MHB Determining one to one and finding formula for inverse help needed

[Snicklefritz]

I have a problem I am trying to solve but no example provided can clarify the steps. I find this to be the problem with math examples, they provide problems but utilize same numbers so people who are not as math savvy such as I cannot figure the steps out as easily as they are provided. Anyway, enough complaining. My problem is determing one to one and also finding the inverse formula.

f(x)= 5x-3 over 2x-1

I am coming up with a final solution of f^-1(x)= x+3 over 5-2x

Is this right?

 

[Opalg]

I have a problem I am trying to solve but no example provided can clarify the steps. I find this to be the problem with math examples, they provide problems but utilize same numbers so people who are not as math savvy such as I cannot figure the steps out as easily as they are provided. Anyway, enough complaining. My problem is determing one to one and also finding the inverse formula.

f(x)= 5x-3 over 2x-1

I am coming up with a final solution of f^-1(x)= x+3 over 5-2x

Is this right?

Hi Snicklefritz, and welcome to MHB!

If you are given a function $y = f(x)$ then to get the inverse function you need to find $x$ in terms of $y$. So in this case, starting from $y = \dfrac{5x-3}{2x-1}$ you get $(2x-1)y = 5x-3$. Then rearrange that as $2xy - 5x = y - 3$, so that $x = \dfrac{y-3}{2y-5}.$ That tells you the inverse function as a function of $y$. If you want it as a function of $x$ then you simply replace the $y$s by $x$s, getting $f^{-1}(x) = \dfrac{x-3}{2x-5}$ – similar to what you had, but not quite the same. (Wondering)

If a function has an inverse then it is automatically one-to-one, so you don't need to do any further work there.

 

[MarkFL]

We are given:

$$f(x)=\frac{5x-3}{2x-1}$$

We can see this is a one-to-one function by observing:

$$f(x)=\frac{1}{2}\left(5-\frac{1}{2x-1}\right)$$

This is simply a transformation of the function:

$$g(x)=\frac{1}{x}$$

which we know to be one-to-one on its domain.

To check to see if you have found the correct inverse, we can use the functional identity:

$$f\left(f^{-1}(x)\right)=f^{-1}\left(f(x)\right)=x$$

So, using this, we find:

$$f\left(f^{-1}(x)\right)=\frac{5\left(\frac{x+3}{5-2x}\right)-3}{2\left(\frac{x+3}{5-2x}\right)-1}=\frac{5(x+3)-3(5-2x)}{2(x+2)-(5-2x)}=\frac{5x+15-15+6x}{2x+4-5+2x}=\frac{11x}{4x-1}\ne x$$

So, you can see you have not computed the inverse correctly.