Findf-1(x) for the one-to-one function

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- Feb 21st 2009, 11:11 PMkeadyjrPrecalculus
Find

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

http://www.webassign.net/www24/symIm...ed28536236.gif - Feb 22nd 2009, 01:01 AMGrandadInverse of a function
Hello keadyjrWhat you have to do here is to let $\displaystyle y = f(x) = \frac{4x-9}{7}$, and then make $\displaystyle x$ the subject of the formula.

So $\displaystyle 7y = 4x-9$

$\displaystyle \Rightarrow 4x = 7y+9$

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

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

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

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

I hope that clears it up OK.

Grandad