Let $In(t) = -(t-1)^2 + 5$ with its natural or implied domain.

Graph of $y = Out(x) =2|x-3|+1$

The graph of $Out$ has a “V”-shape, opening up, with a corner at $(3,1)$.
This illustrates that $Out$ decreases on $(-\infty , 3]$ and increases on $[3, \infty )$. $Out$ has a minimum value of $1$, which occurs at $3$.

If we think of tracing the domain from left to right on the graph, the corresponding values of $Out$ decrease from $\infty$ down to $1$, then reverse direction and increase from $1$ back to $\infty$.

The only values that can come out of the composition, $Out \circ In$, are the values of $Out$. Therefore, the values of $Out \circ In$ are inside the interval $[1, \infty )$.

Graph of $z = In(t) = -(t-1)^2 + 5$

Since $In$ is a quadratic function, its graph has the shape of a parabola, opening down. $In$ increases on $(-\infty , 1]$ and decreases on $[1, \infty )$. $In$ has a maximum value of $5$, which occurs at $1$.

If we think of tracing the domain from left to right on the graph, the values of $In$ increase from $-\infty$ up to $5$, then reverse direction and decrease from $5$ back to $-\infty$.

The output from the function $In$ will become the input to $Out$. Therefore, in the composition, the new input to $Out$ will be the interval $(-\infty , 5]$, which is the output of $In$. This interval is traced twice: the values going into $Out$ will begin at $-\infty$, go up to $5$, then turn around and go back down to $-\infty$. The values of $Out$ will similarly repeat themselves in reverse.

Now to examine the composition:

$\blacktriangleright$ First, the inside function: $In(t) = -(t-1)^2 + 5$

The natural domain of $In(t)$ is the whole real line. As we move from left to right along the real line, the domain numbers increase from $-\infty$ to $\infty$.

The corresponding movement in the range has the function values increasing from $-\infty$ to $5$. The maximum function value $5$ occurs at $1$ in the domain. As we keep moving beyond $1$ in the domain, the corresponding movement in the range is for the function values to decrease from $5$ to $-\infty$.

This path in the range of $In(t)$ becomes the path in the domain of $Out(x)$.

$\blacktriangleright$ Movement inside the domain of $Out(x)$.

Inside the natural domain of $Out(x) =2|x-3|+1$, the domain numbers will increase from $-\infty$ to $5$. Then they will decrease from $5$ back down to $-\infty$.

Following our graph of $Out(x)$, we will move from the far left toward the right until we reach $5$. Then we will turn around and move toward the left.

$\blacktriangleright$ Put those together.

We have the input to $Out$ moving from $-\infty$ up to $5$ and then back down to $-\infty$. What are the corresponding function values for $Out$?

• As the domain numbers move through $(-\infty , 5]$, the values of $Out$ decrease from $\infty$ to $1$ and then increase from $1$ to $5$. In the graph, we can see this switch as a corner in the graph. The graph comes down to the right, hits the corner, then moves back up to the right.
  

  So, we really should view $(-\infty , 5]$ as $(-\infty , 3] \cup [3,5]$ and highlight the change in behavior.

  
  • On $(-\infty , 3]$, $Out$ decreases from $\infty$ to $1$.
  • On $[3,5]$, $Out$ increases from $1$ to $5$

  That was on our first pass through $(-\infty , 5]$.
  

  The input to $Out$ continues beginning at $5$ and moving towards $-\infty$.
  

  As the overall domain to the whole compostion continues to cover $[5, \infty )$, The function $In$ reverses this interval and sends the reversed interval into $Out$. As the overall composition domain continues to move through $[5, \infty )$, the input to $Out$ traces backwards from $5$ to $-\infty$. The overall composition graph keeps moving to the right, but it is retracing the left side in reverse, mirroring it.

• Moving backwards through $(-\infty , 5]$, The values of $Out$ first decrease from $5$ to $1$, arriving again at the corner that occurs at $3$. Again, we really should look at $(-\infty , 5]$ as $(-\infty , 3] \cup [3,5]$.
  • As the domain moves from $5$ to $3$, $Out$ decreases from $5$ to $1$.
  • As the domain moves from $3$ to $-\infty$, $Out$ increases from $1$ to $\infty$.

Since we reverse our travelling direction inside the domain, we hit the corner twice. Plus, right at our reversal, we will create a hill in the graph made by our own retracing of the domain steps.

All together now...

Graph of $w = Out(In(m)) =2 \, |(-(m-1)^2 + 5)-3|+1 = 2 \, |-(m-1)^2 + 2| + 1$

Notice:

$\blacktriangleright$ The function values of $Out \circ In$ are all inside the interval $[1, \infty )$, which was the range of $Out$. Thus, the composition has no zeros, because $Out$ has no zeros.

$\blacktriangleright$ The graph of $Out \circ In$ has two corners, because the corner of the graph of $Out$ was included twice.

The corner in the graph of $Out$ occurs when the input to $Out$ is $3$. When is the output of $In$ equal to $3$?

There is a corner at $1 - \sqrt {2} \approx -0.414$ and one at $1 + \sqrt {2} \approx 2.414$.

$\blacktriangleright$ The graph of $In$ has a hill at $1$, the function values reverse themselves. Therefore, the composition also has its values reversing at $1$.

$\blacktriangleright$ $m=1$ is an axis of symmetry, because the input to $Out$ (which is the output of $In$) reverses itself.

$\blacktriangleright$ The formula for the composition looks like

$$(Out \circ In)(m) = Out(In(m)) = 2 \, |-(m-1)^2 + 2| + 1$$

This is includes an absolute value. We can remove the absolute value symbol by using a piecewise defined formula.

We need to know where the inside of the absolute value bars equals $0$.

These are the same places as the corners in the graph.

The inside is negative on $(-\infty , 1 - \sqrt {2}]$ and on $[1 + \sqrt {2}, \infty )$. On these intervals, the formula will be

$$2 \, |-(m-1)^2 + 2| + 1 = 2 \, (-1)(-(m-1)^2 + 2) + 1 = 2 \, ((m-1)^2 - 2) + 1 = 2 \, (m-1)^2 - 3$$

On the interval $(1 - \sqrt {2}, 1 + \sqrt {2})$, the formula is

$$2 \, |-(m-1)^2 + 2| + 1 = 2 \, (-(m-1)^2 + 2) + 1 = -2 \, (m-1)^2 + 5$$

$$Out(In(m)) = \begin {cases} 2 \, (m-1)^2 -3 & \text { on } \, (-\infty , 1 - \sqrt {2}] \\ -2 \, (m-1)^2 + 5 & \text { on } \, (1 - \sqrt {2}, 1 + \sqrt {2}) \\ 2 \, (m-1)^2 -3 & \text { on } \, [1 + \sqrt {2}, \infty ) \end {cases}$$

The middle piece is a quadratic with a negative leading coefficient. Its graph would be a parabola opening down, which is what we see in the graph.

Composition

Let $Out(x) = -2|x-3|+2$ with its natural or implied domain.
Let $In(t) = 2 \sin (t)+2$ with its natural or implied domain.

Graph $w = (Out \circ In)(r) = Out(In(r))$

Graph of $y = Out(x) = -2|x-3|+2$

The natural domain of $Out$ is $\mathbb {R}$.

If we think of moving through the domain from $-\infty$ to $\infty$, then the graph illustrates that $Out$ will increase on $(-\infty , 3]$ and decrease on $[3, \infty )$. The function $Out$ has a maximum value of $\answer {2}$, which occurs at $\answer {3}$.

However, the output from $In$ is not $\mathbb {R}$, so we will not get the full output of $Out$. In addition, the output of $In$ will oscillate, which means the input values coming into $Out$ will osciallate. This will make the output of $Out$ oscillate.

Graph of $z = In(t) = 2 \sin (t)+2$

The input into $In(t)$ is the whole real line. We naturally think of moving left to right, from $-\infty$ to $\infty$. As we do this, the outputs from $In(t)$ oscillate between $0$ and $\answer {4}$. Therefore, the inputs into $Out(x)$ oscillate back and forth inside the interval $\left [ \answer {0}, \answer {4} \right ]$.

The output of $In(x)$ keeps going back and forth across the interval $[0,4]$.

This becomes the input into $Out$.

Therefore, the input into $Out(x)$ keep going back and forth across the interval $[0,4]$.

What part of the graph of $Out$ does this oscillation run across?

The only part of the graph of $Out$ that is used is on the interval $[0, 4]$.

Therefore, that part of the graph of $Out$ just keeps repeating back and forth.

Except, the “$+2$” in $In(t) = 2 \sin (t) + 2$ will become the input for $Out$. This will cause a horizontal shift for $Out$.

The smooth rounding of the sine curve will round out the corners of the absolute value graph and the $2\pi$ period of $\sin (t)$ will force the same period for the composition.

Notice:

We are repeating the piece of $Out$ on the interval $[0,4]$.

$\blacktriangleright$ The minimum value of $Out$ occurs at $0$. When does the oputput of $In$ equal $0$?

$In(t) = 2 \sin (t)+2 = 0$ when $sin(t) = -1$, which is when $t = -\frac {\pi }{2} + 2k\pi$ where $k \in \mathbb {N}$. We can see this as the lower valleys in the graph of the composition.

$\blacktriangleright$ The maximum value of $Out$ occurs at $3$. When does the oputput of $In$ equal $3$?

$In(t) = 2 \sin (t)+2 = 3$ when $sin(t) = \frac {1}{2}$, which is when $t = \frac {\pi }{6} + 2k\pi$ and $t = \frac {5\pi }{6} + 2k\pi$ where $k \in \mathbb {N}$. We can see this as the two peaks in the graph of the composition.

Graphically, we have to keep our eyes on several things at once. We watch the original input into $In$, then we watch the output of $In$ and picture that as the new input into $Out$, then we watch the output coming from $Out$.