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 $$f(x)= \frac{(x-3)}{2}$$
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$\begingroup$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.
- $f(x)=y$
- $\dfrac{x-3}{2}=y$
- $x-3=2y$
- $x=2y+3$
Therefore, the inverse of $f$ is the unique function $g : \mathbb{R} \rightarrow \mathbb{R}$ defined as follows.
$$g(y)=2y+3$$
$\endgroup$ 3 $\begingroup$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.
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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.
$\endgroup$ 1 $\begingroup$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 $f^{-1}$: $$f^{-1}(x)=2x+3$$
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