Algebra question involving function? How would I solve the following problem
suppose $h(x)=\sqrt{6x-3}$
find two functions f and g such that $(f\cdot g)(x)=h(x)$
How would I solve it do I need to find two functions that mutltiply to $h(x)$.
 A: If you mean multiplication of functions, then 
Let $f(x)g(x) = f^2(x) = \sqrt{6x - 3} = h(x).$ Then $f(x) = (6x - 3)^{\large\frac 14}$.
Then $$f(x)\cdot g(x) = f^2(x) = \left((6x - 3)^{\large\frac 14}\right)^2 = (6x - 3)^{\large\frac 12} = \sqrt{6x - 3} = h(x)$$
If you mean Function composition: 
How about $g(x) = 6x$ and $f(x) = \sqrt{x - 3}$.
Then $$(f\circ g)(x) = f(g(x)) = f(6x) = \sqrt{6x-3} = h(x)$$
A: If you mean function composition:
$$(f \circ g)(x) = \sqrt{6x-3}$$
Let $f(x) = \sqrt{x}$ and $g(x) = 6x - 3$. Then $f(g(x)) = \sqrt{6x-3}$.
Or you could use $g(x) = x$ and then simply $f(x) = \sqrt{6x-3}$. This answer is completely valid, if not a little fresh.
There are actually infinitely many solutions, so I assume you want one. In the event you get a multiple choice question asking for a possible solution, just plug in $g(x)$ as $x$ in $f(x)$ and see which one gives $h(x)$. 
A: I assume that $(f*g)(x)=h(x)$  is the same notation as $fg(x)$ or $$f(g (x))$$
So I will say $f(x) = \sqrt{x}$ and $g(x) = 6x-3$ because
$$f(g (x)) = f(6x-3) = \sqrt{6x-3}$$
