91Ó°ÊÓ

Composite functions combine two or more functions to create a new function. This is usually expressed as

• \( (f \circ g)(x) = f(g(x)) \)

For this, the first function \( g(x) \) is applied to \( x \), and then \( f(x) \) is applied to \( g(x) \). Visualize it as a sequence of steps, where the output of one function becomes the input of another. By handling functions this way, you can transform and manipulate data more effectively, as seen in the problem above.

Let's break it down further. Assume function \( r(x) = 6x + 2 \) (as given) is applied first (to \( x \)), then the function \( v(x) = -7x - 5 \) comes next. For part (a) of the exercise, applying \( v(x) \) to the result of \( r(x) \) leads us to

• \( (v \circ r)(x) = v(r(x)) = -42x - 19 \)

Substitution is a method where one function is plugged into another, replacing the variable. With simple algebra, this becomes quite useful to solve composite functions.

• Start by determining which function is the innermost or first function being solved.
• Replace every instance of the variable in the outer function with the result of the inner function.

For instance, to find \( (v \circ r)(x) \), replace \( x \) in \( v(x) \) with \( r(x) = 6x + 2 \). Thus, you calculate:

• \( v(r(x)) = v(6x + 2) = -42x - 19 \)
  

Similarly, for \( (r \circ v)(x) \), replace \( x \) in \( r(x) \) with \( v(x) = -7x-5 \). This gives you:

• \( r(v(x)) = r(-7x-5) = -42x - 28 \)

The method also extends seamlessly when specific values replace the variable, as seen in finding \((r \circ v)(2)\). You directly substitute 2 for \( x \) in the resulting composite function.

Polynomial functions are expressions involving variables raised to a power, multiplied by coefficients, and added or subtracted together, like \( r(x) = 6x + 2 \) or \( v(x) = -7x - 5 \).

• These functions are composed of terms where each part has a coefficient and a variable with an exponent (which is often positive integers like 1, 2, 3).
• For example, \( 6x \) is a term with a coefficient of 6 and exponent of 1 for \( x \).

Both functions in the given exercise are linear polynomial functions. The highest power of \( x \) is one, indicating a straight-line relationship when graphed. The beauty of dealing with polynomials is their simplicity and predictability in behavior. Once you know the basics of arithmetic, substitutions, and distributions, solving composite polynomial functions becomes straightforward.

Understanding polynomial structures helps in anticipating the kind of results you're working with during substitutions or compositions. As a result, no surprising complexities arise, making polynomials particularly friendly to work with in these mathematical problems.