# Math Help - Precalculus

1. ## Precalculus

Find f -1(x) for the one-to-one function

2. ## Inverse of a function

Find f -1(x) for the one-to-one function

What you have to do here is to let $y = f(x) = \frac{4x-9}{7}$, and then make $x$ the subject of the formula.

So $7y = 4x-9$

$\Rightarrow 4x = 7y+9$

$\Rightarrow x = \frac{7y+9}{4} = g(y)$, say.

So what now have is that $f:x \rightarrow y$ (in other words, $f$ is the function that maps $x$ onto $y$), and $g:y \rightarrow x$. So $g$ maps $y$ back onto $x$. Therefore $g$ is therefore the inverse of $f$; i.e. $g = f^{-1}$

Now it doesn't matter what letter is used for the parameter that's fed into a function, and it's usual to use $x$ rather than $y$. So we would usually write:

$f^{-1}(x) = \frac{7x+9}{4}$

I hope that clears it up OK.