}$
We can replace the given value for $x$ to calculate the value of the function.
$g(0)=\frac{0}{\sqrt{0+1}}=0$
$g(3)=\frac{3}{\sqrt{3+1}}=\frac{3}{2}$
$5g(a)=5\times\frac{a}{\sqrt{a+1}}=\frac{5a}{\sqrt{a+1}}$
$\frac{1}{2}g(4a)=\frac{1}{2}\times\frac{4a}{\sqrt{4a+1}}=\frac{2a}{\sqrt{4a+1}}$
$g(a^2)=\frac{a^2}{\sqrt{a^2+1}}$
$[g(a)]^2=\left[\frac{a}{\sqrt{a+1}}\right]^2=\frac{a^2}{a+1}$
$g(a+h)=\frac{a+h}{\sqrt{a+h+1}}$
$g(x-a)=\frac{x-a}{\sqrt{x-a+1}}$