I am a little confused on the subject of inverse functions and the methods used to do the transformation from function to inverse. How do you make an inverse? Just so i can avoid any ambiguity in my question, let's change it to the following: Would anyone on this fantastic website be kind enough to list the steps of "inversing", may be the word, the simple function of $$f(x)= \frac{(x-3)}{2}$$

  • $\begingroup$ my answer is $$f(x)= 2x + 3$$ $\endgroup$ – Diamond Louis XIV Feb 17 '14 at 2:26

I would do it by solving for $x$ first, then switching $x$ and $y$. I will use your function $f(x)=\dfrac{x-3}{2}$. First we solve for $x$ like this: $$f(x)=\dfrac{x-2}{3}$$ $$x-3=2f(x)$$ $$x=2f(x)-3$$ $$f^{-1}(x)=2x-3$$

Let's try another example. If $f(x)=x^2-1$, and I want to find $f^{-1}(x)$, I would do this. $$f(x)=x^2-1$$ $$x^2=f(x)+1$$ $$x=\pm\sqrt{f(x)+1}$$ Switching $f(x)$ and $x$... $$f^{-1}(x)=\pm\sqrt{x+1}$$ Believe it or not, the inverse is wrong. Why? For a function to have an inverse, it must pass the horizontal line test. If the function has two or more x-values for one y-value, it does not have an inverse. But we can restrict the domain of $x^2$ to be $x \ge 0$. Now it has an inverse; $f^{-1}(x)=\sqrt{x+1}$, $x \ge 0$.
Extra information about inverse of functions: $$f^{-1}(f(x))=x$$ $$f(f^{-1}(x))=x$$ The graph of $f^{-1}(x)$ is the graph of $f(x)$ reflected in the line $y=x$
Common inverses of functions:

If $f(x)=\dfrac{1}{x}$, then $f^{-1}(x)=\dfrac{1}{x}$, $x \neq 0$.

If $f(x)=e^x$, then $f^{-1}(x)=\ln(x)$, $x > 0$.

If $f(x)=a^x$, then $f^{-1}(x)=\log_{a}(x)$, $a, \ x > 0$.

If $f(x)=x^2$, then $f^{-1}(x)=\sqrt{x}$, $x \ge 0$.

If $f(x)=\sin(x)$, then $f^{-1}(x)=\sin^{-1}(x)$, $-\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$.

If $f(x)=\cos(x)$, then $f^{-1}(x)=\cos^{-1}(x)$, $-\dfrac{\pi}{2}\leq x \leq \dfrac{\pi}{2}$.

If $f(x)=\tan(x)$, then $f^{-1}(x)=\tan^{-1}(x)$, $-\pi\leq x \leq \pi$.

Sometimes, a function will never have an inverse, because the function is unsolvable for $x$. For example, take the function $f(x)=2x+\cos(x)$. How are you going to solve for $x$? I don't think you can. Therefore this function does not have an inverse.
Hope I helped.


I know there's already an accepted answer, but I'm not a fan of how its laid out, and it also has some errors in the algebra. So, here's my version.

Please comment with any questions you may have.

Let $f$ denote the unique function $\mathbb{R} \rightarrow \mathbb{R}$ defined as follows. $$f(x)=\frac{x−3}{2}$$

Problem. Find the inverse of $f$.

Solution. Let $x,y \in \mathbb{R}$ be fixed but arbitrary. Then TFAE.

  1. $f(x)=y$
  2. $\dfrac{x-3}{2}=y$
  3. $x-3=2y$
  4. $x=2y+3$

Therefore, the inverse of $f$ is the unique function $g : \mathbb{R} \rightarrow \mathbb{R}$ defined as follows.


  • 1
    $\begingroup$ Thank you for the extra answer, I believe that you definitely put it in proper form, only one issue, It is so correct that i have never seen anything like it. Firstly, I'm guessing that by "$g : \mathbb{R} \rightarrow \mathbb{R}$" you just mean that $ \mathbb{R} $ is the inverse of the fuction? secondly and lastly, i am wondering why you did not switch the $x$ and $y$ values within the equation, why is this so? $\endgroup$ – Diamond Louis XIV Feb 18 '14 at 5:00
  • $\begingroup$ @DiamondLouisXIV, before I address your questions, lets just make sure we're on the same page here. The phrase "$f : X \rightarrow Y$ is a function" means that to every $x \in X$, $f$ associates a particular element of $Y$, that we denote $f(x)$. For example, if we write: let "$f : \mathbb{N} \rightarrow \mathbb{R}$ denote a function," we're basically saying that to every natural number $n$, $f$ associates a special real number that we denote $f(n)$. So in this case, $f$ is like a sequence of real numbers. $\endgroup$ – goblin Feb 18 '14 at 8:01
  • $\begingroup$ Thus to define a function $f$ that maps $X$ into $Y$, or in other words, to define a function $f : X \rightarrow Y,$ we have to specify the value of $f$ at each $x \in X$. For example, "let $f : \mathbb{N} \rightarrow \mathbb{R}$ denote the unique function $f$ such that $f(n)=\sqrt{n}$." What this is basically saying is that $f$ is the unique function such that for every natural number $n$, $f$ returns the real number $\sqrt{n}$. Are you following so far? $\endgroup$ – goblin Feb 18 '14 at 8:01

First of all, to type 3/over 2 in latex you should enter "\frac{3}{2}". For more informations you could just search Latex and there will be a lot of supporting instructions you can look at.

Next, once you have a function, say $y=f(x)$ and you wish to find the inverse function of it. The simplest idea is you treat $y$ as a given number and you find the number of $x$ in terms of $y$. Hence you will have a function such that $x=g(y)$ where $g(y)$ is the inverse function you are looking for. Of course, there are many situations that you can not explicitly solve $x$ in terms of $y$ but I think you won't meet those cases in your homework, exams, etc.

Of couse, $f(x) = 2x+3$ is the correct answer.

  • $\begingroup$ In regards to the first sentence, with only 2 terms in the fraction, you can write it as \frac32 & save yourself 4 keystrokes :D. Similarly, with a single digit in the numerator, \frac2{35} would give you $\frac2{35}$, saving you 2 keystrokes. $\endgroup$ – Kyle Kanos Feb 23 '14 at 18:48

If a function has an inverse then the roles of $x$ and $y$ are switched. For the function you gave $f(x)=\frac{x-3}{2}$ follow the steps.

1) Replace $f(x)$ by $y$: $$y=\frac{x-3}{2}$$

2) Switch the variables $x$ and $y$: $$x=\frac{y-3}{2}$$

3) Solve for $y$: $$2x=y-3 \Longrightarrow y=2x+3$$

4) Replace $y$ by $f^{-1}$: $$f^{-1}(x)=2x+3$$

  • $\begingroup$ You're right. Just fixed. Thanks! $\endgroup$ – NumBee Feb 17 '14 at 13:52

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