To show that $f$ is 1-1, you could show that $$f(x)=f(y)\Longrightarrow x=y.$$ So, for example, for $f(x)=
I'll leave showing that $f(x)=<
Alternatively, to show that $f$ is 1-1, you could show that $$x\ne y\Longrightarrow f(x)\ne f(y).$$
Or, to show that a differentiable $f$ is 1-1, you could show that its derivative $f'$ is either always positive or always negative.
You would discover that a function $g$ is not 1-1, if, when using the first method above, you find that the equation is satisfied for some $x\ne y$. For example, take $g(x)=1-x^2$. Then
$$ \eqalign< &g(x)=g(y)\cr \iff&<1-x^2>= <1-y^2> \cr \iff&-x^2= -y^2\cr \iff&x^2=y^2\cr> $$ The above equation has $x=1$, $y=-1$ as a solution. So, there is $x\ne y$ with $g(x)=g(y)$; thus $g(x)=1-x^2$ is not 1-1.
Of course, to show $g$ is not 1-1, you
need only find two distinct values of the input value $x$ that give $g$ the same output value.
Although you rightfully point out that the graphical method is unreliable; it is still instructive to consider the methods used and why they work:
Graphically, you can use either of the following:
Use the "Horizontal Line Test":
$f$ is 1-1 if and only if every horizontal line intersects the graph of $f$ in at most one point. Note that this is just the graphical interpretation of "if $x\ne y$ then $f(x)\ne f(y)$"; since the intersection points of a horizontal line with the graph of $f$ give $x$ values for which $f(x)$ has the same value (namely the $y$-intercept of the line).
Use the fact that a continuous $f$ is 1-1 if and only if $f$ is either strictly increasing or strictly decreasing. This, of course, is equivalent to the derivative being always positive or always negative in the case where $f$ is differentiable. (Note this method applies to only the green function below.)