Difficulty: Foundational to Intermediate | Prerequisites: Basic algebra, exponent rules
This material covers the core toolkit of precalculus: how to work with functions (their domains, symmetry, inverses, and compositions), how to handle polynomials (degree, zeros, end behaviour), and how to set up and solve equations involving lines and quadratics. If you are comfortable with basic algebra and exponent rules, you are ready for this. These topics form the foundation for everything that follows in calculus, so getting fluent here pays off quickly.
Precalculus functions boil down to knowing what goes in (domain), what symmetry the function has (even, odd, or neither), how to reverse it (inverse), and how to chain functions together (composition). Polynomials are about finding zeros, reading multiplicity, and predicting end behaviour from the leading term. Linear and quadratic functions round things out with slope-intercept form and completing the square.
Domain
The set of all input values (x-values) for which a function produces a real output. In simple terms, it is the collection of numbers you are allowed to plug in.
Even function
A function where f(−x) = f(x) for every x in its domain. Think of it as a function whose graph is symmetric about the y-axis.
Odd function
A function where f(−x) = −f(x) for every x in its domain. In simple terms, it has rotational symmetry of 180° about the origin.
Inverse function
A function f⁻¹(x) that reverses the action of f(x), so that f(f⁻¹(x)) = x. Think of it as "undoing" whatever f did.
Function composition
The operation (f ∘ g)(x) = f(g(x)), which means feeding g's output into f. In simple terms, you run two functions back-to-back.
Difference quotient
The expression [f(x₀ + h) − f(x₀)] / h, which measures the average rate of change of f over a small interval. This is the precursor to the derivative in calculus.
Slope-intercept form
y = mx + b, where m is the slope (rise over run) and b is the y-intercept. The standard way to write a linear equation.
Vertex form
f(x) = a(x − h)² + k, where (h, k) is the vertex of the parabola. You reach it from standard form by completing the square.
Degree of a polynomial
The highest power of x when all factors are multiplied out. It determines how many zeros (counted with multiplicity) the polynomial can have and what its end behaviour looks like.
Multiplicity
The number of times a particular zero appears as a factor. A zero of multiplicity 2 means the graph touches the x-axis and turns around rather than crossing.
End behaviour
What happens to f(x) as x → +∞ or x → −∞. It is controlled entirely by the leading term of the polynomial.
Sign chart
A number-line diagram that tracks where a rational expression is positive, negative, or zero. The method for solving rational inequalities.
Finding the domain means identifying every restriction on the input. There are two main restrictions at this level:
Square roots require the expression under the radical to be ≥ 0.
Denominators cannot equal zero.
Apply both restrictions together when a function has both features.
Example 1: f(x) = √(x − 12)
The expression under the root must be non-negative: x − 12 ≥ 0, so x ≥ 12. Domain: [12, ∞).
Example 2: f(x) = √x / ln x
Two restrictions apply here. The square root requires x ≥ 0, giving x ∈ [0, ∞). The denominator ln x cannot be zero, and ln 1 = 0, so x = 1 must be excluded. Also, ln x is undefined for x = 0 (and √0 / ln 0 is undefined), though √0 = 0 in the numerator. The domain is x ∈ (0, 1) ∪ (1, ∞).
Always check for both types of restriction and combine them. If logarithms are involved, the argument of the logarithm must also be strictly positive.
The algebraic test is straightforward: compute f(−x) and compare it to f(x).
If f(−x) = f(x) for all x, the function is even (symmetric about the y-axis).
If f(−x) = −f(x) for all x, the function is odd (symmetric about the origin).
If neither holds, the function is neither even nor odd.
Do not attempt to determine this by sketching. Use algebra.
Example: g(x) = x⁶ + x² + sin x
Compute g(−x):
g(−x) = (−x)⁶ + (−x)² + sin(−x) = x⁶ + x² − sin x
Compare: g(−x) = x⁶ + x² − sin x, but g(x) = x⁶ + x² + sin x. These are not equal, so g is not even. Check whether g(−x) = −g(x): −g(x) = −x⁶ − x² − sin x, which is also not equal to g(−x). The function is neither even nor odd.
The key observation: the polynomial part (x⁶ + x²) is even, but sin x is odd. Mixing even and odd components usually produces a function that is neither.
To find f⁻¹(x), follow these steps:
Write y = f(x).
Swap x and y.
Solve for y.
Replace y with f⁻¹(x).
Example: f(x) = 4e^(3x−9)
Step 1: y = 4e^(3x−9)
Step 2: x = 4e^(3y−9)
Step 3: Divide both sides by 4, then take ln of both sides:
x/4 = e^(3y−9)
ln(x/4) = 3y − 9
ln(x/4) + 9 = 3y
y = [ln(x/4) + 9] / 3
Step 4: f⁻¹(x) = [ln(x/4) + 9] / 3
The method always works the same way. The tricky part is usually the algebra in Step 3, which depends on what operations f performs (exponentials need logarithms to undo, and vice versa).
Composition means plugging one function into another. (f ∘ g)(x) = f(g(x)): evaluate g first, then feed the result into f.
Example: f(x) = x² − 4, g(x) = √(x + 4)
(f ∘ f)(x): f(f(x)) = f(x² − 4) = (x² − 4)² − 4 = x⁴ − 8x² + 16 − 4 = x⁴ − 8x² + 12
(g ∘ f)(x): g(f(x)) = g(x² − 4) = √((x² − 4) + 4) = √(x²) = |x|
Note that √(x²) simplifies to |x|, not just x (a common slip).
Composition with exponentials and logarithms:
Given f(x) = ½ ln(x + 2) and g(x) = eˣ:
(g ∘ f)(x) = g(½ ln(x + 2)) = e^(½ ln(x+2)) = e^(ln √(x+2)) = √(x + 2)
The exponential and logarithm cancel because e^(ln u) = u. The final answer should contain neither exponentials nor logarithms.
A linear function has the form y = mx + b. To find the equation from two points, compute the slope first, then solve for b.
Example: Find the linear function with f(−3) = −7 and f(5) = −11.
Slope: m = (−7 − (−11)) / (−3 − 5) = 4 / (−8) = −1/2
Substitute one point to find b:
−7 = (−1/2)(−3) + b
−7 = 3/2 + b
b = −7 − 3/2 = −14/2 − 3/2 = −17/2
The equation is y = −(1/2)x − 17/2.
Remember: rise over run means (y₂ − y₁) / (x₂ − x₁). Keep the order consistent for both coordinates.
To convert f(x) = ax² + bx + c into vertex form f(x) = a(x − h)² + k, complete the square.
Example: f(x) = x² − 4x + 5
Take half the coefficient of x (which is −4), square it: (−4/2)² = 4.
Add and subtract this value inside the expression:
f(x) = x² − 4x + 4 − 4 + 5 = (x − 2)² + 1
Vertex form: f(x) = (x − 2)² + 1, so the vertex is at (2, 1).
The vertex is the minimum point when a > 0 (parabola opens upward) or the maximum when a < 0 (opens downward). This technique also helps when solving quadratic equations and analysing graph transformations.
For a polynomial in factored form, you can read off three things directly: the degree, the zeros (with multiplicity), and the end behaviour.
Example: g(x) = −(12x − 7)²(34x + 89)³
Degree: The first factor contributes degree 2, the second contributes degree 3, giving degree 5 overall.
Zeros and multiplicity:
12x − 7 = 0 gives x = 7/12 with multiplicity 2 (the graph touches the axis and bounces back).
34x + 89 = 0 gives x = −89/34 with multiplicity 3 (the graph crosses the axis with a flattening).
End behaviour: Examine the leading term by taking the highest-degree term from each factor:
g(x) ∼ −(12x)²(34x)³ = −144 × 39304 × x⁵ = −5,659,776x⁵
Since the leading coefficient is negative and the degree is odd:
As x → +∞, g(x) → −∞
As x → −∞, g(x) → +∞
The sign of the leading coefficient combined with odd/even degree tells you everything about end behaviour. Odd degree means the two ends go in opposite directions. Even degree means they go in the same direction.
To solve an inequality like 2(x − 1) / [(x + 1)(x − 3)] ≤ 0, use a sign chart.
Find where the numerator is zero: x = 1.
Find where the denominator is zero: x = −1, x = 3. These are excluded from the domain.
Place these critical values on a number line, dividing it into intervals.
Test the sign of each factor in each interval.
Intervals and signs for 2(x − 1) / [(x + 1)(x − 3)]:
(−∞, −1): numerator (−), denominator (−)(−) = (+), overall: negative
(−1, 1): numerator (−), denominator (+)(−) = (−), overall: positive
(1, 3): numerator (+), denominator (+)(−) = (−), overall: negative
(3, ∞): numerator (+), denominator (+)(+) = (+), overall: positive
We want ≤ 0, so we take the negative intervals and include points where the expression equals zero (x = 1), but exclude points where it is undefined (x = −1, x = 3).
Solution: x ∈ (−∞, −1) ∪ [1, 3)
The difference quotient [f(x₀ + h) − f(x₀)] / h measures the average rate of change. This is the algebraic precursor to the derivative.
Example: f(x) = ax² + bx + c
Expand f(x₀ + h):
f(x₀ + h) = a(x₀ + h)² + b(x₀ + h) + c = ax₀² + 2ahx₀ + ah² + bx₀ + bh + c
Subtract f(x₀) = ax₀² + bx₀ + c:
f(x₀ + h) − f(x₀) = 2ahx₀ + ah² + bh
Divide by h:
[f(x₀ + h) − f(x₀)] / h = (h(2ax₀ + ah + b)) / h = 2ax₀ + ah + b
The h in the numerator always cancels. If it does not cancel, something has gone wrong in the algebra. This is a standard exam problem, and the approach is always the same: expand, subtract, simplify, then cancel h.
Slope formula: m = (y₂ − y₁) / (x₂ − x₁)
Slope-intercept form: y = mx + b
Vertex form: f(x) = a(x − h)² + k, vertex at (h, k)
Completing the square: x² + bx = (x + b/2)² − (b/2)²
Quadratic formula: x = [−b ± √(b² − 4ac)] / (2a)
Difference quotient: [f(x₀ + h) − f(x₀)] / h
Even function test: f(−x) = f(x)
Odd function test: f(−x) = −f(x)
Inverse function procedure: write y = f(x), swap x and y, solve for y
Students often think √(x²) = x. It does not. √(x²) = |x|. This matters in composition problems.
When finding domains, students forget to exclude points where a denominator is zero after already restricting for the square root. Both restrictions apply simultaneously.
Students sometimes try to determine even/odd by looking at the graph or guessing from the formula. The exam expects the algebraic test: compute f(−x) and compare.
In the difference quotient, students frequently expand incorrectly or forget to distribute the negative sign when subtracting f(x₀). Careful algebra is the whole point of the exercise.
⚠️ Domain questions appear on nearly every precalculus final. Expect at least one function combining a square root with a logarithm or a fraction.
⚠️ The even/odd test is a standard short-answer question. Show all algebra, not just the conclusion.
⚠️ Inverse function problems frequently involve exponentials and logarithms. You must be comfortable moving between the two.
⚠️ The difference quotient is direct preparation for calculus. Expect a general quadratic version.
True or false: the domain of f(x) = √(x − 5) / (x − 7) is [5, ∞). ___
Fill in the blank: if f(−x) = −f(x), the function is called ___.
True or false: to find an inverse, you swap x and y and then solve for x. ___
Fill in the blank: a zero of multiplicity 2 means the graph ___ the x-axis rather than crossing it.
True or false: the end behaviour of a degree-5 polynomial with positive leading coefficient has both ends going to +∞. ___
Answers: 1. False (must also exclude x = 7). 2. Odd. 3. False (you solve for y, not x). 4. Touches (or bounces off). 5. False (odd degree means opposite ends).
Q: Find the domain of f(x) = √(2x + 6) / (x² − 9).
A: The square root requires 2x + 6 ≥ 0, so x ≥ −3. The denominator x² − 9 = (x − 3)(x + 3) is zero at x = 3 and x = −3. Combining: x ∈ (−3, 3) ∪ (3, ∞).
Q: Is h(x) = x³ − 2x even, odd, or neither?
A: h(−x) = (−x)³ − 2(−x) = −x³ + 2x = −(x³ − 2x) = −h(x). The function is odd.
Q: Find the inverse of f(x) = 3x + 7.
A: Set y = 3x + 7. Swap: x = 3y + 7. Solve: y = (x − 7)/3. So f⁻¹(x) = (x − 7)/3.
Q: Given f(x) = x² + 1 and g(x) = 2x − 3, find (f ∘ g)(x).
A: f(g(x)) = f(2x − 3) = (2x − 3)² + 1 = 4x² − 12x + 9 + 1 = 4x² − 12x + 10.
Q: Convert f(x) = x² + 6x + 2 to vertex form.
A: f(x) = x² + 6x + 9 − 9 + 2 = (x + 3)² − 7. Vertex at (−3, −7).
The difference quotient connects directly to the definition of the derivative in calculus. If you can simplify this expression cleanly, you already understand the mechanics behind differentiation.
Function composition reappears constantly in the chain rule. Getting comfortable with nested functions now saves time later.
Polynomial end behaviour and sign charts lay the groundwork for curve sketching in calculus, where you will combine these ideas with information about critical points and concavity.
Domain restrictions, range, vertical line test, horizontal line test, one-to-one functions, bijection, even symmetry, odd symmetry, y-axis symmetry, origin symmetry, inverse function test, composition of functions, nested functions, slope, y-intercept, point-slope form, standard form of a quadratic, vertex of a parabola, axis of symmetry, leading coefficient, degree of polynomial, zeros of a polynomial, roots, x-intercepts, multiplicity of a root, end behaviour of polynomials, sign chart, sign diagram, test intervals, rational inequality, difference quotient, average rate of change, precalculus final exam review