# Functions and range

$$a \colon \mathbb{R}\setminus\{0\} \to \mathbb{R} \;\text{ defined by }\; a(x)= 6/x \\ b \colon \mathbb{Z} \to \mathbb{R} \;\text{ defined by }\; b(x) = 3x + 1$$ a) State the range of the functions and if they are "onto".
b) State if they are one to one.
c) Does the composition $a \circ b$ exist , and if it does give the function, domain, codomain and range.

My work so far:
a) Range of $a$ is all elements of $\mathbb{R}$ except $0$ - not onto because the range is not the same as the codomain. Range of $b$ is all elements of $\mathbb{R}$ - "onto" because the range is the same as the codomain.

b) Both are one to one because one element of $y$ has one $x$.

c) This is where I'm stuck.

• is the domain for b is $\mathbb Z$, the set of integers? Then it is not onto and the range is only $\{3n+1| n \in \mathbb Z\}$ . Apr 22 '15 at 4:04
• nope the domain is R - {0} Apr 22 '15 at 4:07
• For the second equation? You had written Z. Apr 22 '15 at 4:08
• sorry, i meant b where the function is $3x+1$. Apr 22 '15 at 4:08
• Tried to work this out and got that it does exist because the range of $b$ is within the domain of f $a$ as there are no range values of $b$ that are 0. meaning $a(b(x)) = 6/(3x+1)$ domain is Z co domain is R Apr 22 '15 at 14:56

You were given that $a: \mathbb{R} - \{0\} \to \mathbb{R}$ is defined by $$a(x) = \frac{6}{x}$$ and $b: \mathbb{Z} \to \mathbb{R}$ is defined by $b(x) = 3x + 1$.
The function $a \circ b$ is defined provided that the range of $b$ is contained in the domain of $a$. Since $3x + 1 \neq 0$ for any integer $x$, the range of $b$ is contained in $\mathbb{R} - \{0\}$, which is the domain of $a$. Thus, $$(a \circ b)(x) = a(b(x)) = a(3x + 1) = \frac{6}{3x + 1}$$ exists, as you concluded.
The domain of $a \circ b$ is $\mathbb{Z}$, the domain of $b$, and its codomain is $\mathbb{R}$, the range of $a$. The range of $a \circ b$ is the set $$R_{a \circ b} = \left\{\frac{6}{3n + 1}~\bigg|~n \in \mathbb{Z}\right\}$$