$\sqrt{4x -3}$ injective? Bijective? Inverse? I'm shown part of the function $g(x) = \sqrt{4x-3}$. Is it injective? I said yes as per definition if $f(x) = f(y)$, then $x =y$. Is this right? 
Under what criteria is $g(x)$ bijective? For what domain and co domain does $g(x)$ meet this criteria? I'm totally lost on this one! 
Find the inverse of $g(x)$, sketch the graph and explain how $g(x)$ and $g^{-1}(x)$ relate geometrically. 
Any help on this is greatly appreciated. 
 A: It really makes no sense to ask whether a function is surjective if you don't have the codomain.
That is: you have something like $f(x)=\sqrt{4x-3}$. If you are working with real numbers, $x$ must be $\ge\frac34$ in order to the square root makes sense. Furthermore, the values that takes $f(x)$ are nonnegative.
So we can, in most cases, assume that the domain of $f$ is "as greatest as possible"; in this case, $[\frac34,\infty)$. But a function is not only an expression and a domain. It's a codomain, too.
The codomain is the set in which the values of $f$ lie. But this set can have more elements that $f$ never reaches.
For example: for our $f(x)=\sqrt{4x-3}$, we can set $f:[\frac34,\infty)\to [0,\infty)$ and if $f$ is defined this way, $f$ is surjective. Because any element from $[0,\infty)$ (this is the codomain) is reached by $f$.
But if we set $f:[\frac34,\infty)\to\Bbb R$, then $f$ is not surjective, because $f$ does never take negative values.
A: A function is bijective if it is surjective and injective. A function is surjective if the range is the codomain, i.e. if every value in the codomain is the output of the function. 
An inverse function is the function reflected trough $y=x$. 
