Composite Functions (HSC SSCE Mathematics Advanced): Revision Notes

Worked Example 1: Forming a composite function

Question: Given $f(x) = 3x - 2$ and $g(x) = x^2 + 1$, determine $f(g(x))$.

Solution:

To find $f(g(x))$, we substitute $g(x)$ into $f(x)$.

$f(x) = 3x - 2$

$f(g(x)) = 3(x^2 + 1) - 2$

$f(g(x)) = 3x^2 + 3 - 2$

$f(g(x)) = 3x^2 + 1$

Steps:

• Write the function $f(x)$
• Substitute $g(x) = x^2 + 1$ in place of $x$
• Use the distributive property to expand
• Simplify by evaluating the arithmetic

Worked Example 2: Evaluating at specific values

Question: If $f(x) = -2x - 3$ and $g(x) = -2x - 6$, evaluate:

• a) $f(7)$ and $g(7)$
• b) $f(g(7))$
• c) $g(f(7))$
• d) Determine if $g(f(x)) = f(g(x))$ for all $x$

Solution:

Part a) Substitute $x = 7$ into both functions.

For $f(7)$:

For $g(7)$:

Therefore, $f(7) = -17$ and $g(7) = -20$.

Part b) Use the result for $g(7)$ from part (a) and substitute it into $f(x)$.

From part (a), $g(7) = -20$.

Part c) Use the result for $f(7)$ from part (a) and substitute it into $g(x)$.

From part (a), $f(7) = -17$.

Part d) Compare the results from parts (b) and (c).

From part (b), $f(g(7)) = 37$.

From part (c), $g(f(7)) = 28$.

Since $g(f(7)) = 28 \neq 37 = f(g(7))$, we can conclude:

$g(f(7)) \neq f(g(7)) \rightarrow 28 \neq 37$

Since $g(f(7)) \neq f(g(7))$, it is not true that $g(f(x)) = f(g(x))$ for all $x$.

This demonstrates that the order of composition matters. Applying the functions in different orders produces different results.

Worked Example 3: Domain and range with restrictions

Question: Given $f(x) = 3x - 1$ and $g(x) = \sqrt{x}$, determine:

• a) The domain of $g(f(x))$
• b) The range of $g(f(x))$

Solution:

Part a) Find the domain of $g(f(x))$.

First, form the composite function $g(f(x))$.

$g(x) = \sqrt{x}$

$g(f(x)) = \sqrt{3x - 1}$

For the domain, $g(x) = \sqrt{x}$ requires $x \geq 0$. Therefore, we need $f(x) \geq 0$.

Set up the inequality:

The domain is [$\left[\frac{1}{3}, \infty\right)$].

Part b) Find the range of $g(f(x))$.

For the range, $g(f(x)) = \sqrt{3x - 1}$ produces non-negative values because the square root function outputs $y \geq 0$.

Since $3x - 1 \geq 0$ for $x \geq \frac{1}{3}$, the range is [$[0, \infty)$].

The square root function takes non-negative inputs and produces non-negative outputs. As $x$ increases from $\frac{1}{3}$ towards infinity, $3x - 1$ increases from 0 towards infinity, and therefore $\sqrt{3x - 1}$ also increases from 0 towards infinity.

Worked Example 4: Using graphs to find range

Question: The graphs of functions $f(x)$ and $g(x)$ are shown below. What is the range of the composite function $f(g(x))$?

Solution:

We need to observe the range of $g(x)$ from its graph, intersect it with the domain of $f(x)$, and determine the output values of $f(x)$ over this interval using its graph.

Step 1: From the graph of $g(x)$, the function has a minimum at $(0, 0)$ and approaches $y = 1$ as $x$ increases or decreases without bound. This gives the range [$[0, 1)$].

Step 2: The graph of $f(x)$ shows a vertical asymptote at $x = 0$, so its domain is $(0, \infty)$. When we intersect the range of $g(x)$, which is $[0, 1)$, with the domain of $f(x)$, we get $(0, 1)$.

Step 3: On the graph of $f(x)$, for inputs $x \in (0, 1)$, the output ranges from $-\infty$ as $x$ approaches 0, to 0 as $x$ approaches 1, excluding 0.

The range of $f(g(x))$ is $(-\infty, 0)$.