1.1 Review of Functions

Big idea

Calculus does not study numbers; it studies functions. Every later idea in this course - limit, derivative, integral - is an operation that takes a function in and hands a function or a number back. So the first job is to be precise about what a function is and fluent with the notation, because a limit written as $\lim_{x \to a} f(x)$ is unreadable until $f$ and $f(x)$ mean different things to you.

The definition is deliberately narrow. A function pairs each allowed input with exactly one output. That uniqueness is what lets you write $f(3)$ and expect a single number, and it is what the vertical line test checks on a graph. A rule that sometimes returns two values, like “the number whose square is $x$”, is not a function until you choose one branch.

Two sets travel with every function: the inputs it accepts and the outputs it produces. Domain questions look like bookkeeping now, but they decide where a derivative can exist and which endpoints an integral is allowed to use. When a later problem says a function is differentiable on an interval, the interval came from a domain computation exactly like the ones in this section.

The last theme is reading structure off a graph. Intercepts tell you where the output is zero and what happens at the input zero. Symmetry tells you that half the graph determines the other half, which cuts work in half on integrals over intervals centered at the origin and explains why some answers collapse to zero without any computation. These are habits worth rebuilding now, quickly, before the machinery arrives.

Decoder

A function from a set of inputs to a set of outputs is a rule that assigns to each input exactly one output.

The load-bearing word is “each” and the load-bearing word is “exactly one”. “Each” says no input in the domain may be left unassigned, which is why the domain is part of the function and not an afterthought. “Exactly one” says no input may be sent to two different outputs. Nothing forbids two inputs from sharing an output: $f(x) = x^2$ sends both $-3$ and $3$ to $9$ and is a perfectly good function.

The classic mistake is treating $f$ and $f(x)$ as the same object. The symbol $f$ names the rule; $f(x)$ names the output that the rule produces at $x$, a number. That distinction is why $f$ can be differentiated and $f(2)$ cannot: $f(2)$ is already a constant.

A second habit worth fixing early is reading $f(x+h)$ correctly. It means “run the rule on the input $x+h$”, not “add $h$ to $f(x)$”. Substituting whole expressions into a function, parentheses and all, is the single most used skill of the next chapter, since the definition of the derivative is built on it.

Definitions and results

Function, domain, range. A function $f$ assigns to each element $x$ of its domain a single value $f(x)$. The domain is the set of legal inputs. The range is the set of values actually attained, that is, all $f(x)$ as $x$ runs over the domain.

Notation and evaluation. To evaluate $f$ at an expression, replace every occurrence of the variable by that expression, wrapped in parentheses. The variable name is a placeholder: $f(t) = t^2 + 1$ and $f(x) = x^2 + 1$ define the same function.

Natural domain conventions. Unless a domain is stated, take all real inputs for which the formula produces a real number. In practice that means three restrictions: no division by zero, no even root of a negative number, and no logarithm of a nonpositive number. Intersect the restrictions when a formula has several.

The vertical line test. A set of points in the plane is the graph of a function of $x$ exactly when no vertical line meets it more than once. A circle fails; the upper half of a circle passes.

Zeros and intercepts. The $x$-intercepts are the inputs with $f(x) = 0$, also called the zeros or roots. The $y$-intercept is $f(0)$, and there is at most one of them, since $0$ has exactly one output.

Even and odd symmetry. A function is even when $f(-x) = f(x)$ for every $x$ in the domain, and odd when $f(-x) = -f(x)$. An even graph is unchanged by reflection across the $y$-axis; an odd graph is unchanged by a half-turn about the origin. Most functions are neither.

Consequences of symmetry. An odd function whose domain contains $0$ must satisfy $f(0) = -f(0)$, so $f(0) = 0$ and the graph passes through the origin. Sums of even functions are even, sums of odd functions are odd, and a product of two odd functions is even.

The difference quotient. For $h \ne 0$,

$$ \frac{f(x+h) - f(x)}{h} $$

measures the average rate of change of $f$ between $x$ and $x+h$. Simplifying it until the $h$ in the denominator cancels is the computation the derivative is built from.

Worked examples

Evaluating at an expression

Let $f(x) = 2x^2 - x$. Then $f(3) = 18 - 3 = 15$ and $f(-2) = 8 + 2 = 10$.

Now build the difference quotient. Substitute the whole input $x+h$:

$$ f(x+h) = 2(x+h)^2 - (x+h) = 2x^2 + 4xh + 2h^2 - x - h $$

Subtract $f(x) = 2x^2 - x$. The terms without $h$ cancel, leaving $4xh + 2h^2 - h$. Every surviving term carries a factor of $h$, so the quotient simplifies:

$$ \frac{f(x+h) - f(x)}{h} = \frac{h(4x + 2h - 1)}{h} = 4x + 2h - 1 $$

Check it on numbers. Between $x = 3$ and $x = 4$ (so $h = 1$) the formula predicts $4(3) + 2 - 1 = 13$, and directly $f(4) - f(3) = 28 - 15 = 13$.

Domain and range of a shifted root

Let $g(x) = 3 - \sqrt{x + 4}$.

The only restriction is the square root, which needs $x + 4 \ge 0$, so the domain is $x \ge -4$, written $[-4, \infty)$.

For the range, work outward from the root. As $x$ runs over the domain, $\sqrt{x+4}$ takes every value in $[0, \infty)$, starting at $0$ when $x = -4$. Negating gives every value in $(-\infty, 0]$, and adding $3$ shifts that to $(-\infty, 3]$. The largest output is $g(-4) = 3$, and the outputs decrease without bound from there.

A rational function and its intercepts

Let $h(x) = \dfrac{x + 1}{x^2 - 9}$.

The denominator factors as $(x-3)(x+3)$, so it vanishes at $x = 3$ and $x = -3$. The domain is every real number except those two.

The $y$-intercept is $h(0) = 1/(-9) = -1/9$. For $x$-intercepts, a fraction is zero exactly when its numerator is zero and its denominator is not, so $x + 1 = 0$ gives the single intercept $x = -1$. Note that $-1$ is a legal input, so the point is genuinely on the graph.

Testing symmetry

Let $p(x) = x^4 - 4x^2$. Replace $x$ by $-x$: $(-x)^4 = x^4$ and $(-x)^2 = x^2$, so $p(-x) = x^4 - 4x^2 = p(x)$. The function is even.

Its zeros come from factoring: $x^2(x^2 - 4) = 0$ gives $x = 0$ and $x = \pm 2$, and the $y$-intercept is $p(0) = 0$. The zeros land symmetrically about the $y$-axis, which is what evenness predicts.

Compare $q(x) = x^3 - 4x$. Here $q(-x) = -x^3 + 4x = -q(x)$, so $q$ is odd, and its zeros $0$ and $\pm 2$ are again symmetric - this time by the half-turn. Testing a value confirms the difference in behavior: $p(1) = -3$ and $p(-1) = -3$, while $q(1) = -3$ and $q(-1) = 3$.

Practice

Start with the notation itself. These drills ask you to evaluate a function at numbers and at expressions, and to simplify what comes out.

Next, the two sets that travel with a function. Find the legal inputs by ruling out division by zero and even roots of negatives, then reason outward from the inside of the formula to describe the outputs.

Last, structure you can read from an equation or a graph: where the output is zero, what happens at the input zero, and whether replacing $x$ by $-x$ changes anything.

Quiz

Six items on evaluating function notation, determining domain and range, and identifying intercepts and symmetry.