Mathematics · 10 / 23
Functions: an input and its image
High school foundations · Worked example and 4 exercises with solutions, at your pace.
The principle, simply
A function gives each input in its domain one output. f(x) is the output at x; a preimage of y is an input producing y. Different inputs may share an output. A graph point has coordinates (x, f(x)). An affine function has a constant rate of change.
- 01Input x
- 02Rule f
- 03Output f(x)
- 04Point on the graph
WORKED EXAMPLE
From principle to reasoning
For f(x) = 2x + 1, the output at 3 is 7. To find an input producing 9, solve 2x + 1 = 9, giving x = 4.
Solve a problem in this subject
- Define the objects
State unknowns, domain and data. Real square roots need nonnegative arguments; denominators must be nonzero.
- Choose a transformation
Name the rule: expansion, equivalent equations, factoring, differentiation. Check its conditions. Squaring an equation requires testing candidates in the original.
- Show the reasoning
Each line should follow from the previous one. Keep exact values where possible and distinguish equality, approximation and implication.
- Test and conclude
Substitute into the original expression. Check graph points and domain boundaries. State a solution set or the required unit.
YOUR TURN
01 · Try before checking
For f(x) = x², find the output at −2 and the inputs producing 4.
One hint
Evaluate to find an output; solve an equation to find inputs.
Reveal and understand the solution
f(−2) = 4. Both −2 and 2 produce 4. The horizontal line y = 4 meets the parabola twice.
Compare method, conditions and answer; identify the exact part to revisit.
IMPROVE THROUGH PRACTICE
Three additional challenges with solutions
Start with the first challenge. Write your method before opening a hint; use the solution to find the first step that needs work.
For f(x) = 2x² − 3, find f(−2).
A hint to get started
Put −2 in parentheses before squaring.
Step-by-step solution
- f(−2) = 2 × (−2)² − 3.
- The square is 4, giving 8 − 3 = 5.
- An image is an output value, not a preimage.
Which step did you miss? Retry without help, then explain why the method works.
Find the preimages of 5 under f(x) = x² + 1.
A hint to get started
Solve f(x) = 5.
Step-by-step solution
- x² + 1 = 5, so x² = 4.
- x = −2 or x = 2.
- One output can have multiple preimages.
Which step did you miss? Retry without help, then explain why the method works.
Find the real domain of g(x) = √(x − 3)/(x − 5).
A hint to get started
Satisfy two conditions at once.
Step-by-step solution
- The root requires x ≥ 3.
- The denominator requires x ≠ 5.
- Domain: [3,5) ∪ (5,+∞).
Which step did you miss? Retry without help, then explain why the method works.
The trap to avoid
f(3) is not three times f; the parentheses specify the input.
Without notes, can you…
- Explain the principle in your own words?
- Rebuild the diagram and its connections?
- Solve the exercise and justify the method?
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