Practice Problems on Function Inversion

For each of these functions f, find a candidate inverse function g (if possible!) and prove that the g=f^{-1}.

1. f:\mathbb N\to \mathbb N. f(x)=2x+1

2. f:\mathbb R\to \mathbb R. f(x)=x^2

3. f:\mathbb N\to \mathbb N. f(x)=x^2

4. f:\mathbb R\to \mathbb R. f(x)=x^3+x

5. f:\mathbb R\to \mathbb R. f(x)=x^3-x

6. f:\mathbb Z\to N,

f(x)=\begin{cases} -2z & z\le 0\\ 2z-1 & z > 0 \end{cases}
7. Find an example of a function function that is self-inverse (i.e. f=f^{-1}).

Worked sample problem:

f:\mathbb Z\to N,

f(x)=\begin{cases} -2z & z\le 0\\ 2z-1 & z > 0 \end{cases}

As discussed in class, this function maps the negative numbers to the even numbers and the positive numbers to the odd numbers. Here's a candidate inverse:

g:\mathbb Z\to N,

g(n)=\begin{cases} -\frac{n}2 & n\bmod 2=0\\ \frac{n+1}2 & n \bmod 2=1 \end{cases}

To show that g=f^{-1}, let z\in \mathbb Z and n\in \mathbb N. We'll then show f(z)=n\equiv z=g(n).

"\rightarrow" Let f(z)=n.

• Case I: z\le 0. Then n=-2z, so n is even, thus g(n)=-(-2z)/2=z.

• Case II: z> 0. Then n=2z-1, so n is odd, thus g(n)=(2z-1 + 1)/2=z.

Thus g(n)=z in both cases.

"\leftarrow" Let z=g(n).

• Case I: n odd. Then z=(n+1)/2, so z>0, thus f(z)=2((n+1)/2)-1=n.

• Case I: n even. Then z=-n/2, so z\le 0, thus f(z)=-2(-n/2)=n.

Thus f(z)=n in both cases.

This concludes the proof of everything required by the characterization of the inverse function.

Teaching/DiscreteMathFall2011/FunctionInversion (last edited 2011-11-16 07:32:19 by AndreasKloeckner)