Review question

# Can we find the ranges of $f$ and $g$, and the function $f \circ g$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7607

## Solution

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 $h(x)+f(x)=0.$

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 $y=f(x)$:

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 $\mathbb{R}_+$):

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

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.

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.

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

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 $h^{-1}(x)+f^{-1}(x)\ne 0$.