Let $A = \mathbb R \setminus \{−1/2\}$ and $B =\mathbb R \setminus \{2\}$. Define $f : A \to B$ by the rule $$f(x) = \frac{4x − 3}{2x+1}$$ for all $x \in A$. Show that $f$ is one to one and onto. Find a formula for $f^{-1}: B \to A$.
Which mathematical tools can I research to begin solving a problem of this nature?
N.B. My posted question is not asking for the answers to the question.
A function is one to one: If you give two different inputs for your function, you get two different outputs. Thus, if you plug in $a$ and $b$ in $f$, then you should get two different outputs (if $a$ is not equal to $b$).
To see if your function is one to one, pretend to plug in two values: value $a$ and value $b$. The function gives you then two answers: $\frac{4a-3}{2a+1}$ and $\frac{4b-3}{2b+1}$. Now we want to see when those values are equal to each other. To do so you could substract both 'numbers' from each other and set this equal to $0$:
$$\frac{4a-3}{2a+1} - \frac{4b-3}{2b+1}=0$$ 'Solving' this (do you know what I mean?) gives you an solution for when the two inputs $a$ and $b$ give the same output.
A function is onto if for every element in the range of your function, thus some output, we can give an input from the domain that gives this output. Thus if you take, for example, the number $5\in B$, then you have to show that you can pick a number $x\in A$ such that $f(x)=5$. To show this in general: Pick a number $y\in B$, then solve $$\frac{4x-3}{2x+1}=y$$
The inverse of a function $f$ is a function that, if you give it the 'output' of your function $f$, you get the input back. The output of $f$, for some $x$, is $\frac{4x-3}{2x+1}$. Thus if you have some output $y$, you know that this is in the form $\frac{4x-3}{2x+1}=y$ for some $x$. Do you know how to find the inverse now?
Also, do you know what one to one and onto have to do with the inverse?