Parent functions, shifts, reflections, stretches, even and odd symmetry, function arithmetic, composition, decomposition, one-to-one functions, and inverse functions — the complete toolkit for manipulating and reversing any function family
Chapter 1 established the function as the central object of precalculus. You learned how to read functions as equations, tables, graphs, and real-world models. You practiced finding domain and range, interpreting intercepts and zeros, analyzing increasing and decreasing behavior, and computing average rates of change.
Chapter 2 deepens that foundation. Instead of simply reading functions, you will learn to transform them, combine them, compose them, and reverse them. This chapter has three interlocking goals. The first is to understand transformations — how shifting, reflecting, stretching, and compressing a parent function changes its graph in predictable, algebraic ways. The second is to understand function operations — how two functions can be added, subtracted, multiplied, divided, or composed into new functions. The third is to understand inverse functions — how some functions can be reversed so that their outputs become inputs and their inputs become outputs.
All three ideas will return throughout this book. In Chapters 4 and 5, you will apply transformations directly to sine, cosine, and tangent graphs. In Chapter 9, you will study exponential and logarithmic functions as inverses of each other. In Chapter 13, you will see how inverse relationships connect to rates of change in calculus. Master the ideas in this chapter, and every function family that follows will be easier to understand, graph, and apply.
A sound engineer is adjusting an audio signal. The original signal is described by a function. She shifts the signal forward in time, doubles its volume, and inverts its phase. Each of those operations corresponds to a mathematical transformation. Later, she wants to reverse a filtering process — to recover the original signal from a processed version. That reversal corresponds to an inverse function. After studying Chapter 2, you will understand exactly what is happening at each step.
Every function family has a parent — a simple, unmodified version of the function that carries the essential behavior of the whole family. When you study transformations, you are studying how to build new functions by modifying parent functions in systematic ways. Recognizing parent functions lets you read any transformed graph or equation quickly. Instead of treating each function as something entirely new, you learn to see it as a familiar parent that has been shifted, stretched, reflected, or compressed.
| Name | Formula | Domain | Range | Key Feature |
|---|---|---|---|---|
| Constant | f(x) = c | All reals | {c} | Horizontal line at height c |
| Identity | f(x) = x | All reals | All reals | Diagonal line through origin, slope 1 |
| Absolute Value | f(x) = |x| | All reals | [0, ∞) | V-shape, vertex at origin |
| Quadratic | f(x) = x² | All reals | [0, ∞) | Upward parabola, vertex at origin |
| Cubic | f(x) = x³ | All reals | All reals | S-shaped curve through origin |
| Square Root | f(x) = √x | [0, ∞) | [0, ∞) | Starts at origin, bends right |
| Cube Root | f(x) = ³√x | All reals | All reals | S-curve through origin, symmetric |
| Reciprocal | f(x) = 1/x | x ≠ 0 | y ≠ 0 | Two-branch hyperbola; asymptotes at axes |
| Exponential | f(x) = bˣ, b > 1 | All reals | (0, ∞) | Through (0,1); x-axis is asymptote |
| Logarithmic | f(x) = log_b(x) | (0, ∞) | All reals | Through (1,0); y-axis is asymptote |
| Sine | f(x) = sin(x) | All reals | [−1, 1] | Periodic wave; period 2π |
| Cosine | f(x) = cos(x) | All reals | [−1, 1] | Periodic wave; period 2π; starts at 1 |
You do not need to memorize every detail of every parent right now. What matters is the general shape, the domain, the range, and the location of any critical points or asymptotes. Each parent will be reinforced again and again as you encounter its transformed versions throughout this book.
For each description, name the parent function, write its formula, and state its domain and range.
When you add a constant to the output of a function, you shift its entire graph up or down. The shape of the graph does not change. Only the height of every point changes. If f(x) is a parent function and k is a real number, then g(x) = f(x) + k shifts the graph of f vertically. If k > 0, the graph shifts upward by k units. If k < 0, the graph shifts downward by |k| units.
The reason is direct: adding k to f(x) adds k to every output value. Every point (x, y) on the original graph moves to (x, y + k) on the shifted graph. The x-coordinates are completely unchanged.
f(x) = |x| to g(x) = |x| − 3, and describe the graph of g.g(x) = |x| − 3 is the same as g(x) = f(x) − 3. Since 3 is subtracted outside the absolute value, this is a vertical shift downward by 3 units.f(x) = |x|, vertex at (0, 0). Every point on f will shift down by 3.g(0) = |0| − 3 = −3. This matches the shifted vertex at (0, −3). ✓Draw the parent V-shape y = |x| with vertex at (0, 0) in a lighter style. Then draw the shifted V-shape y = |x| − 3 with vertex at (0, −3) in a heavier style. Label the vertex of each curve and mark the y-axis intercept of the shifted graph at (0, −3).
When a constant is added inside the function's argument — directly to the variable — it is a horizontal shift, not a vertical shift. For example, |x + 3| is a horizontal shift, while |x| + 3 is a vertical shift. This distinction is covered next and is one of the most important in the chapter.
For each, identify the parent function, describe the vertical shift, and state the new range.
g(x) = x² + 4g(x) = √x − 2g(x) = |x| + 7g(x) = x³ − 5g(x) = (1/x) + 1g(x) = √x + 0.5When you add a constant inside the argument of a function — directly to the input variable — you shift the graph left or right. The shape does not change, but the horizontal position of every point does. If f(x) is a parent function and h is a real number, then g(x) = f(x − h) shifts the graph of f horizontally. If h > 0, the graph shifts to the right by h units. If h < 0, the graph shifts to the left by |h| units.
Many students expect f(x − h) with h > 0 to shift the graph left, because they see a subtraction. The reason it shifts right is this: to get the same output that f produces at input x = 0, the shifted version g(x) = f(x − h) needs input x = h. The graph reaches its original value later — shifted to the right.
g(x) = f(x − h) shifts the graph of f right by h units (when h > 0).
g(x) = f(x + h) shifts the graph of f left by h units (when h > 0).
Memory key: the shift direction is opposite the sign seen inside the argument. Write it in the form (x − h) to read h as the rightward shift directly.
f(x) = x² to g(x) = (x + 2)² and describe the resulting graph.(x + 2)² as (x − (−2))². Here h = −2. Since h is negative, the shift is 2 units to the left.f(x) = x², vertex at (0, 0). Every point shifts 2 units left.g(−2) = (−2 + 2)² = 0² = 0. Vertex at (−2, 0) confirmed. ✓Draw the parent parabola y = x² with vertex at (0, 0) in a lighter style. Draw the shifted parabola y = (x + 2)² with vertex at (−2, 0) in a heavier style. Label both vertices and show that the curves have identical shapes, differing only in horizontal position.
For each function, identify the parent, describe the direction and amount of horizontal shift, and state the new location of the key point (vertex, starting point, or intersection with axes).
g(x) = (x − 5)²g(x) = |x + 4|g(x) = √(x − 3)g(x) = (x + 1)³g(x) = 1/(x − 2)g(x) = √(x + 6)A reflection flips a graph across an axis. There are two types of reflections that appear most often in precalculus.
g(x) = −f(x) reflects the graph across the x-axis. Every (x, y) becomes (x, −y). Points above move below; points below move above.
g(x) = f(−x) reflects the graph across the y-axis. Every (x, y) becomes (−x, y). Points to the right move to the left.
f(x) = √x to g(x) = −√x.g(4) = −√4 = −2. Matches (4, −2). ✓f(x) = √x to g(x) = √(−x).x ≤ 0.g(−4) = √(−(−4)) = √4 = 2. Matches (−4, 2). ✓Draw y = √x starting at the origin and curving up-right. Draw y = −√x starting at the origin and curving down-right. Draw y = √(−x) starting at the origin and curving up-left. Label all three curves. Show the x-axis and y-axis as mirror lines for each reflection.
g(x) = −x²g(x) = −|x|g(x) = (−x)³g(x) = −√xg(x) = √(−x)g(x) = −(1/x)For each: state whether the reflection is across the x-axis or y-axis, describe the resulting graph, and state the domain and range.
g(x) = a · f(x)
If a > 1: vertical stretch by factor a. Each y-value is multiplied by a, pulling the graph away from the x-axis.
If 0 < a < 1: vertical compression by factor a. Each y-value is multiplied by a, pushing the graph toward the x-axis.
g(x) = f(bx)
If b > 1: horizontal compression by factor 1/b. Features are closer together.
If 0 < b < 1: horizontal stretch by factor 1/b. Features are farther apart.
Multiplying the input by something larger than 1 compresses the graph horizontally, not stretches it. Think of it this way: a larger multiplier on the input means the graph reaches its values faster — the features are compressed closer together horizontally. The direction is always opposite to vertical effects.
f(x) = x² to g(x) = 3x².g(2) = 3(2)² = 12. ✓f(x) = |x| to g(x) = |2x|.g(1) = |2(1)| = 2. Confirms (1, 2). ✓Identify the parent, type of stretch or compression, and factor. Describe how the graph changes.
g(x) = 4x²g(x) = (1/3)x²g(x) = |3x|g(x) = √(x/2)g(x) = 2√xg(x) = (1/2)|x|Real functions often involve multiple transformations applied together. To read or graph a transformed function correctly, you must apply the transformations in the right order. Given a function of the general form:
g(x) = a · f(b(x − h)) + k
Apply transformations in this order, reading from inside out:
1. Horizontal shift right by h (factor out b first to find h clearly)
2. Horizontal compression/stretch by factor 1/|b|
3. Reflection across y-axis if b < 0
4. Vertical stretch/compression by factor |a|
5. Reflection across x-axis if a < 0
6. Vertical shift by k (always last)
f(x) = x² to g(x) = −2(x − 3)² + 1.f(x) = x², upward parabola, vertex (0, 0).Step 1 — Shift right 3: vertex moves to (3, 0)
Step 2 — Stretch by 2: parabola becomes narrower
Step 3 — Reflect x-axis: parabola opens downward, vertex still at (3, 0)
Step 4 — Shift up 1: vertex moves to (3, 1)
g(3) = −2(0)² + 1 = 1. Vertex at (3, 1) confirmed. ✓ Also: g(4) = −2(1)² + 1 = −1. Point (4, −1) lies on the graph. ✓Draw the parent parabola y = x² opening upward at the origin. Then draw g(x) = −2(x − 3)² + 1 as a narrower, downward-opening parabola with vertex at (3, 1). Label the vertex, axis of symmetry x = 3, and mark the point (4, −1) on the curve.
When you see g(x) = f(2x − 6), do not read the horizontal shift as 6. Factor first: g(x) = f(2(x − 3)). Now you can see the horizontal shift is 3 units right, and there is a horizontal compression by 1/2. Reading the shift without factoring is one of the most common errors in this chapter.
For each function, identify the parent, list all transformations in order, state the new location of the key point, and describe the graph.
g(x) = (x + 4)² − 3g(x) = −|x − 1| + 2g(x) = 2√(x + 3)g(x) = −3(x − 2)² + 5g(x) = (1/2)|x + 1| − 4g(x) = −√(x − 5) + 1Some functions have a special property: their graphs are symmetric. For functions, there are two important types of symmetry. Symmetry about the y-axis occurs when the left half of the graph mirrors the right half — this is associated with even functions. Symmetry about the origin occurs when rotating the graph 180 degrees about the origin produces the same graph — this is associated with odd functions.
| Type | Algebraic Condition | Graph Symmetry | Examples |
|---|---|---|---|
| Even | f(−x) = f(x) | y-axis | x², |x|, cos(x), x⁴ |
| Odd | f(−x) = −f(x) | Origin | x, x³, 1/x, sin(x) |
| Neither | Neither condition holds | None | x² + x, √x, 2ˣ |
Substitute −x for x and simplify. Then compare f(−x) to f(x) and to −f(x). If f(−x) = f(x): even. If f(−x) = −f(x): odd. If neither: neither. Always simplify completely before comparing.
f(x) = x³ − x is even, odd, or neither.f(−x) = (−x)³ − (−x) = −x³ + x
Compare to f(x) = x³ − x:
f(−x) = −x³ + x ≠ x³ − x = f(x) → not even
Compare to −f(x) = −(x³ − x) = −x³ + x:
f(−x) = −x³ + x = −f(x) → odd ✓
A function that is both even and odd must satisfy f(x) = 0 for all x. Only the zero function f(x) = 0 qualifies. Most functions are neither. Do not confuse "neither" with an error — most real functions are simply neither even nor odd.
Determine whether each function is even, odd, or neither. Show your work by computing f(−x).
f(x) = x⁴ − 3x²f(x) = x³ + xf(x) = x³ + x²f(x) = x/(x² + 1)f(x) = |x| − 4f(x) = x² − xf(x) = 2x⁵ − x³ + xf(x) = x² + 3Two functions can be combined using the standard arithmetic operations to create new functions. Given functions f and g with domains D_f and D_g, the resulting functions and their domains are defined as follows.
Sum: (f + g)(x) = f(x) + g(x), domain: D_f ∩ D_g
Difference: (f − g)(x) = f(x) − g(x), domain: D_f ∩ D_g
Product: (f · g)(x) = f(x) · g(x), domain: D_f ∩ D_g
Quotient: (f/g)(x) = f(x)/g(x), domain: D_f ∩ D_g, g(x) ≠ 0
f(x) = √x and g(x) = x − 4. Find (f + g)(x) and (f/g)(x), and state the domain of each.Sum:
(f + g)(x) = √x + (x − 4) = √x + x − 4
Domain: [0, ∞)
Quotient:
(f/g)(x) = √x / (x − 4)
Domain: [0, 4) ∪ (4, ∞)
Let f(x) = x² − 1 and g(x) = x + 3. Find each and state its domain.
(f + g)(x)(f − g)(x)(f · g)(x)(f/g)(x)(f + g)(2)(f · g)(0)(f/g)(5)(f − g)(−3) — is (f/g)(−3) defined? Why?Function composition is the process of applying one function to the output of another. Instead of combining outputs arithmetically, composition feeds the output of one function into the input of another. The composition of f and g, written (f ∘ g)(x), means: first apply g to x, then apply f to the result.
(f ∘ g)(x) = f(g(x))
Read as "f composed with g" or "f of g of x." Apply the inner function g first, then the outer function f. The order matters. In general, f ∘ g ≠ g ∘ f.
Domain of (f ∘ g): all x in the domain of g such that g(x) is in the domain of f.
f(x) = x² + 1 and g(x) = 2x − 3. Find (f ∘ g)(x) and (g ∘ f)(x).(f ∘ g)(x) = f(g(x)) = f(2x − 3)
= (2x − 3)² + 1
= 4x² − 12x + 9 + 1
= 4x² − 12x + 10
(g ∘ f)(x) = g(f(x)) = g(x² + 1)
= 2(x² + 1) − 3
= 2x² + 2 − 3
= 2x² − 1
f(x) = √x and g(x) = x − 5. Find (f ∘ g)(x) and state its domain.(f ∘ g)(x) = f(g(x)) = √(x − 5)
Domain: need x − 5 ≥ 0, so x ≥ 5. Domain: [5, ∞)(f ∘ g)(x) = f(g(x)) means one function is applied after the other. (f · g)(x) = f(x) · g(x) means the outputs are multiplied. These are fundamentally different operations. The symbol ∘ denotes sequential application, not multiplication.
Let f(x) = 2x + 1 and g(x) = x² − 3.
(f ∘ g)(x)(g ∘ f)(x)(f ∘ g)(2)(g ∘ f)(−1)Composition builds a new function from two simpler ones. Decomposition reverses this: you start with a complex function and identify what simpler functions could have been composed to produce it. There is usually one decomposition that is most natural. The goal is to identify an outer function f and an inner function g such that h(x) = f(g(x)).
A useful strategy: identify the last operation that would be performed when computing h(x). That last operation is the outer function f. Everything fed into it is the inner function g.
h(x) = (3x + 7)⁵ as a composition f(g(x)).Let g(x) = 3x + 7 (inner function)
Let f(x) = x⁵ (outer function)
Then f(g(x)) = (3x + 7)⁵ = h(x) ✓
h(x) = √(x² − 4) as a composition.Let g(x) = x² − 4 and f(x) = √x
f(g(x)) = √(x² − 4) = h(x) ✓Decompose each function into the form f(g(x)) by identifying an inner function g and outer function f.
h(x) = (x + 5)⁴h(x) = √(3x − 2)h(x) = 1/(x² + 1)h(x) = |2x − 7|h(x) = (x³ − 4)²h(x) = √(x + 9)h(x) = (4x − 1)³h(x) = 1/√(x − 3)Not every function can be reversed. To reverse a function reliably — to know that every output came from exactly one input — the function must be one-to-one. A function is one-to-one if no two different inputs produce the same output. Formally: f is one-to-one if f(a) = f(b) implies a = b.
A function is one-to-one if and only if no horizontal line intersects its graph more than once.
Compare to the vertical line test, which tests whether a relation is a function at all (each input has at most one output). The horizontal line test tests whether the function is one-to-one (each output has at most one input).
| Function | One-to-One? | Reason |
|---|---|---|
f(x) = x³ | Yes | Every horizontal line hits the cubic curve at most once. |
f(x) = x² | No | y = 4 is produced by both x = 2 and x = −2. |
f(x) = sin(x) | No | Many horizontal lines intersect the sine curve infinitely often. |
f(x) = 3x − 5 | Yes | A non-horizontal line is hit at most once by any horizontal line. |
f(x) = |x| | No | y = 3 is produced by both x = 3 and x = −3. |
Non-one-to-one functions can still have inverses if their domain is restricted appropriately. For example, f(x) = x² restricted to [0, ∞) is one-to-one, and its inverse is f⁻¹(x) = √x. This strategy is used in Chapters 4 and 6 to define inverse trigonometric functions: arcsine, arccosine, and arctangent.
For each function, determine whether it is one-to-one. Justify your answer using either the definition or the horizontal line test. For non-one-to-one functions, suggest a domain restriction that would make it one-to-one.
f(x) = 2x + 7f(x) = x² − 4f(x) = x³ + 1f(x) = |x − 3|f(x) = 1/xf(x) = √xIf f is a one-to-one function, then its inverse function f⁻¹ reverses the input-output relationship of f. If f takes x to y, then f⁻¹ takes y back to x: f(x) = y ⟺ f⁻¹(y) = x. The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.
Two functions f and g are inverse functions of each other if and only if composing them in either order returns the original input: f(g(x)) = x for all x in the domain of g, and g(f(x)) = x for all x in the domain of f.
1. Write y = f(x).
2. Swap x and y.
3. Solve the resulting equation for y.
4. Write y = f⁻¹(x).
5. Verify: confirm that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x.
f(x) = 3x − 7.Step 1: y = 3x − 7
Step 2: Swap: x = 3y − 7
Step 3: Solve: x + 7 = 3y
y = (x + 7)/3
Step 4: f⁻¹(x) = (x + 7)/3
f(f⁻¹(x)) = f((x+7)/3) = 3·(x+7)/3 − 7 = (x+7) − 7 = x ✓
f⁻¹(f(x)) = f⁻¹(3x−7) = (3x−7+7)/3 = 3x/3 = x ✓
f(x) = ³√(x − 2).Step 1: y = ³√(x − 2)
Step 2: Swap: x = ³√(y − 2)
Step 3: Cube both sides: x³ = y − 2
Solve: y = x³ + 2
Step 4: f⁻¹(x) = x³ + 2
f(f⁻¹(x)) = ³√((x³+2) − 2) = ³√(x³) = x ✓f(x) = (x − 4)/(x + 2) and state its domain.Step 1: y = (x − 4)/(x + 2)
Step 2: x = (y − 4)/(y + 2)
Step 3: x(y + 2) = y − 4
xy + 2x = y − 4
xy − y = −4 − 2x
y(x − 1) = −4 − 2x
y = (−4 − 2x)/(x − 1) = −2(x + 2)/(x − 1)
Step 4: f⁻¹(x) = −2(x + 2)/(x − 1)
f⁻¹(x) is the inverse function. It is not 1/f(x). The superscript −1 on a function name means "inverse," not "raise to the power −1." The reciprocal of f(x) is written [f(x)]⁻¹ or 1/f(x). These are entirely different. A student who computes f⁻¹(x) = 1/f(x) is making a fundamental error — every time.
Determine whether each function is one-to-one. If yes, find the inverse algebraically and verify using composition.
f(x) = 5x − 2f(x) = x³ + 4f(x) = 2/(x + 1)f(x) = x² + 3 (all real numbers — is this one-to-one? What domain restriction makes it so?)f(x) = x² + 3 restricted to [0, ∞) — now find the inverse.f(x) = √(x − 3)If (a, b) is a point on the graph of f, then (b, a) is a point on the graph of f⁻¹. Swapping x and y in every point is exactly the geometric transformation of reflecting across the line y = x. This means: the graph of f⁻¹ is the reflection of the graph of f across the line y = x.
To graph an inverse from the graph of the original function, fold the coordinate plane along the diagonal line y = x. The reflected curve is the graph of the inverse. No formula is required — you only need to mirror the points.
f(x) = x³ passes through (−2,−8), (−1,−1), (0,0), (1,1), (2,8). Describe the graph of f⁻¹.(−2, −8) → (−8, −2)
(−1, −1) → (−1, −1)
(0, 0) → (0, 0)
(1, 1) → (1, 1)
(2, 8) → (8, 2)
Draw y = x³ as an S-shaped curve through the labeled points. Draw the dashed line y = x as a reference. Draw y = ³√x as the reflection of the cubic through the same key points (with coordinates swapped). Label both curves and show that the dashed line acts as the mirror axis between them.
Domain of f⁻¹ = Range of f = [0, 9]
Range of f⁻¹ = Domain of f = [−2, 5]Correction: Changes outside the function (e.g., f(x) − 3) affect the output and shift the graph vertically. Changes inside the argument (e.g., f(x − 3)) affect the input and shift the graph horizontally — in the opposite direction of the sign. Ask yourself: where is the change? Outside = vertical. Inside = horizontal (opposite sign).
Correction: Factor first: g(x) = (2(x − 3))². The horizontal shift is 3 units right, and there is a horizontal compression by 1/2. Always factor out the coefficient from the argument before identifying the shift.
Correction: Work from inside out. Horizontal transformations (inside the argument) come before vertical ones (outside). Within each group, apply stretch/compression before shift for accuracy. When listing transformations on the vertex: apply them in the correct inside-out sequence.
Correction: f(x) = x² on all reals is not one-to-one — both x = 2 and x = −2 produce y = 4. Before finding an inverse, verify the function is one-to-one with the horizontal line test. If it fails, restrict the domain. f⁻¹(x) = √x is only valid when f is restricted to [0, ∞).
Correction: f⁻¹(x) is the inverse function, found by swapping inputs and outputs. It is NOT 1/f(x). The notation is a naming convention, not an exponent. The reciprocal is written [f(x)]⁻¹ or 1/f(x). Confusing these is a fundamental error.
Correction: Always verify both compositions. A small algebraic error in swapping or solving is easy to make but hard to spot — and the composition check catches it immediately. Make this check a mandatory final step, not optional.
Correction: Start with the domain of g. Then further restrict to exclude any x-values where g(x) falls outside the domain of f. The domain of the composition is always a subset of the domain of g — never the domain of f alone.
Correction: The composition symbol ∘ means "apply one after the other": (f ∘ g)(x) = f(g(x)). Multiplication of functions uses a dot or juxtaposition: (f · g)(x) = f(x) · g(x). These produce entirely different results for most function pairs.
Answer in complete sentences. No computation required unless noted.
For each function, identify the parent, list all transformations, and describe the resulting graph.
g(x) = x² + 5g(x) = (x − 4)²g(x) = −|x|g(x) = |x + 3| − 2g(x) = √x + 1g(x) = √(−x)g(x) = −√(x − 1)g(x) = 4x²g(x) = (x/2)²g(x) = −3(x + 2)² + 7g(x) = 1/(x − 3)g(x) = 1/x + 4g(x) = 2|x − 1| − 3g(x) = −³√(x + 5)Let f(x) = x² − 1 and g(x) = x + 3.
Let f(x) = 2x + 1 and g(x) = x² − 3.
Decompose each into f(g(x)).
h(x) = (x + 5)⁴h(x) = √(3x − 2)h(x) = 1/(x² + 1)h(x) = |2x − 7|h(x) = (x³ − 4)²For each, determine if it is one-to-one. If yes, find the inverse and verify using composition.
f(x) = 5x − 2f(x) = x³ + 4f(x) = 2/(x + 1)f(x) = x² + 3 (all reals)f(x) = x² + 3 on [0, ∞)f(x) = √(x − 3)f(x) = (x − 4)/(x + 2)Each problem shows a student's work containing an error. Identify the mistake and write the correct solution.
What is the error? What is the correct transformation?
For f(x) = x + 2 and g(x) = x². What is the error? What is the correct answer?
What is the error? Find the correct inverse and verify it.
The student tested f(−x) = −x³ + 2 against −f(x). What is −f(x) for this function? Is the student's conclusion correct? Explain carefully.
What is the error? What is the correct domain?
g(x) = 3|x + 4| − 1 differs from the parent f(x) = |x|.Chapter 2 ends Part I of this book. You have built a complete toolkit for working with functions: transformations, symmetry, operations, composition, and inverses. These are not isolated precalculus skills — they are the language of everything that follows.
In Part II — Trigonometric Foundations — each of these tools returns immediately and repeatedly. When you encounter y = 3 sin(2x − π) + 1 in Chapter 5, you will recognize it as a sine parent that has been vertically stretched, horizontally compressed, horizontally shifted, and vertically shifted — exactly the skills from Sections 2.5 and 2.6. The inverse trigonometric functions (arcsine, arccosine, arctangent) introduced in later chapters require the domain-restriction strategy from Section 2.11: restricting a non-one-to-one function to a domain where it becomes one-to-one. The even and odd function facts from Section 2.7 apply directly: cosine is even (cos(−x) = cos(x)) and sine is odd (sin(−x) = −sin(x)). Function composition appears throughout trigonometric identities, and decomposition instinct helps you choose which substitution to make.
The habits you built in this chapter — reading inside-to-outside, factoring before interpreting, always verifying — are the habits of a precise mathematical thinker. They will serve you in every chapter of this book and into calculus.