Can we find the ranges of f and g, and the function f ∘ g?

Can we find the ranges of $f$ and $g$ , and the function $f \circ g$ ?

Question

Functions

$f$

and

$g$

are defined by

\begin{align*} f:{}&x \to \log_a x,\quad&&(x \in \mathbb{R}_+, a>1),\\ g:{}&x\to \frac{1}{x},&&(x \in \mathbb{R}_+). \end{align*}

State the ranges of $f$ and $g$ , and show that if $h$ denotes the composite function $f \circ g$ , then $h(x)+f(x)=0.$

Explain briefly why the composite function $g \circ f$ cannot be properly defined unless the domain is restricted to a subset of $\mathbb{R}_+$ , and state a possible subset which would be suitable.

Define fully the inverses of $f$ and $g$ , and determine whether or not $h^{-1}(x)+f^{-1}(x)=0$ .

Notational note

Note that the symbol $\mathbb{R}$ refers to the set of all real numbers, that is, all the numbers on the number line.

Similarly, we denote the integers $(...,-2,-1,0,1,2,...)$ by $\mathbb{Z}$ , and the rationals (numbers $a/b$ , where $a$ and $b$ are whole numbers, $b \neq 0$ ) by $\mathbb{Q}$ .

Adding a small subscript “ $+$ ” after any of these symbols means we only include the positive members of the relevant set. So, for example, $\mathbb{R}_+$ refers to the positive real numbers. Similarly, a small subscript “ $-$ ” would mean include only the negative numbers in this set.

An alternative notation which is also quite common is to use a superscript “ $+$ ” instead, for example $\mathbb{R}^+$ .

Solution

The range of $f$ is $\mathbb{R}$ and the range of $g$ is $\mathbb{R}_+$ .

It is often helpful for domain and range questions to think about the graph of the relevant functions; here is the graph of $y=f(x)$ :

Graph of f of x. The function is only defined for positive x. It tends to minus infinity as x tends to 0, and to infinity as x tends to infinity, with decreasing gradient.

We can see that the graph gives every $y$ value for some $x$ value, so the range is the whole of $\mathbb{R}$ .

Here is the graph of $y=g(x)$ (confining the domain to only $\mathbb{R}_+$ ):

Graph of g of x for positive x. It tends to infinity as x tends to 0, and tends to 0 as x tends to infinity.

Then it’s clear that the values of $y$ which the graph passes through are those greater than zero, so the range of $g$ is $\mathbb{R}_+$ .

We have

\begin{align*} h(x)&=(f\circ g)(x)\\ &=f(g(x))\\ &=\log_a\left(\frac{1}{x}\right)\\ &=\log_a(x^{-1})\\ &=-\log_a x \end{align*}

So $h(x)+f(x)=-\log_a x+\log_a x=0$ .

For the composition $f_2 \circ f_1$ of two functions, $f_1$ and $f_2$ , to be properly defined, we must have the range of $f_1$ contained within the domain of $f_2$ . (Otherwise, for some $x$ in the domain of $f_1$ , we would have $f_1(x)$ not in the domain of $f_2$ , so $f_2(f_1(x))$ will not be defined.)

In this case, we are considering $g\circ f$ so $f_2=g$ and $f_1=f$ . Therefore we require the range of $f$ to be contained within the domain of $g$ .

But the domain of $g$ is $\mathbb{R}_+$ while the range of $f$ is $\mathbb{R}$ . So the range of $f$ is not contained within the domain of $g$ , thus $g \circ f$ is not properly defined unless we restrict the domain to a subset of $\mathbb{R}_+$ .

For a suitable subset, we need the range of $f$ with this restricted domain to be a subset of $\mathbb{R}_+$ . We see that $f(x)>0$ when $x>1$ , so any subset of $\{x:x>1\}$ (the real numbers greater than $1$ ) would suffice.

When defining a function, it is vital to specify the domain.

Recall that for any function $f$ , the domain of $f^{-1}$ is the range of $f$ : every possible output of $f$ has to be an allowable input to $f^{-1}$ to get back to the original value.

As $f(x)=\log_a x$ , to find the inverse, we can write $x=f(y)$ and rearrange to find $y$ in terms of $x$ .

What we’re really doing is saying that $f\bigl(f^{-1}(x)\bigr)=x$ and then writing $y=f^{-1}(x)$ . So we are using the result that $f\circ f^{-1}$ and $f^{-1}\circ f$ are both the identity function.

So in our case,

\begin{align*} &&x&=f(y)&&\quad\\ \implies&&x&=\log_a y\\ \implies&&a^x&=y \end{align*}

So $f^{-1}(x)=a^x$ for $x$ in $\mathbb{R}$ .

For

$g(x)$

, we have

\begin{align*} &&x&=g(y)&&\quad\\ \implies&&x&=\frac{1}{y}\\ \implies&&y&=\frac{1}{x} \end{align*}

So $g^{-1}(x)=1/x$ for $x$ in $\mathbb{R}_+$ .

For the final part, we start by finding

$h(x)$

\begin{align*} &&x&=h(y)&&\quad\\ \implies&&x&=-\log_a y\\ \implies&&-x&=\log_a y\\ \implies&&a^{-x}&=y \end{align*}

so $h^{-1}(x)=a^{-x}$ .

Thus $h^{-1}(x)+f^{-1}(x)=a^{-x}+a^x>0$ for all $x$ , so $h^{-1}(x)+f^{-1}(x)\ne 0$ .

Can we find the ranges of $f$ and $g$ , and the function $f \circ g$ ?

Question

Functions

$f$

and

$g$

are defined by

\begin{align*} f:{}&x \to \log_a x,\quad&&(x \in \mathbb{R}_+, a>1),\\ g:{}&x\to \frac{1}{x},&&(x \in \mathbb{R}_+). \end{align*}

State the ranges of $f$ and $g$ , and show that if $h$ denotes the composite function $f \circ g$ , then $h(x)+f(x)=0.$

Explain briefly why the composite function $g \circ f$ cannot be properly defined unless the domain is restricted to a subset of $\mathbb{R}_+$ , and state a possible subset which would be suitable.

Define fully the inverses of $f$ and $g$ , and determine whether or not $h^{-1}(x)+f^{-1}(x)=0$ .

Notational note

Note that the symbol $\mathbb{R}$ refers to the set of all real numbers, that is, all the numbers on the number line.

Similarly, we denote the integers $(...,-2,-1,0,1,2,...)$ by $\mathbb{Z}$ , and the rationals (numbers $a/b$ , where $a$ and $b$ are whole numbers, $b \neq 0$ ) by $\mathbb{Q}$ .

An alternative notation which is also quite common is to use a superscript “ $+$ ” instead, for example $\mathbb{R}^+$ .

Solution

The range of $f$ is $\mathbb{R}$ and the range of $g$ is $\mathbb{R}_+$ .

It is often helpful for domain and range questions to think about the graph of the relevant functions; here is the graph of $y=f(x)$ :

Graph of f of x. The function is only defined for positive x. It tends to minus infinity as x tends to 0, and to infinity as x tends to infinity, with decreasing gradient.

We can see that the graph gives every $y$ value for some $x$ value, so the range is the whole of $\mathbb{R}$ .

Here is the graph of $y=g(x)$ (confining the domain to only $\mathbb{R}_+$ ):

Graph of g of x for positive x. It tends to infinity as x tends to 0, and tends to 0 as x tends to infinity.

Then it’s clear that the values of $y$ which the graph passes through are those greater than zero, so the range of $g$ is $\mathbb{R}_+$ .

We have

\begin{align*} h(x)&=(f\circ g)(x)\\ &=f(g(x))\\ &=\log_a\left(\frac{1}{x}\right)\\ &=\log_a(x^{-1})\\ &=-\log_a x \end{align*}

So $h(x)+f(x)=-\log_a x+\log_a x=0$ .

In this case, we are considering $g\circ f$ so $f_2=g$ and $f_1=f$ . Therefore we require the range of $f$ to be contained within the domain of $g$ .

When defining a function, it is vital to specify the domain.

Recall that for any function $f$ , the domain of $f^{-1}$ is the range of $f$ : every possible output of $f$ has to be an allowable input to $f^{-1}$ to get back to the original value.

As $f(x)=\log_a x$ , to find the inverse, we can write $x=f(y)$ and rearrange to find $y$ in terms of $x$ .

So in our case,

\begin{align*} &&x&=f(y)&&\quad\\ \implies&&x&=\log_a y\\ \implies&&a^x&=y \end{align*}

So $f^{-1}(x)=a^x$ for $x$ in $\mathbb{R}$ .

For

$g(x)$

, we have

\begin{align*} &&x&=g(y)&&\quad\\ \implies&&x&=\frac{1}{y}\\ \implies&&y&=\frac{1}{x} \end{align*}

So $g^{-1}(x)=1/x$ for $x$ in $\mathbb{R}_+$ .

For the final part, we start by finding

$h(x)$

\begin{align*} &&x&=h(y)&&\quad\\ \implies&&x&=-\log_a y\\ \implies&&-x&=\log_a y\\ \implies&&a^{-x}&=y \end{align*}

so $h^{-1}(x)=a^{-x}$ .

Thus $h^{-1}(x)+f^{-1}(x)=a^{-x}+a^x>0$ for all $x$ , so $h^{-1}(x)+f^{-1}(x)\ne 0$ .

Can we find the ranges of fff and ggg, and the function f∘gf \circ gf∘g?

Question

Notational note

Solution

Can we find the ranges of fff and ggg, and the function f∘gf \circ gf∘g?

Question

Notational note

Solution

Can we find the ranges of $f$ and $g$ , and the function $f \circ g$ ?

Can we find the ranges of $f$ and $g$ , and the function $f \circ g$ ?