Curriculum · Course · Week 1 · فارسی
A function assigns exactly one output to each allowed input. The domain lists allowed inputs; the range lists outputs reached on that domain. A vertical line through the graph can meet it at most once.
For f(x) = 3 − 2x on [0, 4], the endpoint outputs are 3 and −5. The slope is −2, so the function decreases throughout the interval and its range is [−5, 3].
For x ≠ 2, (x² − 4)/(x − 2) simplifies to x + 2. Draw the line and leave an open circle at (2, 4). The zero denominator excludes x = 2 even after cancellation.
Graph of a line with a hole at x = 2
Find the domain and range of g(x) = x² on [−2, 1]. The domain is [−2, 1] and the range is [0, 4]. Checking the endpoints alone would miss the minimum at zero.
A formula tells you what operation to perform, a table shows selected input–output pairs, and a graph shows how the outputs change across the domain. Two inputs can have the same output. One input with two different outputs fails the function test. For instance, x² gives the same output for −2 and 2, while the circle x² + y² = 4 gives y = 2 and y = −2 at x = 0.
| Expression | Domain check | Example |
|---|---|---|
| Polynomial | Every real input is allowed | x² − 3 |
| Quotient | Exclude zeros of the original denominator | 1/(x − 3) excludes 3 |
| Square root | The radicand must be nonnegative | √(x − 2) requires x ≥ 2 |
For f(x) = 2x + 1 on [−1, 2], evaluate f(−1) = −1, f(0) = 1, f(1) = 3 and f(2) = 5. Plot these points and join them with a straight segment. Both endpoints are included. The graph crosses the y-axis at (0, 1) and the x-axis at (−0.5, 0). Its range is [−1, 5]. The slope tells you that an increase of 1 in x produces an increase of 2 in y.
Answers. 1. −1, x = 4, and [−1, 3]. 2. x ≤ 5 and g(1) = 2 if g(x) is that square-root function. 3. Yes. The added pair assigns a second output to input 1.
The blue line models a sensor whose output rises from 1 V to 5 V as its input increases from 0 to 2. Read the change in output, 4 V, and divide by the change in input, 2. The slope is 2 V per input unit. The starting value is 1 V. Restricting the input to [0, 2] restricts the output to [1, 5].
Read the graph. Start at the left endpoint and trace the rise and run. The horizontal guide measures a run of 2 and the vertical guide measures a rise of 4 V. At the midpoint the output is 3 V. If the same sensor started at 2 V and kept the same slope, the whole line would shift up by 1 V. Its output at input 2 would then be 6 V.
After this lesson you should be able to state a domain with reasons, distinguish range from codomain, interpret a slope with units, and justify a graph from an equation rather than its appearance. These are prerequisites for limits, not optional vocabulary.
Writing f: D → C specifies a domain D and a codomain C. The range is the subset of C that the function actually reaches. For f(x)=x² with real inputs and real codomain, the range is [0,∞). Changing the allowed inputs to [−1,2] changes the range to [0,4]. A formula alone does not tell you a physical operating range.
For a sensor, x might be temperature and f(x) a voltage. A mathematically valid input can still lie outside the calibrated temperature interval. Record the mathematical restriction and the physical restriction separately, then take their intersection. Use interval brackets for included endpoints and parentheses for excluded endpoints.
Find the real domain of h(x)=√(x+2)/(x−1). The root requires x+2≥0, hence x≥−2. The denominator requires x≠1. Both must hold, giving [−2,1) ∪ (1,∞). At −2 the denominator is nonzero, so that endpoint is included. At 1 the numerator is defined but the quotient is not. Testing one condition and forgetting the other gives an incorrect answer.
A temperature sensor reads 0.5 V at 0 °C and 2.5 V at 100 °C. Assume a linear response only over this interval. The slope is m=(2.5−0.5)/(100−0)=0.020 V/°C. Therefore v(T)=0.020T+0.5 for 0≤T≤100. At 35 °C the predicted voltage is 1.20 V. The range is [0.5,2.5] V. A reading of 2.9 V lies outside this model’s calibrated range. Extrapolating to 120 °C is an additional assumption, not a consequence of the two calibration points.
Between two distinct inputs a and b, the secant slope is (f(b)−f(a))/(b−a). For f(x)=x² from 1 to 3 it is (9−1)/(3−1)=4. From 1 to 2 it is 3. A straight line has the same slope between every pair of distinct inputs, while a quadratic generally does not. This difference will motivate the derivative later. No derivative rule is needed here.
Show the restriction, calculation and interpretation for each answer. Problems 1–6 are core work; 7–8 are optional extensions within spare study time.
Common mistakes: calling the codomain the range, restoring a cancelled forbidden input, dropping units from a slope, and treating an extrapolation as a measured fact.