Question

Find f^-1(x) for the following one-to-one function f:
f(x)= 5/ x-2

Answer 1

Hint: Let \(y=\dfrac{5}{x-2}\). Then swap x with y to get \(x=\dfrac{5}{y-2}\). Then solve for y to get \(f^{-1}(x)\).

Can you take things from here? :-)

Answer 2

no i have no idea :/ im looking through my notes and cant find them for this chapter ugh

Answer 3

Alright. The first thing to know is that graphically speaking, the function \(\large f(x)\) and it's inverse \(\large f^{-1}(x)\) are symmetric about the line \(\large y=x\). Hence, whenever you try to compute \(\large f^{-1}(x)\) given \(\large f(x)\) algebraically, that's why we can swap x with y and then solve for y to get the value of \(\large f^{-1}(x)\).

Now, in your problem, you were given that \(\large f(x) = y = \dfrac{5}{x-2}\). Thus, if you swap x with y, we end up with \(\large x=\dfrac{5}{y-2}\). If we now solve for y, we get the inverse function \(\large f^{-1}(x)\) that we're looking for. So, we see that
\[\large \begin{aligned} x = \frac{5}{y-2} &\implies x(y-2) = 5\quad \text{(multiplied both sides by $y-2$)}\\ &\implies y-2 = \frac{5}{x}\quad \text{(divided both sides by $x$)}\\ &\implies y=f^{-1}(x)=\frac{5}{x}+2\quad\text{(added $2$ to both sides)}.\end{aligned}\] In conclusion, we see that \(\large f^{-1}(x)=\dfrac{5}{x}+2\).

Does this clarify things? I hope this makes sense! :-)

Answer 4

thank you so much for your help i really appreciate it! im not that great with math and i have a final tomorrow and i dont get this stuff at all so im trying to understand it and practice haha