Practice Work

1

(a)

$$\begin{array}{rl}f\circ f& =\{\begin{array}{ll}f(n-1)& \text{if }n>5\\ f(n+1)& \text{if }0\le n\le 5\end{array}\\ & =\{\begin{array}{ll}(n-1)-1& \text{if }n>5\text{ and }n-1>5\\ (n-1)+2& \text{if }n>5\text{ and }0\le n-1\le 5\\ (n+1)-1& \text{if }0\le n\le 5\text{ and }n+1>5\\ (n+1)+2& \text{if }0\le n\le 5\text{ and }0\le n+1\le 5\end{array}\\ & =\{\begin{array}{ll}(n-1)-1& \text{if }n>6\\ (n-1)+2& \text{if }n=6\\ (n+1)-1& \text{if }n=5\\ (n+1)+2& \text{if }n\le 4\end{array}\end{array}$$

title:
Correct answer.

(b)

$$\begin{array}{rl}f\circ g& =f(g(n))\\ & =f(3n+2)\\ & =\{\begin{array}{ll}3n+2-1& \text{if }3n+2>5\\ 3n+2+1& \text{if }0\le 3n+2\le 5\end{array}\\ & =\{\begin{array}{ll}3n+1& \text{if }3n>3\\ 3n+3& \text{if }0\le 3n+2\le 5\end{array}\\ & =\{\begin{array}{ll}3n+1& \text{if }n>1\\ 3n+3& \text{if }0\le 3n+2\le 5\end{array}\end{array}$$

title:
Correct answer.

(c)

$$\begin{array}{rl}g\circ f& =g(f(n))\\ & =\{\begin{array}{ll}3(n-1)+2& \text{if }n>5\\ 3(n+1)+2& \text{if }0\le n\le 5\end{array}\\ & =\{\begin{array}{ll}3n-3+2& \text{if }n>5\\ 3n+3+2& \text{if }0\le n\le 5\end{array}\\ & =\{\begin{array}{ll}3n-1& \text{if }n>5\\ 3n+5& \text{if }0\le n\le 5\end{array}\end{array}$$

title:
Correct answer.

2

1. No, there isn’t a possible function that is 1:1.
2. Function defined: $$f(a)=1,f(b)=2,f(c)=3,f(d)=4,f(e)=4$$
3. Function defined as above.
4. Function defined: $$f(a)=f(b)=f(c)=f(d)=f(e)=1$$ The range of this function is $f(x)=1$.

title:
Correct answers.