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SASpecialist MathematicsSyllabus dot point

When can a function be reversed, and how do composition and inversion interact with domain and range?

Form composite functions and find inverse functions, attending to domain and range and the condition for an inverse to exist.

Building composite functions and tracking their domains, the one-to-one condition for an inverse to exist, finding inverse functions algebraically, and the reflection relationship in y equals x.

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  1. What this dot point is asking
  2. Composite functions
  3. When does an inverse exist?
  4. Finding an inverse algebraically
  5. The graphical relationship

What this dot point is asking

You need to form composite functions with correct domains, determine when an inverse function exists, find inverse functions, and relate a function and its inverse graphically.

Composite functions

The composite (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)) applies gg first and feeds the result into ff. Order matters: in general f∘gβ‰ g∘ff \circ g \ne g \circ f.

For example, if f(x)=xf(x) = \sqrt{x} and g(x)=xβˆ’4g(x) = x - 4, then (f∘g)(x)=xβˆ’4(f \circ g)(x) = \sqrt{x - 4}, defined only for xβ‰₯4x \ge 4, even though gg itself is defined for all real xx.

When does an inverse exist?

An inverse function fβˆ’1f^{-1} reverses ff: if f(a)=bf(a) = b then fβˆ’1(b)=af^{-1}(b) = a. For this reversal to be a function, each output must come from exactly one input.

Restricting the domain can create an inverse: f(x)=x2f(x) = x^2 on xβ‰₯0x \ge 0 is one-to-one and has inverse fβˆ’1(x)=xf^{-1}(x) = \sqrt{x}.

Finding an inverse algebraically

The mechanical recipe: write y=f(x)y = f(x), swap xx and yy, then solve for yy. The domain of fβˆ’1f^{-1} is the range of ff, and the range of fβˆ’1f^{-1} is the domain of ff.

The graphical relationship

The graph of fβˆ’1f^{-1} is the reflection of the graph of ff in the line y=xy = x. Consequently a point (a,b)(a, b) on ff corresponds to (b,a)(b, a) on fβˆ’1f^{-1}, and any intersection of ff with its inverse lies on the line y=xy = x (for increasing functions). Composing a function with its inverse returns the input:

fβˆ’1(f(x))=xandf(fβˆ’1(x))=x,f^{-1}(f(x)) = x \quad \text{and} \quad f(f^{-1}(x)) = x,

each on the appropriate domain.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2018 SACE Stage 22 marksThe graph of g(x) = sin(x) + cos(x) is drawn for -pi <= x <= pi. Explain why g(x) is a function, but does not have an inverse function.
Show worked answer β†’

g(x) is a function because every value of x in the domain -pi <= x <= pi maps to exactly one value of g(x); the graph passes the vertical line test. [1 mark]

However, g(x) does not have an inverse function because it is not one-to-one over this domain. The curve rises then falls (it has a maximum), so a horizontal line such as y = 1 cuts the graph in more than one place, meaning two different x values give the same output. [1 mark]

For an inverse function to exist the original must be one-to-one (it must pass the horizontal line test), so the domain would need to be restricted to an interval on which g is strictly increasing or strictly decreasing.

2018 SACE Stage 21 marksf(x) = sin(x) + cos(x) is the same rule as g(x) but restricted to a domain on which it is one-to-one. Explain why this restricted function f(x) does have an inverse function.
Show worked answer β†’

Restricting the domain to an interval on which f is strictly monotonic (here, strictly increasing, since sin(x) + cos(x) = sqrt(2) sin(x + pi/4) rises across the chosen interval) makes the function one-to-one. [1 mark]

Because each output value is now produced by exactly one input, the graph passes the horizontal line test, so the mapping can be reversed: every y in the range corresponds to a single x. A one-to-one function always has an inverse function, hence f(x) has an inverse.