The formal definition of a function, function notation and evaluation, domain and range in interval notation, graphs and the vertical line test, zeros and intercepts, building new functions by arithmetic and by composition, and even and odd symmetry — the vocabulary every later section of calculus is stated in.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 1 — Functions and Graphs
Review of Functions
Objectives
Five outcomes. None of them is new; all of them are about to be used on every page for the rest of the course.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 8-29 — the section these objectives are drawn from
Warm-up
You have been using functions since your first algebra class. This section makes the definition precise, because calculus leans on the precise version.
Discussion prompt
The area of a square is determined by its side length. Which is the input and which is the output, and why can we call area a function of side length but not the other way round for a rectangle?
Hint: Ask whether one input could honestly produce two different outputs.
Answer:
\[ A(s) = s^2, \quad s > 0 \]
Side length is the input, area the output. Every side length gives exactly one area, so area is a function of side length.
For a rectangle the same is not true: an area of 12 could come from 3 by 4, or 2 by 6, or 1 by 12. One input, many outputs — so the rectangle's side length is not a function of its area. That asymmetry is exactly what the definition on the next slide is protecting.
Concept
A function consists of a set of inputs, a set of outputs, and a rule assigning each input to exactly one output. The set of inputs is the domain; the set of outputs is the range. Everything else in this section is machinery for working with that one sentence.
function — A rule that assigns to each element of a set of inputs exactly one element of a set of outputs. The set of inputs is the domain, and the set of outputs is the range.
\[ f : D \to R, \qquad x \mapsto f(x) \]
Note carefully what is not forbidden. Two different inputs may share an output — squaring sends both 3 and negative 3 to 9, and that is a perfectly good function. What is forbidden is one input with two outputs.
Figure (svg): A function drawn as an input-output machine, with two different inputs allowed to share one output but no input allowed two outputs
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 8-9
Section
Section 1
Concept
We write the output as f of x. The letter f names the rule; x names the input; f of x names the number that comes out. The parentheses are not multiplication, and this is the single most common misreading in a first calculus course.
\[ f(x) = 3x^2 + 1 \;\Longrightarrow\; f(2) = 3(2)^2 + 1 = 13 \]
The input does not have to be a number. Substituting an expression, or another function's output, works exactly the same way: replace every x in the rule with whatever sits inside the parentheses, in brackets.
Figure (svg): The parts of function notation labelled: the name, the input variable, and the rule
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 9-11 — function notation and the independent variable
Picture it
The machine does not care what you feed it, as long as it is in the domain.
Figure (svg): A function drawn as an input-output machine, with two different inputs allowed to share one output but no input allowed two outputs
Evaluating at a number, at an expression, and at a second function's output are the same single act of substitution. Only the thing being substituted changes.
Worked example
Example 1.1. Substitute in brackets, every time.
\[ \text{For } f(x) = 3x^2 + 2x - 1, \text{ find } f(-2) \text{ and } f(a + h). \]
Replace every x with the input, in brackets
Why: Brackets keep the negative attached to the input, which is where sign errors start.
\[ f(-2) = 3(-2) ^{2} + 2(-2) - 1 \]
Evaluate the power before the products
Why: Negative 2 squared is positive 4, because the bracket squares the sign too.
\[ = 3(4) - 4 - 1 \]
Finish the arithmetic
Why: Twelve minus 4 minus 1.
\[ f(-2) = 7 \]
Now substitute the expression a + h the same way
Why: Nothing changes procedurally; the input is just longer.
\[ f(a + h) = 3(a + h) ^{2} + 2(a + h) - 1 \]
Expand and collect
Why: The square of the binomial is what generates the cross term.
\[ = 3 a ^{2} + 6 a h + 3 h ^{2} + 2 a + 2 h - 1 \]
Figure (svg): The solution to Worked example evaluating at a number and an expression shown as a ladder of expressions, one row per legal move
\[ f(-2) = 7, \qquad f(a+h) = 3a^2 + 6ah + 3h^2 + 2a + 2h - 1 \]
Verify: set h to zero
Why: Putting h equal to 0 in the second answer should return f of a, and it does: the terms with h vanish and 3a squared plus 2a minus 1 is left, which is the original rule with a in place of x. That one substitution checks the whole expansion in a single line, and it is worth doing every time — the expression f(a+h) is about to become the numerator of every derivative you compute in Chapter 3.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 10-10
Matching
Substitute in brackets each time.
Match the pairs
Why: The second and fourth are the ones worth checking twice. f(0) is the constant term, which is negative 1 — a fast sanity check on any polynomial. And g(-1) is 1 over negative 2, not 1 over 2: the input's sign survives the substitution.
Worked example
The same substitution, with a function as the input.
\[ \text{For } f(x) = x^2 + 1 \text{ and } g(x) = \frac{1}{x - 1}, \text{ find } f(g(3)). \]
Work from the inside out
Why: The inner function runs first, because its output is what f will receive.
\[ g(3) = \frac{1}{3 - 1} \]
Evaluate the inner function
Why: Three minus 1 is 2, so g of 3 is one half.
\[ g(3) = \frac{1}{2} \]
Feed that number to the outer function
Why: Now substitute one half for x in the rule for f.
\[ f(\frac{1}{2}) = (\frac{1}{2}) ^{2} + 1 \]
Finish
Why: A quarter plus 1 is five quarters.
\[ f(g(3)) = \frac{5}{4} \]
Figure (svg): The solution to Worked example evaluating one function at another shown as a ladder of expressions, one row per legal move
\[ f(g(3)) = \tfrac{5}{4} \]
Verify: do it in the other order and watch it differ
Why: Reversing gives g(f(3)) = g(10) = 1/9, which is not 5/4. That is not a mistake in either computation — it is the fact that composition does not commute. If your two orders ever agree by accident, check a second input before believing it.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 22-23
Trap
\[ f(x) = 3x + 2 \]
Read the parentheses as multiplication
Why: The student treats f as a number multiplying x.
\[ f(2) = f \cdot 2 = 2f \quad \text{(wrong)} \]
This produces an expression containing a meaningless symbol f instead of the number 8, and every later step inherits the confusion.
\[ f(x) = 3x + 2 \;\Longrightarrow\; f(2) = 3(2) + 2 = 8 \]
Read f as a NAME and the parentheses as 'evaluated at'
Why: f is not a quantity; it is the label on the rule. Only f of x is a number.
A reliable habit: say the notation out loud as f of 2, never as f times 2. The distinction matters enormously later, when you meet expressions like f prime of x and the argument is itself a whole expression.
Fill the middle
The step every derivative in Chapter 3 will begin with.
Fill in the blanks
f(x) = x^2 \;\Longrightarrow\; f(a + h) = (a+h)^2 = a^2 + 2ah + h^2
Why: Squaring a plus h gives a squared plus 2ah plus h squared. The cross term 2ah is the one that survives division by h and becomes the derivative; forgetting it is the single most expensive slip in the definition of the derivative.
Ranking
Evaluating f(g(x)) at a number.
Put in order
Why: Step one is the one that is skipped, and it is where the errors are. In f(g(x)) the inner function is g, even though f is written first — the notation reads outside-in but evaluates inside-out.
Prediction
Commit before reasoning.
Predict first
For f(x) = x^2, what is f(a + h)?
Correct: (a + h)^2, which expands to a squared plus 2ah plus h squared.
\[ f(a+h) - f(a) = (a+h)^2 - a^2 = 2ah + h^2 \]
Divide that by h and you have 2a plus h. Let h shrink to nothing and you have 2a — the derivative of x squared, three chapters early.
Why: The input is the whole expression a plus h, so the whole expression gets squared. Distributing f across the addition is the classic error: functions are not linear operators in general, and f(a+h) is almost never f(a) + f(h). The gap between them is exactly 2ah, and that gap is what the derivative measures.
Section
Section 2
Concept
Unless a domain is stated, the domain of a function given by a formula is every real number for which the formula makes sense. In practice only two things break: dividing by zero, and taking an even root of a negative number.
domain — The set of inputs a function accepts. When only a formula is given, the domain is understood to be every real number for which that formula produces a real output.
\[ f(x) = \frac{1}{x-1} \;\Longrightarrow\; D = (-\infty, 1) \cup (1, \infty) \]
The range is harder, because it asks what actually comes out rather than what may go in. For simple functions, read it off the graph's vertical shadow, or reason from the algebra: a square is never negative, so x squared plus 1 never dips below 1.
Figure (svg): A graph with its domain marked as a span on the horizontal axis and its range as a span on the vertical axis
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 9-13 — domain, range, and interval notation
Picture it
The square root function, with both shadows drawn.
Figure (svg): A graph with its domain marked as a span on the horizontal axis and its range as a span on the vertical axis
The horizontal shadow is the domain and the vertical shadow the range. Sliding a graph up and down changes its range and leaves its domain alone; sliding it left and right does the reverse.
Worked example
Example 1.2. Find what breaks, then find what comes out.
\[ \text{Find the domain and range of } f(x) = \sqrt{3 - 2x}. \]
Demand the radicand be non-negative
Why: An even root of a negative number is not real, so the inside must be zero or more.
\[ 3 - 2 x \ge 0 \]
Solve the inequality
Why: Subtracting 3 and dividing by negative 2 reverses the inequality sign.
\[ x \le \frac{3}{2} \]
Write the domain in interval notation
Why: Everything from negative infinity up to and including three halves.
\[ D = (-\infty, \frac{3}{2}) \]
Find the range from the output side
Why: A square root produces zero or a positive number, and 3 - 2x runs over all of those as x runs over the domain.
\[ R = [0, \infty] \]
Figure (svg): The solution to Worked example domain and range of a root function shown as a ladder of expressions, one row per legal move
\[ D = \left(-\infty, \tfrac{3}{2}\right], \qquad R = [0, \infty) \]
Verify: test the endpoint and one point outside
Why: At x equal to three halves the radicand is 0 and f is 0, so the endpoint belongs and the bracket is square. At x equal to 2 the radicand is negative 1, which has no real root, so everything above three halves is correctly excluded. Testing exactly one point on each side of a boundary catches a reversed inequality immediately.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 12-13
Sorting
Every restriction in this course comes from one of two causes.
Sort into buckets
Sort each function by what limits its domain.
Polynomials having no restrictions at all is worth internalising: it is why they are the easiest functions in calculus, and why the whole of Chapter 3 starts with them.
Worked example
Checkpoint 1.2. Now the break is a zero denominator.
\[ \text{Find the domain and range of } f(x) = \frac{3}{x - 2}. \]
Demand a non-zero denominator
Why: Division by zero is undefined, so x minus 2 must not be 0.
\[ x - 2 \ne 0 \]
Solve for the excluded input
Why: The single forbidden input is 2.
\[ x \ne 2 \]
Write the domain as a union of two intervals
Why: The number line with one point punched out.
\[ D = (-\infty, 2) U(2, \infty) \]
Find the range
Why: A non-zero numerator over anything can be made as large or as small as you like, but never exactly zero.
\[ R = (-\infty, 0) U(0, \infty) \]
Figure (svg): The solution to Worked example domain of a quotient shown as a ladder of expressions, one row per legal move
\[ D = (-\infty, 2) \cup (2, \infty), \qquad R = (-\infty, 0) \cup (0, \infty) \]
Verify: try to solve f(x) = 0
Why: Setting 3 over x minus 2 equal to zero gives 3 equal to 0, which is false, so no input produces the output 0 — confirming that 0 really is missing from the range. This is a general and useful move: to test whether a value is in the range, set the formula equal to it and see whether the equation has a solution.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 13-13
Error analysis
A student finds the domain of the square root of 3 minus 2x.
Annotate
On: \( 3 - 2x \ge 0 \;\Longrightarrow\; -2x \ge -3 \;\Longrightarrow\; x \ge \tfrac{3}{2} \)
The check that catches this in one line: substitute x equal to 2, which the wrong answer admits. The radicand becomes negative 1, so 2 cannot be in the domain and the inequality must point the other way.
Fill the middle
Interval notation, with the right kind of bracket.
Fill in the blanks
f(x) = \sqrt3/2 \;\Longrightarrow\; D = \left(-\infty, ___\right]
Why: The radicand is zero at three halves, and the square bracket says that endpoint is included — the root of zero is a perfectly good output. A round bracket there would have wrongly excluded the one input where f equals 0.
Two truths and a lie
All three are about domain and range.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The range of 3 over x minus 2 is everything except zero, which is a union of two intervals, not one. Ranges can also be a scatter of isolated values — the range of a function that only outputs integers, for instance. Assuming a range must be a single interval leads to writing down an answer that quietly includes values nothing maps to.
Prediction
Commit before reasoning.
Predict first
For f(x) = sqrt(x - 4), is 4 in the domain?
Correct: Yes — the radicand is zero, the root of zero is zero, so f(4) = 0.
\[ f(4) = \sqrt{4 - 4} = \sqrt{0} = 0 \;\Longrightarrow\; D = [4, \infty) \]
Why: The condition on an even root is that the radicand be non-negative, not positive: zero is allowed and produces the output zero. So the domain is written with a square bracket at 4. This matters more than it looks — an endpoint wrongly excluded is exactly the kind of small error that turns into a missing endpoint on a closed interval when you hunt for extreme values in Section 4.3.
Section
Section 3
Concept
The graph of a function is the set of points whose first coordinate is an input and whose second is the matching output. Because each input has exactly one output, no vertical line can meet the graph twice — which turns the definition into something you can check by eye.
vertical line test — Given a function, every vertical line drawn meets its graph no more than once. If some vertical line meets a set of points more than once, that set is not the graph of a function.
\[ \text{graph of } f = \{(x, f(x)) : x \in D\} \]
A zero of f is an input where the output is zero, so it is where the graph meets the horizontal axis. The y-intercept is a different thing entirely: it is the single output f of 0. A function can have many zeros but at most one y-intercept.
Figure (svg): A parabola passing the vertical line test beside a circle failing it, with the vertical lines drawn in
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 13-16 — graphs, the vertical line test, and zeros
Picture it
A parabola and a circle, with vertical lines laid across both.
Figure (svg): A parabola passing the vertical line test beside a circle failing it, with the vertical lines drawn in
The circle is a perfectly respectable curve; it simply is not the graph of a function of x, because the vertical line at x equal to 1 meets it twice. Splitting it into an upper and a lower half gives two functions.
Worked example
Example 1.3, worked exactly as the book does it.
\[ \text{For } f(x) = -4x + 2, \text{ find the zeros and the } y\text{-intercept, then sketch.} \]
Set the output equal to zero
Why: A zero is an input whose output is 0, so this is an equation in x.
\[ -4 x + 2 = 0 \]
Solve for x
Why: Subtracting 2 and dividing by negative 4.
\[ x = \frac{1}{2} \]
Evaluate at input zero for the y-intercept
Why: The y-intercept is the output when the input is 0.
\[ f(0) = 2 \]
Sketch through the two points
Why: A linear function needs only two points, and these are the two easiest to find.
\[ \text{line through } (\frac{1}{2}, 0)\text{ and } (0, 2) \]
Figure (svg): The line f of x equals negative 4x plus 2 with its zero and its y-intercept both marked
\[ \text{zero at } x = \tfrac{1}{2}, \qquad y\text{-intercept } (0, 2) \]
Verify: check the slope between the two points
Why: From (0, 2) to (1/2, 0) the rise is negative 2 over a run of one half, giving a slope of negative 4 — which matches the coefficient in the rule. When the slope you compute from your two points disagrees with the coefficient, one of the two points is wrong.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 14-15
Discrimination
Apply the vertical line test to each description.
Sort into buckets
Sort each curve.
Worked example
Checkpoint 1.3. Now the equation is quadratic, so expect up to two.
\[ \text{Find the zeros of } f(x) = x^2 - 5x + 6. \]
Set the output to zero
Why: Same first move as before; only the equation's degree has changed.
\[ x ^{2} - 5 x + 6 = 0 \]
Factor the trinomial
Why: Two numbers multiplying to 6 and adding to negative 5 are negative 2 and negative 3.
\[ (x - 2) (x - 3) = 0 \]
Apply the zero product property
Why: A product is zero exactly when one of its factors is.
\[ x = 2\text{ or } x = 3 \]
Note the y-intercept separately
Why: It is f of 0, the constant term.
\[ f(0) = 6 \]
Figure (svg): The solution to Worked example a function with two zeros shown as a ladder of expressions, one row per legal move
\[ x = 2 \text{ and } x = 3, \qquad y\text{-intercept } (0, 6) \]
Verify: substitute both zeros back
Why: f(2) = 4 - 10 + 6 = 0 and f(3) = 9 - 15 + 6 = 0, so both are genuine. A second check comes free from the structure: for x squared plus bx plus c the zeros multiply to c and add to negative b, and 2 times 3 is 6 while 2 plus 3 is 5. Both relations hold.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 15-16
Trap
\[ f(x) = -4x + 2 \]
Report the y-intercept as the zero
Why: The student sees the constant 2 and calls it the zero.
\[ \text{zero at } x = 2 \quad \text{(wrong)} \]
Substituting confirms the failure: f(2) is negative 8 plus 2, which is negative 6, not 0.
\[ \text{zero: solve } f(x) = 0 \;\Longrightarrow\; x = \tfrac{1}{2} \]
\[ y\text{-intercept: evaluate } f(0) = 2 \]
Keep the two questions apart by what is set to zero
Why: For a zero, the OUTPUT is zero. For the y-intercept, the INPUT is zero.
They are different points on the graph and they lie on different axes. A function may have no zeros, one, or many, but it has at most one y-intercept — precisely because it is a function.
Matching
Set each output to zero and solve.
Match the pairs
Why: The third is the instructive one: x squared plus 1 is never zero for real x, so its graph never touches the horizontal axis. A function is under no obligation to have any zeros at all, and Section 4.9 will spend its time hunting for the ones that exist but resist algebra.
Notation
Four statements about the same function, each saying something different.
Annotate
On: \( f(2) = 0, \quad f(0) = 2, \quad f(x) = 0, \quad f = 0 \)
The first two differ only in the order of two symbols, and they describe points on different axes. Reading the notation slowly is not pedantry; it is the difference between answering the question asked and a different one.
Prediction
Commit before reasoning.
Predict first
How many y-intercepts can the graph of a function have?
Correct: At most one — and none when 0 is not in the domain.
\[ f(x) = \tfrac{1}{x}: \; 0 \notin D \;\Longrightarrow\; \text{no } y\text{-intercept} \]
Why: The y-intercept is the output at input zero. If 0 is in the domain there is exactly one such output, because a function assigns exactly one; if 0 is not in the domain, as for 1 over x, there is no y-intercept at all. This is the vertical line test applied to the particular vertical line x equals 0, which is why the answer is at most one rather than any number.
Section
Section 4
Concept
Two functions can be added, subtracted, multiplied and divided by doing that arithmetic to their outputs at each shared input. Composition is different in kind: instead of combining two outputs, it feeds one function's output into the other as an input.
composite function — Given functions f and g, the composite f composed with g is the function whose value at x is f evaluated at g of x. Its domain is those x in the domain of g for which g of x lies in the domain of f.
\[ (f \circ g)(x) = f(g(x)) \]
Every combination inherits its domain from its parts. For the four arithmetic operations the domain is the overlap of the two domains, and division carries the extra demand that the denominator function not be zero.
Figure (svg): The four algebraic combinations of two functions, each with its domain condition
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 20-24 — combining functions and composition
Picture it
Composition is a pipeline, and pipelines have a direction.
Figure (svg): A composition drawn as two machines in series, with the inner function feeding the outer one
The notation is read outside-in but evaluated inside-out, which is why the order trips people. With f squaring and g subtracting, doing them in the two orders gives genuinely different functions.
Worked example
Example 1.9. The domain question is the real content here.
\[ \text{For } f(x) = \frac{1}{x-1} \text{ and } g(x) = \sqrt{x},\; \text{find } (f \circ g)(x) \text{ and its domain.} \]
Substitute g into f
Why: Every x in the rule for f is replaced by the whole expression for g.
\[ f(g(x)) = \frac{1}{\sqrt{x} - 1} \]
Require x to be in the domain of the inner function
Why: The square root needs a non-negative input.
\[ x \ge 0 \]
Require g(x) to be in the domain of the outer function
Why: f cannot accept 1, so the root must not equal 1.
\[ \sqrt{x} \ne 1 \]
Solve that second condition
Why: The square root equals 1 exactly when x is 1.
\[ x \ne 1 \]
Combine both conditions
Why: Non-negative, with 1 removed.
\[ D = [0, 1] U(1, \infty) \]
Figure (svg): The solution to Worked example a composition and its domain shown as a ladder of expressions, one row per legal move
\[ (f \circ g)(x) = \frac{1}{\sqrt{x} - 1}, \qquad D = [0, 1) \cup (1, \infty) \]
Verify: test the two boundary inputs
Why: At x equal to 0 the value is 1 over negative 1, which is negative 1 — defined, so 0 belongs and the bracket is square. At x equal to 1 the denominator is 0, so 1 is genuinely excluded. Both conditions had to be imposed: dropping the second would have wrongly admitted x equal to 1, and the formula alone gives no warning.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 23-24
Ranking
Finding the domain of f composed with g.
Put in order
Why: Steps b and c are both mandatory and only c is visible in the final formula, which is exactly why b gets skipped. Intersecting at the end, rather than reading the answer off the simplified expression, is what makes the method reliable.
Worked example
Checkpoint 1.9. The same two functions, the other way round.
\[ \text{With the same } f \text{ and } g, \text{ find } (g \circ f)(x) \text{ and its domain.} \]
Substitute f into g this time
Why: Now f runs first and its output goes under the root.
\[ g(f(x)) = \sqrt{\frac{1}{x - 1}} \]
Require x to be in the domain of f
Why: The denominator must not vanish.
\[ x \ne 1 \]
Require f(x) to be in the domain of g
Why: The root needs a non-negative input, so the fraction must be at least 0.
\[ \frac{1}{x - 1} \ge 0 \]
Solve that inequality
Why: A positive numerator over x minus 1 is non-negative only when x minus 1 is positive; it is never 0.
\[ x > 1 \]
Combine
Why: The condition x greater than 1 already excludes 1.
\[ D = (1, \infty) \]
Figure (svg): The solution to Worked example composition does not commute shown as a ladder of expressions, one row per legal move
\[ (g \circ f)(x) = \sqrt{\frac{1}{x-1}}, \qquad D = (1, \infty) \]
Verify: compare the two answers
Why: The two compositions have different formulas AND different domains — one is [0,1) union (1, infinity), the other is (1, infinity). Comparing at a shared input settles it: at x equal to 4, f(g(4)) is 1 over 1, which is 1, while g(f(4)) is the root of one third, about 0.577. Different numbers, so the order genuinely matters.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 24-24
Error analysis
A student composes f(x) = 1/(x-1) with g(x) = sqrt(x) and states the domain.
Annotate
On: \( (f \circ g)(x) = \frac{1}{\sqrt{x} - 1}, \qquad D = \{x : \sqrt{x} \ne 1\} = (-\infty, 1) \cup (1, \infty) \)
A composition carries TWO conditions, and only one of them is visible in the final formula. Always go back to the inner function: the simplified expression forgets where its ingredients were allowed to live.
Fill the middle
Take f to be the squaring-plus-one rule and g the reciprocal of one less than the input.
Fill in the blanks
f(g(2)) = f\left(1\right) = 1^2 + 1 = 2
Why: g of 2 is 1 over 2 minus 1, which is 1. Feeding 1 to f gives 1 squared plus 1, or 2. Doing it the other way, g(f(2)) is g(5), which is one quarter — a different number, confirming again that these two operations do not commute.
Sorting
Read what is being done to the outputs.
Sort into buckets
Sort each expression.
The visual tell is what sits inside the outer parentheses: an x means arithmetic, another function means composition. That distinction is what the chain rule in Section 3.6 exists to handle.
Prediction
Commit before reasoning.
Predict first
Is f(g(x)) ever equal to g(f(x)) for all x?
Correct: Yes, for special pairs — inverse functions being the important case.
\[ f(x) = 2x, \; g(x) = \tfrac{x}{2} \;\Longrightarrow\; f(g(x)) = x = g(f(x)) \]
Why: In general the two orders differ, but they coincide for particular pairs. The important one is a function and its inverse, where both compositions collapse to x itself: that is the definition of an inverse, and it is the whole content of Section 1.4. So the honest statement is not that composition never commutes but that it does not commute in general, which is a much more useful thing to know.
Section
Section 5
Concept
A function is even when replacing the input by its negative leaves the output unchanged, and odd when it flips the output's sign. Everything else is neither, which is the commonest case by a wide margin.
even and odd functions — A function f is even if f of negative x equals f of x for every x in the domain, and odd if f of negative x equals the negative of f of x. Even graphs are symmetric about the vertical axis; odd graphs are unchanged by a half turn about the origin.
\[ \text{even: } f(-x) = f(x); \qquad \text{odd: } f(-x) = -f(x) \]
The names come from powers: x to an even power is an even function, x to an odd power is an odd function. That mnemonic is reliable for pure powers and misleading for anything else, so the substitution test is what to trust.
Figure (svg): An even function mirrored across the vertical axis beside an odd function rotated a half turn about the origin
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 24-27 — symmetry of functions
Picture it
The two symmetries side by side, with a chord drawn on each.
Figure (svg): An even function mirrored across the vertical axis beside an odd function rotated a half turn about the origin
On the even graph the chord joining the two points is horizontal; on the odd graph it passes through the origin. Those are the two pictures, and they are what the algebra is describing.
Worked example
Example 1.10. Substitute negative x and compare with the original.
\[ \text{Classify } f(x) = -5x^4 + 7x^2 - 2 \text{ as even, odd, or neither.} \]
Substitute negative x for x
Why: Every occurrence, in brackets.
\[ f(-x) = -5(-x) ^{4} + 7(-x) ^{2} - 2 \]
Simplify the even powers
Why: A negative raised to an even power is positive, so both signs disappear.
\[ = -5 x ^{4} + 7 x ^{2} - 2 \]
Compare with the original
Why: The result is identical to f of x.
\[ f(-x) = f(x) \]
State the conclusion
Why: Matching the original exactly is the definition of even.
Figure (svg): The solution to Worked example classifying by substitution shown as a ladder of expressions, one row per legal move
\[ f(-x) = f(x) \;\Longrightarrow\; f \text{ is even} \]
Verify: test one pair of inputs numerically
Why: f(1) is negative 5 plus 7 minus 2, which is 0; f(-1) is also 0. Trying a second pair, f(2) is negative 80 plus 28 minus 2, which is negative 54, and f(-2) is the same. A numerical test cannot prove evenness, but a single mismatched pair would disprove it instantly, which makes it a cheap and effective check.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 26-26
Sorting
Substitute negative x and compare with both f and negative f.
Sort into buckets
Sort each function.
The constant function 5 catches most people. It is even, because 5 equals 5 for every input; its graph is a horizontal line, which is certainly symmetric about the vertical axis. And 1 over x is odd despite not being a polynomial — the test is about the substitution, not about the shape of the formula.
Worked example
Checkpoint 1.10. Most functions land here.
\[ \text{Classify } f(x) = 4x^3 - 5x + 1. \]
Substitute negative x
Why: In brackets, as always.
\[ f(-x) = 4(-x) ^{3} - 5(-x) + 1 \]
Simplify
Why: An odd power keeps the negative; the middle sign flips too.
\[ = -4 x ^{3} + 5 x + 1 \]
Compare with f(x)
Why: It is not equal to 4x cubed minus 5x plus 1, so not even.
\[ f(-x) \ne f(x) \]
Compare with the negative of f(x)
Why: Negating f gives negative 4x cubed plus 5x minus 1, whose constant is negative 1, not positive 1.
\[ f(-x) \ne - f(x) \]
Conclude
Why: Neither condition holds.
Figure (svg): The solution to Worked example a function that is neither shown as a ladder of expressions, one row per legal move
\[ f(-x) \ne f(x) \text{ and } f(-x) \ne -f(x) \;\Longrightarrow\; \text{neither} \]
Verify: find the term that spoils it
Why: The constant 1 is the culprit. Without it, 4x cubed minus 5x is odd, since both terms have odd degree. A non-zero constant is an even-degree term, so mixing it with odd-degree terms guarantees neither. That diagnosis is faster than the full test once you can see it: a polynomial is even exactly when all its terms have even degree, odd exactly when all have odd degree, and neither the moment the degrees are mixed.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 27-27
Trap
\[ f(x) = x^2 + x \]
Reason that it is not even, so it must be odd
Why: The student treats the two categories as exhaustive.
\[ f(-x) = x^2 - x \;\ne\; -f(x) = -x^2 - x \quad \text{(so not odd either)} \]
The conclusion 'therefore odd' was never justified. Both tests fail, and there is no third condition being satisfied.
\[ f(-x) = x^2 - x \]
Run BOTH tests and allow both to fail
Why: Even and odd are two special properties, not a partition of all functions.
\[ f(-x) \ne f(x) \;\text{ and }\; f(-x) \ne -f(x) \;\Longrightarrow\; \text{neither} \]
Almost every function you meet is neither. The two labels are worth having precisely because they are rare: when one does apply, it halves the work — an even function's graph is determined by the half of it where x is positive.
Fill the middle
Classifying the quartic from the worked example, which has only even-degree terms.
Fill in the blanks
f(-x) = -5(-x)^4 + 7(-x)^2 - 2 = -5x^4 + 7x^2 - 2 = f(x)
Why: Both exponents are even, so both negatives vanish and the expression comes back identical to f of x. That is the definition of an even function, and it means the graph is its own mirror image in the vertical axis.
Two truths and a lie
All three are about symmetry.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the misconception this whole idea guards against. Most functions are neither: x squared plus x fails both tests, and so does almost anything assembled without regard for symmetry. The two labels are useful because they are exceptional, not because they are exhaustive.
Prediction
Commit before reasoning.
Predict first
You know f is even and you have graphed it for x from 0 to 5. How much work remains to graph it from negative 5 to 5?
Correct: None. Reflecting the half you have across the vertical axis completes it.
\[ f \text{ even} \;\Longrightarrow\; \int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx \]
\[ f \text{ odd} \;\Longrightarrow\; \int_{-a}^{a} f(x)\,dx = 0 \]
Why: That is the practical value of the classification: an even function is fully determined by its behaviour on the non-negative inputs, and an odd one by the same half plus a rotation. It saves half the plotting here, and in Section 5.4 it saves an entire integral — the integral of an odd function over an interval symmetric about zero is zero without any computation at all.
Comparison
Fill the blanks. Every term here is used without explanation from Chapter 2 onward.
Comparison matrix
| Term | What it is | How you find it |
|---|---|---|
| Domain | the inputs the rule accepts | exclude zero denominators and negative even radicands |
| Range | the outputs actually produced | read the vertical shadow of the graph |
| Zero | an input whose output is 0 | solve f(x) = 0 |
| y-intercept | the output at input 0 | evaluate f(0) |
| Even | symmetric in the vertical axis | check f(-x) = f(x) |
The third and fourth rows are the pair worth separating in your head. One sets the output to zero and solves for an input; the other sets the input to zero and reports an output.
Pattern
Given a formula and asked to describe the function it defines.
Doing the domain first is not a stylistic preference. Every later step is a statement about inputs, and a step performed at an input outside the domain is not merely inelegant but false.
Stewart, Calculus: Early Transcendentals 8e, §1.1 Four Ways to Represent a Function §1.1, pp. 10-22
Check
Evaluation. Brackets around the input.
Check your understanding
For f(x) = 3x^2 + 2x - 1, find f(-2).
Answer: A
Why: Three times negative 2 squared is 3 times 4, or 12; plus 2 times negative 2, which is negative 4; minus 1. That is 7.
Check
Domain. Ask what would break.
Check your understanding
What is the domain of f(x) = sqrt(x)/(x - 4)?
Answer: A
Why: Two conditions apply at once: the root needs x at least 0, and the denominator forbids x equal to 4.
Check
Symmetry. Run both tests.
Check your understanding
Classify f(x) = x^3 - x as even, odd, or neither.
Answer: A
Why: Substituting gives negative x cubed plus x, which is the negative of x cubed minus x. Both terms have odd degree.
Real world
A shipping company charges by weight. A package up to 1 pound costs 5 dollars; each additional pound or part of a pound adds 2 dollars, up to a 50 pound limit.
Discussion prompt
Is cost a function of weight? Is weight a function of cost? Give the domain and range of whichever ones are functions, and say what the graph looks like.
Hint: Ask the defining question in each direction: could one input honestly produce two different outputs?
Answer:
Cost is a function of weight. Every weight from 0 to 50 pounds determines exactly one price. The domain is the interval from 0 to 50, and the range is the finite set 5, 7, 9, and so on up to 103 — fifty separate values, not an interval.
Weight is not a function of cost. A 1.2 pound package and a 1.9 pound package both cost 7 dollars, so the input 7 would need two outputs. This is the rectangle-area situation from the warm-up, in a different costume.
The graph is a staircase: flat over each pound-wide band, then jumping 2 dollars. It passes the vertical line test at every input, so it is a function — but it is discontinuous at every whole number of pounds, which is exactly the kind of behaviour Section 2.4 exists to describe.
\[ C(w) = 5 + 2\lceil w - 1 \rceil \quad \text{for } 0 < w \le 50 \]
Note what the range being a finite set of values, rather than an interval, tells you: no amount of adjusting the weight will ever produce a cost of 6 dollars.
Commit first
Answer, then rate your confidence honestly.
Predict first
If f(-x) = f(x) for every x in the domain, what must be true of the graph?
Correct: It is unchanged when reflected in the vertical axis.
\[ (x, f(x)) \text{ on the graph} \;\Longrightarrow\; (-x, f(x)) \text{ on the graph} \]
Why: The condition says the outputs at x and at negative x agree, so the points (x, f(x)) and (-x, f(x)) are both on the graph — a mirror pair across the vertical axis. Passing through the origin is the consequence of being ODD, not even, and even then only when 0 is in the domain. The half turn is also the odd condition. And the count of zeros need not be even: x squared is even and has exactly one zero.
Explain it
They can evaluate functions fine but say the vertical line test is an arbitrary rule someone invented.
Discussion prompt
In four sentences or fewer, explain why the vertical line test follows from the definition of a function rather than being an extra rule.
Hint: Ask what a vertical line actually is, in terms of inputs.
Answer:
A vertical line is the set of all points sharing one input value. So asking how many times a curve meets the vertical line at x equal to 3 is exactly asking how many outputs the curve assigns to the input 3.
The definition of a function says that number must be at most one. So the vertical line test is not an additional rule — it is the definition, translated from words about inputs and outputs into a picture about lines and crossings. If a curve fails it, the failure is that some input has been given two outputs.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For interval notation, decide the bracket by asking whether the endpoint itself is a legal input. For compositions, write the two conditions on separate lines before combining them, so the inner one cannot be forgotten. For zeros versus intercepts, say out loud which quantity you are setting to zero. For symmetry, always run both tests and let both fail. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take the four functions f(x) = 1/(x-1), g(x) = sqrt(x), h(x) = x^2 - 1 and k(x) = x^3 - x and give each one a column across the page. In each column write: the domain in interval notation, the reason for any exclusion, the zeros, the value of f at 0 if it is defined, and the even-odd classification with the substitution that proves it. Sketch each graph beneath its column, marking the zeros and the y-intercept. Then, across the bottom, form the compositions f of g and g of f, work out both domains from the two conditions rather than from the simplified formula, and write one sentence saying why they differ. In a margin, write the one sentence that is the definition of a function, and beside it note which of your four columns would change if that definition allowed an input to have two outputs.
If your sketches of h and k look unrelated, look again: h is even and k is odd, so one is a mirror and the other a half turn. Recognising that saved you half the plotting in each case, and it is the first time in this course that a structural fact has done arithmetic for you.
Recap
Five things, and every one of them is assumed without comment from Chapter 2 onward.
| If you see | Then |
|---|---|
| A denominator | Exclude the inputs making it zero |
| An even root | Demand the radicand be non-negative |
| f(x) = 0 | Solve for x: these are the zeros |
| f(0) | Evaluate: this is the y-intercept |
| A composition | Impose both domain conditions, not just the visible one |
| Only even-degree terms | The function is even |
| Mixed degrees | Neither even nor odd |
Section 1.2 sorts the functions you have just been handling into families — linear, polynomial, rational, algebraic, transcendental — and asks what shape each family has. That classification is what lets Chapter 3 give one differentiation rule per family rather than one per function.
OpenStax Calculus Volume 1, §1.1 Review of Functions §1.1, pp. 8-29 — everything on these slides traces back here
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