Can we sketch and describe these composite functions?

Can we sketch and describe these composite functions?

Question

1. The functions $f$ and $g$ are defined by

$$\begin{align*} f&: x\mapsto e^x, &&x\in\mathbb{R}, \\ g&: x\mapsto x-1, &&x\in\mathbb{R}. \end{align*}$$

Sketch in a single diagram the graphs of $f$ and the composite functions $gf$ and $fg$, labelling each graph clearly. State briefly the relationship

1. between the graphs of $f$ and $gf$,
2. between the graphs of $f$ and $fg$.
3. Express $(1-x)(x-3)$ in the form $a-(x-b)^2$, where $a$ and $b$ are constants. State the coordinates of the maximum point on the graph of $y=(1-x)(x-3)$, and state what symmetry the graph possesses. Sketch the graph of $y=e^{(1-x)(x-3)}$.

Suggestion

How do we compose two functions together?

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How does expressing $(1-x)(x-3)$ in the form $a-(x-b)^2$ help us find the maximum value of $y$?

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Things to look out for when sketching a curve:

• can I find the asymptotes?
• can I find the intercepts with the axes?
• can I find the stationary points and their nature?
• what happens to the curve as $x\rightarrow \pm \infty ?$

Solution

First, we’ll find the composite functions $gf$ and $fg$. We have $gf = e^x-1$ and $fg = e^{x-1}.$

Let’s first sketch $y=e^x$. We know that $e^x\to 0$ as $x\to -\infty$, the graph crosses the $y$-axis at $(1,0)$ and it is strictly increasing. So it looks roughly like this:

$y=e^x$

Now let’s add a sketch of $e^x-1$. We know that this is a translation of $y=e^x$ by $-1$ in the $y$-direction, and that $y=e^x$ and $y=e^x-1$ cannot intersect. So our graph now looks like:

$y=e^x$ and $y=e^x -1$

Finally, we add $y=e^{x-1}$. This is a translation of $y=e^x$ by $1$ in the $x$-direction. The graphs of $y = e^x$ and $y=e^{x-1}$ don’t meet, but $y=e^{x-1}$ does meet $y=e^x-1$. So a sketch of the three functions looks like this:

$y=e^x$ and $y=e^x -1$ and $y=e^{(x-1)}$

The graph of $gf$ is a translation of $f$ by one unit in the negative $y$ direction.

The graph of $fg$ is a translation of $f$ by one unit in the positive $x$ direction.

We expand $(1-x)(x-3)$ to obtain $-x^2+4x-3$.

Completing the square, we find that $(1-x)(x-3)=-x^2+4x-3=-(x-2)^2+4-3=1-(x-2)^2.$

We could also do this by expanding both expressions and solving for $a$ and $b$ after comparing coefficients.

From the rewriting of the equation $y=(1-x)(x-3)$ in the form $y=1-(x-2)^2$, we see that the maximum point on the graph occurs when $y=1$ because the expression $-(x-2)^2$ must be negative or zero. The expression $-(x-2)^2$ is $0$ when $x=2$ so the maximum point has coordinates $(2,1)$. The graph possesses line symmetry about $x=2$.

So the graph

• has a maximum at $x=2$, $y=e$;
• has symmetry about $x=2$;
• has its $y$-intercept at $e^{-3}$;
• sees $y\rightarrow0$ as $x\rightarrow\pm\infty$.

The sketch is