Review question

# Can we work with the domain, codomain and range of a bijection? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5966

## Solution

1. 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.

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

1. it is $one-one,$ and
2. 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.

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.

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?’

1. $R'$ 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*}