In this module, we will investigate the behaviour and discuss the properties of functions formed by multiplying or dividing functions.

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

a. Determine the equations of the functions \((f \cdot g)(x)\) and \(\left(\dfrac{f}{g}\right)(x)\) and identify the domain of each.

Solution

\(f \cdot g\) is a polynomial function, as are \(f\) and \(g\), so the domain is \(\left\{x \mid x \in \mathbb{R} \right\}\).

\(\dfrac{f}{g}\) is a rational function which is undefined when \(x^2 - 4 = 0\).

Thus, the domain of \(\dfrac{f}{g}\) is \(\left\{ x \mid x \neq \pm 2, x \in \mathbb{R} \right\}\).

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

b. Identify the coordinates of the point on each function \(f\), \(g\), \(f \cdot g\), and \(\dfrac{f}{g}\) when \(x = -1\). What do you notice?

Solution

From these calculations, we know that \((-1,-1)\) is a point on \(f\) and \((-1,-3)\) is a point on \(g\).

The corresponding point on \(f \cdot g\) is \((-1,3)\). The \(y\)-coordinate of this point, \(3\), can be obtained by multiplying the corresponding \(y\)-coordinates of \(f\) and \(g\); i.e., \(-1\) and \(-3\).

In general, for any given value of \(x\) in the domain of \(f\) and \(g\), the corresponding \(y\)-coordinate on the product function, \((f \cdot g)(x)\), can be obtained by multiplying the corresponding \(y\)-coordinates of \(f(x)\) and \(g(x)\).

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

b. Identify the coordinates of the point on each function \(f\), \(g\), \(f \cdot g\), and \(\dfrac{f}{g}\) when \(x = -1\). What do you notice?

Solution

Similarly, the corresponding point on \(\left(\dfrac{f}{g}\right)(x)\) is \(\left(-1,\dfrac{1}{3}\right)\).

The \(y\)-coordinate of this point, \(\dfrac{1}{3}\), can be obtained by dividing the \(y\)-coordinate of \(f\) by the corresponding \(y\)-coordinate of \(g\). That is, \(\dfrac{-1}{-3}\), which is \(\dfrac{1}{3}\).

In general, for any given value of \(x\) in the domain of \(f\) and \(g\), the corresponding \(y\)-coordinate on the quotient function , \(\left(\dfrac{f}{g}\right)(x)\) , can be obtained by dividing the \(y\)-coordinate of \(f(x)\) by the corresponding \(y\)-coordinate of \(g(x)\), where \(g(x) \neq 0\).

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

First, graph \(f(x) = x\) and \(g(x) = x^2 - 4\) on the same axes. Note that the domain of both functions is the set of all real numbers. By multiplying the corresponding \(y\)-coordinates of points on \(f\) and \(g\), we can determine points on the graph of \(f \cdot g\).

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

The product function, \((f \cdot g)(x) = x^3 - 4x\), is a cubic polynomial function.

The leading coefficient is positive, so \(y \rightarrow -\infty\) as \(x \rightarrow -\infty\), and \(y \rightarrow \infty\) as \(x \rightarrow \infty\).

From the factored form of the equation, we can verify that the zeros are \(-2\), \(0\), and \(2\).

All the zeros are of order (multiplicity) \(1\), so the graph will pass directly through the \(x\)-axis at each zero.

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

Some general observations:

• The zeros of \(f \cdot g\) consist of all the zeros of the two functions \(f\) and \(g\).
• When the graphs of \(f\) and \(g\) are both above the \(x\)-axis, or both below the \(x\)-axis, then the graph of \(f \cdot g\) is above the \(x\)-axis.
• When the graphs of \(f\) and \(g\) lie on opposite sides of the \(x\)-axis, then the graph of \(f \cdot g\) is below the \(x\)-axis.
• When \(f(x) = 1\), \((f \cdot g)(x)\) intersects \(g(x)\) since

• When \(g(x) = 1\), \((f \cdot g)(x)\) intersects \(f(x)\) since

• \(f\) is an odd function, and \(g\) is an even function. The product \(f \cdot g\) is an odd function.

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

To determine the graph of \(\left(\dfrac{f}{g}\right)(x) = \dfrac{x}{x^2 - 4}\), we start with the graph \(f(x) = x\) and \(g(x) = x^2 - 4\).

The domain of \(\dfrac{f}{g}\) is \(\left\{x \mid x \neq \pm 2, x \in \mathbb{R}\right\}\) since \(g(x) \neq 0\).

By dividing the \(y\)-coordinates of \(f\) by the corresponding \(y\)-coordinates of \(g\), we can determine points on the graph of \(\dfrac{f}{g}\).

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

Additional point(s) on \(\dfrac{f}{g}\) can be identified by considering the points of intersection of \(f\) and \(g\). At these points, \(f(x) = g(x)\), so \(\left(\dfrac{f}{g}\right)(x) = 1\).

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

At \(x = \pm 2\), \(g(x) = 0\), but \(f(x) \neq 0\), so \(\left(\dfrac{f}{g}\right)(x)\) has vertical asymptotes \(x = 2\) and \(x = -2\).

Since the degree of \(g(x)\) is greater than the degree of \(f(x)\), then \(\left(\dfrac{f}{g}\right)(x) \rightarrow 0\) as \(x \rightarrow \pm \infty\).

Thus, \(y = 0\) is a horizontal asymptote.

Consider the two functions \(f(x) = x\) and \(g(x) = x^2 - 4\).

c. Using the graphs of \(f(x) = x\) and \(g(x) = x^2 - 4\), sketch the graphs of \(y = (f \cdot g)(x)\) and \(y = \left(\dfrac{f}{g}\right)(x)\).

Solution

When \(x \lt -2\), \(f \lt 0\) (below the \(x\)-axis) and \(g \gt 0\) (above the \(x\)-axis), so \(\dfrac{f}{g} \lt 0\) and its graph will approach the \(x\)-axis from below as \(x \rightarrow -\infty\) and approach \(-\infty\) as \(x \rightarrow -2^{-}\). When \(-2 \lt x \lt 0\), both \(f \lt 0\) and \(g \lt 0\), so \(\dfrac{f}{g} \gt 0\) and the graph approaches \(+ \infty\) as \(x \rightarrow -2^{+}\).

In a similar way, we can determine the behaviour of the graph to the right of the \(y\)-axis. Notice that \(\left(\dfrac{f}{g}\right)(x)\) has odd symmetry.