Composite Functions

Overview

A composite function is formed when one function is applied to the result of another function. In other words, the output of one function becomes the input for another function. The notation for a composite function is written as \( (f \circ g)(x) \), which is read as "f composed with g of x". This means that we first apply \( g(x) \), and then apply \( f \) to the result of \( g(x) \).

Notation and Definition

The composite function \( (f \circ g)(x) \) is defined as:

\( (f \circ g)(x) = f(g(x)) \)

This means that \( g(x) \) is applied first, and then the result of \( g(x) \) is plugged into \( f(x) \). In terms of composition, the order of functions matters:

\( (f \circ g)(x) \neq (g \circ f)(x) \)

How to Compute Composite Functions

To compute the composite function \( (f \circ g)(x) \), follow these steps:

  1. Identify the inner function \( g(x) \).
  2. Apply the inner function \( g(x) \) to the given input \( x \).
  3. Substitute the result of \( g(x) \) into the outer function \( f(x) \).
  4. Simplify the expression if necessary.

Example

Let \( f(x) = 2x + 1 \) and \( g(x) = x^2 - 3 \). The composite function \( (f \circ g)(x) \) is:

\( (f ∘ g)(x) = f(g(x)) = f(x^2 - 3) = 2(x^2 - 3) + 1 = 2x^2 - 6 + 1 = 2x^2 - 5 \)

So, the composite function is \( (f \circ g)(x) = 2x^2 - 5 \).

Inverse of Composite Functions

It’s important to remember that the inverse of a composite function is not simply the composition of the inverses of the individual functions. That is:

\( (f ∘ g)^{-1} ≠ f^{-1} ∘ g^{-1} \)

To find the inverse of a composite function, we must carefully work through the algebraic steps based on the given functions.

Practice Questions

  1. Let \( f(x) = 3x + 4 \) and \( g(x) = x - 5 \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = x - 5 \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(x - 5) = 3(x - 5) + 4 = 3x - 15 + 4 = 3x - 11 \)

    So, \( (f \circ g)(x) = 3x - 11 \).

  2. Let \( f(x) = x^2 \) and \( g(x) = 2x + 1 \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = 2x + 1 \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)^2 = 4x^2 + 4x + 1 \)

    So, \( (f \circ g)(x) = 4x^2 + 4x + 1 \).

  3. Let \( f(x) = \sqrt{x} \) and \( g(x) = x^2 + 3 \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = x^2 + 3 \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(x^2 + 3) = \sqrt{x^2 + 3} \)

    So, \( (f \circ g)(x) = \sqrt{x^2 + 3} \).

  4. Let \( f(x) = 2x + 5 \) and \( g(x) = x^2 - 4x \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = x^2 - 4x \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(x^2 - 4x) = 2(x^2 - 4x) + 5 = 2x^2 - 8x + 5 \)

    So, \( (f \circ g)(x) = 2x^2 - 8x + 5 \).

  5. Let \( f(x) = x + 6 \) and \( g(x) = 3x - 1 \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = 3x - 1 \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(3x - 1) = (3x - 1) + 6 = 3x + 5 \)

    So, \( (f \circ g)(x) = 3x + 5 \).

  6. Let \( f(x) = 2x - 7 \) and \( g(x) = x^3 \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = x^3 \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(x^3) = 2x^3 - 7 \)

    So, \( (f \circ g)(x) = 2x^3 - 7 \).

  7. Let \( f(x) = \cos(x) \) and \( g(x) = x + \frac{\pi}{2} \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = x + \frac{\pi}{2} \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = \cos(x + \frac{\pi}{2}) = -\sin(x) \)

    So, \( (f \circ g)(x) = -\sin(x) \).

  8. Let \( f(x) = \frac{1}{x} \) and \( g(x) = x - 4 \). Find \( (f \circ g)(x) \).
    Solution

    First, apply \( g(x) = x - 4 \). Then substitute \( g(x) \) into \( f(x) \):

    \( (f ∘ g)(x) = f(g(x)) = f(x - 4) = \frac{1}{x - 4} \)

    So, \( (f \circ g)(x) = \frac{1}{x - 4} \).

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