A function is more than an equation with an f in front. It is a rule together with a set of allowed inputs, and that input control determines whether compositions and inverses make mathematical sense.
This guide develops four connected ideas: domain, range, composite functions and inverse functions. The emphasis is on deciding what is allowed before manipulating symbols, then checking the result against the original function.
1. Function notation is an instruction
If f(x) = 2x − 3, then f(5) means apply the rule to the input 5: f(5) = 10 − 3 = 7. Likewise, f(a + 1) = 2(a + 1) − 3 = 2a − 1. The brackets identify the complete input.
A common error is to treat f(x) as multiplication of f and x. It is not. The notation names the output produced by function f when the input is x.
2. Domain: which inputs are permitted?
The domain is the set of allowed inputs. For f(x) = 1/(x − 4), x = 4 is excluded because it makes the denominator zero. For g(x) = √(x + 2) over the real numbers, x must satisfy x + 2 ≥ 0, so x ≥ −2.
Domain restrictions are not optional notes added after solving. They are part of the function itself and can affect compositions and inverses.
3. Range: which outputs can occur?
The range is the set of possible outputs. If f(x) = x² for all real x, then f(x) can never be negative, so the range is y ≥ 0. If the domain is restricted to x ≥ 0, the range remains y ≥ 0, but that restriction becomes important when considering an inverse.
For h(x) = (x − 3)² + 5, the squared quantity is at least zero, so h(x) ≥ 5. The graph interpretation agrees: the turning point is (3,5), and the parabola opens upward.
4. Composite functions: one output becomes another input
Let f(x) = 2x + 1 and g(x) = x². Then fg(x), when used to mean f(g(x)), is f(x²) = 2x² + 1. In the opposite order, gf(x) = g(2x + 1) = (2x + 1)².
These are generally different. Composition is ordered: the function closest to x acts first.
5. Composition also has domain conditions
Suppose f(x) = 1/(x − 2) and g(x) = x + 3. Then f(g(x)) = 1/(x + 1), so x ≠ −1. The restriction arises because g(x) must not produce the forbidden input 2 for f.
Now reverse the order: g(f(x)) = 1/(x − 2) + 3, with x ≠ 2. The two compositions have different formulas and different restrictions.
6. Inverse functions undo a one-to-one function
If f(x) = 3x − 5, set y = 3x − 5 and solve for x: x = (y + 5)/3. Swap the variable labels to write f⁻¹(x) = (x + 5)/3.
Check by composition: f(f⁻¹(x)) = 3[(x + 5)/3] − 5 = x. Likewise, f⁻¹(f(x)) = [(3x − 5) + 5]/3 = x.
7. Why x² needs a restricted domain before inversion
The function f(x) = x² on all real numbers is not one-to-one: f(2) = f(−2) = 4. An inverse function cannot decide whether the input 4 should return 2 or −2.
If we restrict f to x ≥ 0, it becomes one-to-one and f⁻¹(x) = √x for x ≥ 0. If instead the original domain is x ≤ 0, the inverse is f⁻¹(x) = −√x. The restriction determines the branch.
8. Domain and range swap under inversion
For an invertible function, the domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹. Graphically, the graphs of y = f(x) and y = f⁻¹(x) are reflections of each other in the line y = x.
This provides a powerful check. If a claimed inverse has a domain inconsistent with the original range, revisit the restriction before continuing.
9. Worked example with a rational function
Let f(x) = (2x + 3)/(x − 1), x ≠ 1. Find the inverse.
Set y = (2x + 3)/(x − 1). Then yx − y = 2x + 3, so x(y − 2) = y + 3. Therefore x = (y + 3)/(y − 2), where y ≠ 2.
Hence f⁻¹(x) = (x + 3)/(x − 2), x ≠ 2. The original range excludes 2, which becomes the inverse domain restriction.
10. Common mistakes
- Confusing f⁻¹(x) with 1/f(x).
- Reversing the order of a composite function.
- Finding an inverse algebraically without checking one-to-one behaviour.
- Dropping a denominator, square-root or logarithm restriction.
- Forgetting that the original range becomes the inverse domain.
11. Independent practice
- Given f(x)=4x−7, find f(3) and f(a−2).
- State the real domain of g(x)=1/(x+5).
- State the real domain of h(x)=√(2x−6).
- If f(x)=x+2 and g(x)=3x, find f(g(x)) and g(f(x)).
- Find the inverse of f(x)=5x+4.
- Explain why f(x)=x² on all real x has no inverse function.
12. Answers
- f(3)=5; f(a−2)=4a−15.
- x≠−5.
- x≥3.
- f(g(x))=3x+2; g(f(x))=3x+6.
- f⁻¹(x)=(x−4)/5.
- It is not one-to-one because distinct inputs such as 2 and −2 produce the same output.
13. What mastery looks like
A student has strong function control when they can state restrictions before calculation, distinguish domain from range, compose functions in the correct order, decide whether an inverse exists and verify an inverse by composition or graph structure.
Continue through the Additional Mathematics Hub and the A-Math Algebra Guide.