0
$\begingroup$

$f(n) = \begin{cases}-2n,&\text{if }n < 0\\ 2n+1,&\text{if }n\ge 0\end{cases}$

$f^{-1}(n) = \begin{cases}-\frac{n}2,&\text{if }n\text{ is even}\\\\ \frac{n-1}2,&\text{if }n\text{ is odd}\end{cases}$

I would like to know where the $n$ is even and $n$ is odd come from exactly and how to determine the "condition" for similar functions.

  • 0
    yeah it's just$a$mistake2012-11-04

2 Answers 2

1

Look at the outputs of the function. The odd outputs are the $2n+1$ so to get $n$ back you subtract 1 and then divide by 2: input = (output-1)/2. The even outputs are the $-2n$ so to get back $n$ you divide by $-2$; so input = -(output)/2. Now since inverse function reverses inputs and outputs, you get the formula you displayed. It might help to temporarily change the variable names, and then switch back to $n$ for the input variable to $f^{-1}.$

0

Make a little table of $f$: $\matrix{n&-4&-3&-2&-1&0&1&2&3&4\cr f(n)&8&6&4&2&1&3&5&7&9\cr}$ Now, what do you have to do to get back from $8,6,4,2,1,3,5,7,9$ to $-4,-3,-2,-1,0,1,2,3,4$, from $f(n)$ to $n$? For the even numbers, $8,6,4,2$, you just have to divide by $2$, and change the sign: $x\mapsto-x/2$. For the odd numbers, $1,3,5,7,9$, you subtract $1$ and then divide by $2$: $x\mapsto(x-1)/2$.