Multiplying Functions

When you have two functions, say \(f(x)\) and \(g(x)\), you can combine them in several ways: adding, subtracting, multiplying, or dividing. Multiplying functions means building a brand new function whose output at any input \(x\) is simply \(f(x)\) times \(g(x)\). This new function is usually called the "product function," and it behaves just like any other function once it is simplified.

The product of two functions is written as \((f \cdot g)(x)\) or sometimes \((fg)(x)\). Both mean the same thing:

\((f \cdot g)(x) = f(x) \times g(x)\)

To evaluate this, you plug the same value of \(x\) into both \(f\) and \(g\), then multiply the two results together. If you want the product at a specific number, say \(x = 3\), you compute \(f(3) \times g(3)\).

Before multiplying, it helps to know each function's own domain, the set of \(x\)-values each one is allowed to use. If you have not reviewed how to write domain and range, it is worth checking that first, since the same ideas apply here.

When you multiply \(f(x)\) and \(g(x)\), the domain of the new product function is the intersection of the two original domains, meaning only the \(x\)-values that work for both functions at once. If either \(f\) or \(g\) is undefined at a certain \(x\), then \((f \cdot g)(x)\) is undefined there too, even if the simplified product expression looks fine.

Let \(f(x) = x + 3\) and \(g(x) = x - 2\). Find \((f \cdot g)(x)\).

\((f \cdot g)(x) = f(x) \times g(x) = (x + 3)(x - 2)\)

Distribute (or use the FOIL pattern) to expand:

\((x + 3)(x - 2) = x^2 - 2x + 3x - 6 = x^2 + x - 6\)

So \((f \cdot g)(x) = x^2 + x - 6\). Since both \(f\) and \(g\) are polynomials with domain "all real numbers," the product also has domain all real numbers. The graph below shows a similar product of two simple linear pieces, \(h(x) = x(x - 2)\), where the parabola's zeros line up exactly with the zeros of each original factor.

Notice that the product crosses the \(x\)-axis exactly where either original function equals zero. That happens because if \(f(x) = 0\) or \(g(x) = 0\) at some point, then \(f(x) \times g(x) = 0\) there as well.

Using the same \(f(x) = x + 3\) and \(g(x) = x - 2\) from before, find \((f \cdot g)(4)\).

Method 1, evaluate first: \(f(4) = 7\) and \(g(4) = 2\), so \((f \cdot g)(4) = 7 \times 2 = 14\).

Method 2, use the simplified product: \((f \cdot g)(x) = x^2 + x - 6\), so \((f \cdot g)(4) = 16 + 4 - 6 = 14\).

Both methods agree, which is a good way to check your work.

The same product rule applies when \(f(x)\) and \(g(x)\) are rational functions (fractions with polynomials on top and bottom). Multiply the numerators together, multiply the denominators together, then simplify by cancelling common factors when possible.

\(\left(\frac{x+1}{x-3}\right) \times \left(\frac{x-3}{x+2}\right) = \frac{(x+1)(x-3)}{(x-3)(x+2)} = \frac{x+1}{x+2}, \quad x \ne 3\)

Notice the restriction \(x \ne 3\) still applies even after the \((x - 3)\) factor cancels, because \(x = 3\) made the original denominator zero. This is the trickiest part of multiplying rational functions: always track the domain restrictions from before you simplify, not just after.

Multiplying is one of four basic ways to combine two functions. The table below summarizes all four, so you can see how the product fits alongside the others.

Multiplication and division of functions are closely related, and division brings its own extra domain rule since you can never divide by zero. Combining functions this way also connects to ideas used in the difference quotient: applications of functions, where you subtract and divide function values to study rates of change.

To multiply two functions: substitute the same \(x\) into each function, multiply the results, simplify the expression, and restrict the domain to values allowed by both original functions. Whether you are working with polynomials or rational expressions, this same process holds true every time.

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