Can we sketch the inverse of this composite function?

Can we sketch the inverse of this composite function?

Question

The functions

$f$

,

$g$

and

$h$

are defined by

$$\begin{align*} f &: x \mapsto \ln x, & \qquad \left(x\in\mathbb{R}^+\right), \\ g &: x \mapsto \frac{1}{x},& \qquad \left(x\in\mathbb{R}^+\right), \\ h &: x \mapsto x^2, &\quad \big(x\in\mathbb{R}\big). \end{align*}$$

1. Give definitions of each of the functions $fg$ and $f^{-1}$, and state, in each of the following cases, a relationship between the graphs of
2. $f$ and $fg$,
3. $f$ and $f^{-1}$.

The diagram shows a sketch of the graph $y=hf(x)$. State how the sketch shows that $hf$ is not one–one, and prove that, if $\alpha$ and $\beta$, where $0<\alpha <\beta$, are such that $hf(\alpha)=hf(\beta)$, then $\alpha=g(\beta)$.

1. The function $\phi$ is defined by $\phi : x \mapsto hf(x),\qquad (0 < x \leq 1).$ Sketch the graph of $\phi^{-1}$, and give an explicit expression in terms of $x$ for $\phi^{-1}(x)$.

Suggestion

1. Give definitions of each of the functions $fg$ and $f^{-1}$ and state the relationship between the graphs of $f$ and both $fg$ and $f^{-1}$.

The function $\ln x$ is the logarithm to base $e$ of $x$.

Be careful with the order of your functions when composing them! If we take the function $fg$, which order for $f$ and $g$ do we apply here?

Would a sketch help us describe the relationship between $f$ and $fg$? And $f$ and $f^{-1}$?

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1. State how the sketch shows that $hf$ is not one–one…

What does one–one mean? What do graphs of one–one functions look like?

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…and prove that, if $\alpha$ and $\beta$, where $0<\alpha <\beta$, are such that $hf(\alpha)=hf(\beta)$, then $\alpha=g(\beta)$.

What is the function $hf$?

How do we find the inverse of a function? And its graph?

What happens if we sketch $y=\phi(x)$ on a graph, then the line $y=x$?

How do we find the inverse of a function using algebra?

Can we rearrange $y=(\ln x)^2$ to make $x$ the subject? What happens when we take the square root?

Solution

For $fg$: composing $f$ and $g$ we find $fg(x) = f[g(x)]=\ln\left(\frac{1}{x}\right)=-\ln x.$

For $f^{-1}$: rearranging the formula $y=\ln x$ to make $x$ the subject we find $x=e^y$, so $f^{-1}(x)=e^x.$

…(a) state a relationship between the graphs of $f$ and $fg$

Reflection in the $x$–axis.

…(b) state a relationship between the graphs of $f$ and $f^{-1}.$

Reflection in the line $y=x$.

The diagram shows a sketch of the graph $y=hf(x)$. State how the sketch shows that $hf$ is not one–one.

There are horizontal lines that cut the graph twice. In fact, for every positive $y$-value, there are two $x$-values that $hf$ maps to the $y$-value.

… and prove that, if $\alpha$ and $\beta$, where $0<\alpha <\beta$, are such that $hf(\alpha)=hf(\beta)$, then $\alpha=g(\beta)$.

The composition $hf(x)=h(\ln x)=(\ln x)^2$.

$$\begin{align*} hf(\alpha) &= hf(\beta) \\ \Longrightarrow(\ln \alpha)^2 &= (\ln\beta)^2\\ \Longrightarrow \ln \alpha &= -\ln\beta. \end{align*}$$

We must have the minus sign here, since

$\alpha \neq \beta$

. Now we have

$$\begin{align*} \ln \alpha &= -\ln\beta = \ln\left(\frac{1}{\beta}\right)\\ \Longrightarrow \alpha &= \frac{1}{\beta}=g(\beta). \end{align*}$$

From the graph we see that if $\alpha < \beta$, $0<\alpha < 1$ and $\beta > 1$.

$$\begin{align*} y &= (\ln x)^2 \\ \Longrightarrow -\sqrt{y} &= \ln x \end{align*}$$

(since

$0 < x \leq 1$

, we know

$\ln x$

is negative) and exponentiating both sides gives

$$\begin{align*} x &= e^{-\sqrt{y}} \end{align*}$$

Switching $x$ and $y$ we find that the inverse of $\phi$ is $\phi^{-1}(x)=e^{-\sqrt{x}}.$ We can find the graph of the inverse of a one-one function by reflecting in the line $y=x$.

Below are the graphs of both $y=\phi(x)$ and $y=\phi^{-1}(x)$.