Inverse and composite functions: VCE Math Methods Unit 2 Year 11

What this dot point is asking

VCAA wants you to construct composite functions, determine when an inverse function exists, find the inverse algebraically, and recognise the graphical relationship between a function and its inverse.

Composite functions

A composite function applies one function inside another:

Read "f of g of x". Apply $g$ first, then $f$.

The order matters: $f \circ g$ is in general not the same as $g \circ f$.

Domain considerations: the composite $f(g(x))$ requires $x$ to be in the domain of $g$, and $g(x)$ to be in the domain of $f$.

Worked example

$f(x) = \sqrt{x}$, $g(x) = x - 4$.

$(f \circ g)(x) = f(x - 4) = \sqrt{x - 4}$, defined for $x \geq 4$.

$(g \circ f)(x) = g(\sqrt{x}) = \sqrt{x} - 4$, defined for $x \geq 0$.

Different functions with different domains.

Inverse functions

The inverse of $f$ is the function $f^{-1}$ such that:

on appropriate domains.

When does $f^{-1}$ exist?

$f^{-1}$ exists if and only if $f$ is one-to-one: no two inputs map to the same output. Equivalently, $f$ passes the horizontal line test.

Functions that are not one-to-one (parabolas, $\sin$ on $\mathbb{R}$) require a domain restriction to be invertible.

Finding $f^{-1}$ algebraically

1. Write $y = f(x)$.
2. Swap $x$ and $y$.
3. Solve for $y$.
4. Write $f^{-1}(x) =$ (the solved expression).

Domain and range swap

Under inversion:

• Domain of $f^{-1}$ = range of $f$.
• Range of $f^{-1}$ = domain of $f$.

Graphical relationship

The graph of $f^{-1}$ is the reflection of the graph of $f$ in the line $y = x$.

If $(a, b)$ is on $f$, then $(b, a)$ is on $f^{-1}$.

Intersections of $f$ and $f^{-1}$ lie on $y = x$. To find them, solve $f(x) = x$.

Worked examples

Linear

$f(x) = 3x + 2$. Swap: $x = 3y + 2$. Solve: $y = (x - 2)/3$. So $f^{-1}(x) = (x - 2)/3$.

Quadratic with restriction

$f: [0, \infty) \to [0, \infty)$, $f(x) = x^2$. Swap: $x = y^2$. Solve (positive root, since range of $f^{-1}$ is $[0, \infty)$): $y = \sqrt{x}$. So $f^{-1}(x) = \sqrt{x}$ on $[0, \infty)$.

Exponential

$f(x) = 2^x$. The inverse is $f^{-1}(x) = \log_2(x)$ on $(0, \infty)$.

Verifying the inverse

To check $g(x) = f^{-1}(x)$, verify both $f(g(x)) = x$ and $g(f(x)) = x$ on the appropriate domains.

Examples in context

Example 1. Chained discounts. An online store applies a $10\%$ discount $g(p) = 0.9p$, then a $\$5$ voucher $f(p) = p - 5$. The final price is $(f \circ g)(p) = 0.9p - 5$. On a $\$50$ item that is $0.9(50) - 5 = \$40$. To recover the original price from a final price, invert: $y = 0.9p - 5$ gives $p = \frac{y + 5}{0.9}$, so $f^{-1}\text{-style}$ reversal of $\$40$ returns $\frac{45}{0.9} = \$50$.

Example 2. Temperature scale inverse. A sensor reports $F(c) = 1.8c + 32$ (Celsius to Fahrenheit). Its inverse converts back: swap $f = 1.8c + 32$ to get $c = \frac{f - 32}{1.8}$, so $F^{-1}(f) = \frac{f - 32}{1.8}$. A reading of $212^\circ$F gives $\frac{180}{1.8} = 100^\circ$C, confirming $F(100) = 212$.

Try this

Q1. For $f(x) = x + 4$ and $g(x) = 3x$, find $(f \circ g)(x)$ and $(g \circ f)(x)$. [2+2 marks]

• Cue. $(f \circ g)(x) = 3x + 4$; $(g \circ f)(x) = 3(x + 4) = 3x + 12$.

Q2. Find the inverse of $f(x) = 5x - 1$ and verify $f(f^{-1}(2)) = 2$. [3 marks]

• Cue. $f^{-1}(x) = \frac{x + 1}{5}$; $f^{-1}(2) = 0.6$, $f(0.6) = 3 - 1 = 2$.

Q3. State why $f(x) = x^2$ on $\mathbb{R}$ has no inverse, and give a domain restriction that fixes this. [2 marks]

• Cue. Not one-to-one (fails horizontal line test); restrict to $x \ge 0$, giving $f^{-1}(x) = \sqrt{x}$.