# Functions: domain, codomain, range and graph

I have a question in my course.

Let $$A = \{0, 1\}$$, $$B = \{4, 5, 6, 7\}$$ and $$C = \mathbb{N}$$ (natural numbers). Consider the following functions:

$$f:A\to B$$ s.t. $$f(x) = x+5$$

$$g:B\to C$$ s.t. $$g(x) = 2x−2$$

$$h:C \to A$$ s.t. $$h(x) = \begin{cases} 0 & \text{if x < 10} \\ 1 & \text{if x \ge 10} \\ \end{cases}$$

Which of the following compositions are defined? If they are defined, compute their domain, codomain, range and graph.

(a) $$f \circ g$$

(b) $$g \circ f$$

(c) $$h \circ g$$

(d) $$g\circ h$$

• Domain of $$f$$ is $$A = \{0, 1\}$$; domain of $$g$$ is $$B = \{4,5,6,7\}$$; domain of $$h$$ is $$C = \mathbb{N}$$;

• Codomain of $$f$$ is $$B = \{4,5,6,7\}$$; codomain of $$g$$ is $$C = \mathbb{N}$$; codomain of $$h$$ is $$A = \{0,1\}$$.

• Range of $$f$$ is $$\{5, 6\}$$; range of $$g$$ is $$\{4,8,10,12\}$$; range of $$h$$ is $$\{0,1\}$$.

a) $$f \circ g = f(g(x)) = 2x + 5 -2 = 2x + 3$$

b) $$g \circ f = g(f(x)) = 2(x + 5) – 2 = 2x + 8$$

c) $$h \circ g = -2 \text{ if } x < 10;\ h \circ g = 0 \text{ if } x \ge 10$$

d) $$g \circ h = g(h(x)) = -2\text{ if } x < 10; g \circ h = 0 \text{ if } x \ge 10$$

Could I please get a review and how to calculate the graph?

• You need to check whether the compositions are even defined before you develop formulas for them. Also, the range of $g$ is $\{6, 8, 10, 12\}$. Note that $g(4) = 2 \cdot 4 - 2 = 6$. Nov 23 '20 at 21:13
• how can i check if the compositions are defined? ah, thanks for the comment. I missed the 6! Nov 23 '20 at 22:28

The composite function $$f \circ g$$ is only defined if the codomain of $$g$$ is contained in the domain of $$f$$. Depending on the definition you are using, you may need the stronger condition that the codomain of $$g$$ equals the domain of $$f$$.

In this case, the codomain of $$g$$ is $$\mathbb{N}$$ and the domain of $$A = \{0, 1\}$$. Moreover, as noted in the comments, since the range of $$g$$ is $$\{6, 8, 10, 12\}$$, the elements in the range of $$g$$ are not contained in the domain of $$f$$. Thus, the composite function $$f \circ g$$ is undefined.

Since $$f: A \to B$$ and $$g: B \to C$$, the composite function $$g \circ f$$ is defined since the codomain of $$f$$ equals the domain of $$g$$. As you correctly concluded, the function $$g \circ f: A \to C$$ is defined by $$(g \circ f)(x) = g(f(x)) = g(x + 5) = 2(x + 5) - 2 = 2x + 8$$ The graph of the composite function $$g \circ f$$ is the set of all ordered pairs $$(a, c) \in A \times C$$ such that $$(g \circ f)(a) = c$$ for some $$a \in A$$. Since $$(g \circ f)(0) = 8$$ and $$(g \circ f)(1) = 10$$, the graph of the function $$g \circ f$$ is the set $$G_{g \circ f} = \{(0, 8), (1, 10)\}$$

Since $$g: B \to C$$ and $$h: C \to A$$, the composite function $$h \circ g$$ is defined since the codomain of $$g$$ equals the domain of $$h$$. Since $$h: C \to A$$ is defined by $$h(x) = \begin{cases} 0 & \text{if x < 10}\\ 1 & \text{if x \geq 10} \end{cases}$$ and $$g: B \to C$$ is defined by $$g(x) = 2x - 2$$, observe that if $$x \in B$$, then $$(g \circ h)(x) = \begin{cases} 0 & \text{if x < 6}\\ 1 & \text{if x \geq 6} \end{cases}$$ To see this, observe that if $$x < 6$$, $$g(x) = 2x - 2 < 2 \cdot 6 - 2 = 10$$, and that if $$x \geq 6$$, $$g(x) = 2x - 2 \geq 2 \cdot 6 - 2 = 10$$. Since \begin{align*} (h \circ g)(4) & = h(g(4)) = h(6) = 0\\ (h \circ g)(5) & = h(g(5)) = h(8) = 0\\ (h \circ g)(6) & = h(g(6)) = h(10) = 1\\ (h \circ g)(7) & = h(g(7)) = h(12) = 1 \end{align*} the graph of $$h \circ g$$ is $$G_{h \circ g} = \{(4, 0), (5, 0), (6, 1), (7, 1)\}$$

Observe that $$h: C \to A$$ and $$g: B \to C$$, so the codomain of $$h$$ is not contained in the domain of $$g$$ since $$A$$ is not a subset of $$B$$. Thus, the composite function $$g \circ h$$ is not defined.

• thank you, very well answered! I can read well and understand the procedure. Nov 24 '20 at 8:12
• so the codomain of f is B{4,5,6,7} ? Nov 24 '20 at 12:21
• Yes, as you yourself wrote, the codomain of $f$ is $B$, the codomain of $g$ is $C$, and the codomain of $h$ is $A$. When you write the function $f: A \to B$ defined by ..., that means the domain of $f$ is $A$ and the codomain of $f$ is $B$. The rule defining the function $f$ determines its range, which must be a subset of the codomain. Nov 24 '20 at 12:32
• not sure why 𝑔:𝐵→𝐶 is defined by 𝑔(𝑥)=2𝑥−2, observe that if 𝑥∈𝐵, then (𝑔∘ℎ)(𝑥)={0 if 𝑥<6; 1 if 𝑥≥6}. Where does 6 comes from? Nov 24 '20 at 12:41
• ah, ok, you take 6 as it is the lowest value in B range. Nov 24 '20 at 12:47