Can we work with the domain, codomain and range of a bijection?

Question

The function $f$ from $\mathbb{R}$ to $\mathbb{R}$ is given by $f: x\mapsto x^2+2x+2$ . Find the range of $f$ , and state, with a reason, whether or not $f$ is bijective. The range of $f$ is denoted by $S$ , and the function $g$ from $S$ to $\mathbb{R}$ is given by $g: x\mapsto 1/x$ . State the image of $x$ under the composite function $g\circ f$ , and give the range of this composite function.
$R'$ denotes the set of real numbers excluding $0$ and $1$ . Functions $\phi$ and $\psi$ from $R’$ to $R'$ are given by $\phi : x\mapsto \frac{1}{1-x}, \qquad \qquad \psi : x \mapsto 1-x.$ Give in a similar form the definitions of $\phi ^{-1}$ and $(\phi \circ \psi)^{-1}$ .

Suggestion

To explain, the symbol $\mathbb{R}$ denotes the set of all real numbers, that is, the set of all numbers on the number line.

We’re definitely going to need to understand the word bijective for this question. Click on the link in the question above for a careful explanation.

The notation $f \circ g$ means ‘the function $g$ followed by the function $f$ ’, or ‘ $f$ composed with $g$ ’.

Could we complete the square on $x^2+2x+2$ ?

Solution

In the context of this question, we could say a function $f$ is bijective if and only if

it is (one-one,) and
its codomain equals its range.

So for $f$ to be a bijection, no horizontal line cuts the graph of $y = f(x)$ more than once, and if $k$ is a value in the codomain, then the horizontal line $y = k$ cuts $y = f(x)$ exactly once. In this case, we say $f$ is ‘one-one and onto’.

We can rewrite the definition of $f$ so that it will be easier to see the range. Completing the square, we have $f: x\mapsto (x+1)^2 +1.$ Now $(x+1)^2$ can take every value greater than or equal to $0$ , so $f(x)$ must be greater than or equal to $1$ .

Therefore the range of $f$ is the set of all real numbers greater than or equal to $1$ , which can be written as $\{f(x) \in \mathbb{R}, f(x) \geq 1\}$ .

Since $f(0) = f(-2)$ , $f$ is not one-one, and therefore not bijective.

We know by the definition of $g$ that

g\circ f:x \mapsto \frac{1}{f(x)} = \frac{1}{x^2+2x+2},

which is the image of $x$ under $g \circ f$ . We don’t have to worry about the denominator being zero since $f(x) > 0$ .

Since $f(x) \geq 1$ , we have $\dfrac{1}{f(x)} \leq 1$ . Also, since $f(x)>0$ , we have $\dfrac{1}{f(x)} > 0$ .

The function $g \circ f$ therefore takes no value less than or equal to $0$ or greater than $1$ .

Does it take every value in the interval from 0 to 1?

Take any $y$ with $0 < y \leq 1$ . Let $w = \dfrac{1}{y}$ , so $w \geq 1$ . Then $f(z)=w$ for some $z$ (since $f$ takes every value greater than or equal to $1$ ), and so $g \circ f(z) = y$ .

Hence the range of $g \circ f$ is $0 < g \circ f(x) \leq 1$ .

Or we could say that to find the range of a function, we can draw a graph of the function, and ask, ‘What are the possible $y$ -values?’

$R'$ denotes the set of real numbers excluding $0$ and $1$ . Functions $\phi$ and $\psi$ from $R’$ to $R'$ are given by $\phi : x\mapsto \dfrac{1}{1-x}, \qquad \qquad \psi : x \mapsto 1-x.$

Give in a similar form the definitions of $\phi ^{-1}$ and $(\phi \circ \psi)^{-1}$ .

If $y=\dfrac{1}{1-x}$ , can we make $x$ the subject here? We have $y-yx=1$ , and so $x=1-\dfrac{1}{y}$ . Therefore $\phi ^{-1}: x\mapsto 1-\frac{1}{x}.$

Check: does

$(\phi^{-1} \circ \phi)(x)=x$

? If not, we’ve made a mistake! We have

\begin{align*} (\phi^{-1} \circ \phi)(x) &= \phi^{-1} ( \phi(x))\\ &= \phi^{-1} \left(\dfrac{1}{1-x}\right) \\ &= 1-\frac{1}{\left(\dfrac{1}{1-x}\right)} \\ &= 1-(1-x) \\ &= x. \end{align*}

In order to find $(\phi \circ \psi)^{-1}$ we first need to find $\phi \circ \psi$ . We have $\phi (\psi(x)) = \phi(1-x)=\frac{1}{1-(1-x)}=\frac{1}{x}.$ This is clearly a self-inverse function, so $(\phi \circ \psi)^{-1}:x \mapsto \frac{1}{x}.$

Alternatively, we could use the fact that

$(\phi \circ \psi)^{-1} = \psi^{-1} \circ \phi^{-1}$

, and the observation that

$\psi$

is self-inverse (it is its own inverse), so

\begin{align*} (\phi \circ \psi)^{-1}(x) &= \psi^{-1}\left(\phi^{-1}(x)\right) \\ &= \psi^{-1} \left(1 - \dfrac{1}{x}\right) \\ &= 1 - \left(1 - \dfrac{1}{x}\right) \\ &= \dfrac{1}{x}. \end{align*}