Functions, Graphs and Growth Rates

Differential calculus studies how quantities change, and every derivative you will ever compute is the derivative of a function. So before any calculus, you need fluency with the standard library of functions: what each one looks like, where it is defined, how it behaves at the edges of its domain, and how fast it grows. Everything later in the course — limits, asymptotes, optimisation, curve sketching — is a question about the graphs collected here.

Functions, domain and range

A function $f$ assigns to each input $x$ in its domain exactly one output $f(x)$. The range is the set of outputs actually attained. When a formula is given with no domain stated, the natural domain is the largest set of real numbers for which the formula makes sense.

Take the intersection of all the requirements: a formula with three ingredients must satisfy all three at once.

The standard library

Rational functions and asymptotes

Let $f = P/Q$ with $P,Q$ polynomials. Factor both and cancel before drawing conclusions.

• A factor $(x-c)$ that cancels completely gives a hole at $x=c$ (the point is missing, the graph does not blow up). Its height is the value of the reduced formula at $c$.
• A zero $c$ of the reduced denominator gives a vertical asymptote $x=c$. Determine the sign on each side from the signs of the reduced numerator and denominator.
• End behaviour, comparing degrees: $\deg P < \deg Q \Rightarrow$ horizontal asymptote $y=0$; $\deg P = \deg Q \Rightarrow y =$ ratio of leading coefficients; $\deg P = \deg Q + 1 \Rightarrow$ a slant asymptote, obtained by polynomial long division; $\deg P \ge \deg Q + 2 \Rightarrow$ no linear asymptote (the graph follows the polynomial quotient).

A graph may cross a horizontal or slant asymptote — the asymptote only describes behaviour as $x \to \pm\infty$. It can never cross a vertical asymptote.

Transformations

Let $c>0$, $h,k$ real.

Two more: $y = \lvert f(x) \rvert$ flips the part of the graph below the $x$-axis up above it; $y = f(\lvert x \rvert)$ deletes the part with $x<0$ and replaces it with the mirror image of the $x \ge 0$ part.

Horizontal transformations act on the input, so always factor the argument first: $y = f(2x-6) = f\big(2(x-3)\big)$ is a horizontal compression by $\frac{1}{2}$ followed by a shift right 3 — not right 6. Vertical operations act on the output, in the order they are written: $y = c\,f(x)+k$ stretches by $c$ first and then shifts by $k$, which is not the same graph as $y = c\big(f(x)+k\big) = c\,f(x)+ck$. As a book-keeping check, horizontal operations affect only the domain and leave the range alone; vertical operations affect only the range and leave the domain alone.

Even and odd symmetry

These require a domain symmetric about $0$ (if $x$ is in the domain, so is $-x$).

• Even: $f(-x) = f(x)$ for all $x$ in the domain. Graph is symmetric in the $y$-axis. Examples: $x^{2}$, $\cos x$, $\lvert x \rvert$.
• Odd: $f(-x) = -f(x)$ for all $x$ in the domain. Graph has $180^{\circ}$ rotational symmetry about the origin; if $0$ is in the domain then $f(0)=0$. Examples: $x^{3}$, $\sin x$, $\tan x$, $1/x$.

Combination rules: even $\times$ even $=$ even, odd $\times$ odd $=$ even, even $\times$ odd $=$ odd (same for quotients). Even $+$ even $=$ even, odd $+$ odd $=$ odd; a mixed sum is usually neither. The only function that is both even and odd is $f(x)=0$. Most functions are neither — "neither" is a legitimate and common answer.

Composition and inverses

$(f \circ g)(x) = f(g(x))$. Its domain is the set of $x$ in the domain of $g$ such that $g(x)$ lies in the domain of $f$ — read off from the pieces, never from the simplified formula. In general $f \circ g \ne g \circ f$.

$f$ is one-to-one if $f(x_1)=f(x_2)$ forces $x_1=x_2$; graphically, no horizontal line meets the graph more than once (horizontal line test). Every strictly increasing or strictly decreasing function is one-to-one, but the converse fails — a one-to-one function need not be monotone. An inverse $f^{-1}$ exists exactly when $f$ is one-to-one, and it then satisfies $f^{-1}(f(x)) = x \quad \text{for } x \text{ in dom}(f), \qquad f(f^{-1}(y)) = y \quad \text{for } y \text{ in ran}(f),$ with $\text{dom}(f^{-1}) = \text{ran}(f)$ and $\text{ran}(f^{-1}) = \text{dom}(f)$. A function that fails the test can be made invertible by restricting its domain: $\sin$ restricted to $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ gives $\arcsin$, with domain $[-1,1]$.

Exponential and logarithm laws

Take $a>0$, $a \ne 1$. Every exponential is a rescaled $e^x$: $a^{x} = e^{x \ln a}$, and $\log_a x = \dfrac{\ln x}{\ln a}$.

Edge case worth memorising: $\ln(x^{2}) = 2\ln\lvert x \rvert$ for $x \ne 0$. Writing $2\ln x$ silently throws away the whole half $x<0$ of the domain.

Sinusoids: amplitude, period, phase shift

Put every sinusoid in the standard form $y = A\sin\big(B(x-C)\big) + D \qquad (A \ne 0,\ B \ne 0),$ factoring $B$ out of the argument first: $\sin\!\left(3x+\frac{\pi}{2}\right) = \sin\!\left(3\left(x+\frac{\pi}{6}\right)\right)$, a shift of $\frac{\pi}{6}$ left, not $\frac{\pi}{2}$ left.

The same table applies with $\cos$ in place of $\sin$. Shift identities that kill a phase: $\sin\!\left(\theta+\frac{\pi}{2}\right) = \cos\theta$, $\cos\!\left(\theta-\frac{\pi}{2}\right) = \sin\theta$, $\sin(\theta+\pi) = -\sin\theta$.

Comparing long-run growth rates

For functions eventually positive, say $f$ grows faster than $g$ if $\lim_{x \to \infty} \dfrac{f(x)}{g(x)} = \infty$ (equivalently $g/f \to 0$). The three facts you need, valid for every $p>0$, $q>0$ and every $a>1$: $\lim_{x \to \infty}\frac{(\ln x)^{q}}{x^{p}} = 0, \qquad \lim_{x \to \infty}\frac{x^{p}}{a^{x}} = 0, \qquad \lim_{x \to 0^{+}} x^{p}\,\lvert \ln x \rvert = 0.$

To compare two exponentials written differently, put both in the form $e^{cx}$: $2^{x} = e^{x\ln 2}$ with $\ln 2 \approx 0.693$, so $2^{x}$ beats $e^{x/2}$. The practical technique for a limit at infinity is to divide numerator and denominator by the single fastest-growing term that appears. (These facts can be proved later with l'Hopital's Rule, which requires the quotient to be of the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, both functions to be differentiable near the point with the denominator's derivative non-zero there, and the limit of the ratio of derivatives to exist or be $\pm\infty$.)

Curve sketching checklist (no calculus needed)

1. Domain. 2. Intercepts. 3. Symmetry (even, odd, periodic). 4. Vertical asymptotes and holes, with the sign of $f$ on each side. 5. End behaviour: horizontal or slant asymptote, or the dominant term. 6. Sign chart from the fully factored form. 7. Plot two or three extra points and join using the shapes from the standard library.

Worked example 1 — a complete rational sketch

Sketch $f(x) = \dfrac{x^{2}-4}{x^{2}-x-6}$.

Factor. $x^{2}-4 = (x-2)(x+2)$ and $x^{2}-x-6 = (x-3)(x+2)$, so $f(x) = \frac{(x-2)(x+2)}{(x-3)(x+2)}.$ Domain. The denominator vanishes at $x=3$ and $x=-2$, so the domain is $(-\infty,-2) \cup (-2,3) \cup (3,\infty)$.

Reduce. For $x \ne -2$, $f(x) = \dfrac{x-2}{x-3}$. The cancelled factor gives a hole at $x=-2$ of height $\dfrac{-2-2}{-2-3} = \dfrac{-4}{-5} = \dfrac{4}{5}$.

Vertical asymptote. Only $x=3$. As $x \to 3^{-}$ the numerator $\to 1 > 0$ and the denominator $\to 0^{-}$, so $f \to -\infty$; as $x \to 3^{+}$ the denominator $\to 0^{+}$, so $f \to +\infty$.

Horizontal asymptote. Degrees are equal with leading coefficients $1$ and $1$, so $y=1$. Does the graph cross it? Solving $\dfrac{x-2}{x-3} = 1$ gives $x-2 = x-3$, i.e. $-2=-3$: impossible, so it never crosses.

Intercepts. $f(x)=0$ only when $x-2=0$, giving $(2,0)$; and $f(0) = \dfrac{-2}{-3} = \dfrac{2}{3}$, giving $\left(0,\frac{2}{3}\right)$.

Sign. $\dfrac{x-2}{x-3} > 0$ on $(-\infty,2)$ and on $(3,\infty)$, and is negative on $(2,3)$.

Position relative to $y=1$. The sign chart does not settle this — subtract instead: $f(x) - 1 = \frac{x-2}{x-3} - 1 = \frac{(x-2)-(x-3)}{x-3} = \frac{1}{x-3}.$ That is positive exactly when $x>3$, so the curve sits above $y=1$ on $(3,\infty)$ and below $y=1$ on $(-\infty,3)$, with a hole punched out at $\left(-2,\frac{4}{5}\right)$ (and indeed $\frac45 < 1$).

Worked example 2 — a growth rate by substitution

Show that $\lim_{x \to \infty} \dfrac{(\ln x)^{3}}{\sqrt{x}} = 0$.

Substitute $x = e^{t}$, so $t = \ln x$ and $t \to \infty$ as $x \to \infty$ (legitimate because $t \mapsto e^{t}$ is continuous, strictly increasing and maps onto $(0,\infty)$). Then $\sqrt{x} = e^{t/2}$ and $\frac{(\ln x)^{3}}{\sqrt{x}} = \frac{t^{3}}{e^{t/2}} = \frac{t^{3}}{\left(e^{1/2}\right)^{t}}.$ Now apply $\lim_{t\to\infty} \dfrac{t^{p}}{a^{t}} = 0$ with $p=3$ and $a = e^{1/2} \approx 1.6487 > 1$. Hence the limit is $\mathbf{0}$.

This convergence is slow, which is exactly why intuition needs the theorem: at $x = 10^{6}$ the ratio is $\dfrac{(13.8155)^{3}}{1000} \approx 2.64$, still bigger than $1$; at $x = 10^{12}$ it is $\dfrac{(27.631)^{3}}{10^{6}} \approx 0.021$.

Common mistakes

• Wrong: $\sqrt{x^{2}} = x$. Right: $\sqrt{x^{2}} = \lvert x \rvert$; the square root symbol always returns the non-negative root.
• Wrong: the domain of $\sqrt{x^{2}-9}$ is $x \ge 3$. Right: $x^{2} \ge 9$ means $\lvert x \rvert \ge 3$, so the domain is $(-\infty,-3] \cup [3,\infty)$.
• Wrong: $\ln(a+b) = \ln a + \ln b$. Right: only products split: $\ln(ab) = \ln a + \ln b$ for $a,b>0$. $\ln(a+b)$ does not simplify at all.
• Wrong: $y = f(2x-6)$ is "compress horizontally by $\frac{1}{2}$, then shift right $6$". Right: that sequence produces $f(2(x-6)) = f(2x-12)$. Factor first: $f(2(x-3))$ — compress, then shift right $3$ (or shift right $6$ first and then compress).
• Wrong: $f(x) = x^{2}+x$ is even because $x^{2}$ is. Right: $f(-1) = 0$ but $f(1) = 2$, so it is neither even nor odd. One term does not decide.
• Wrong: $\ln x$ is odd because its graph "curls". Right: its domain $(0,\infty)$ is not symmetric about $0$, so $f(-x)$ is undefined and the function is neither.
• Wrong: cancelling $(x+2)$ removes the problem at $x=-2$, so the graph is unbroken there. Right: the original function is undefined at $x=-2$; the graph has a hole at $\left(-2,\frac{4}{5}\right)$.
• Wrong: a graph can never touch its horizontal asymptote. Right: $f(x) = \dfrac{\sin x}{x}$ crosses $y=0$ at every $x = k\pi$ with $k \ne 0$, yet $y=0$ is still its horizontal asymptote.
• Wrong: with $f(x)=x^{2}$ and $g(x)=\sqrt{x}$, $f(g(x)) = x$ for every real $x$. Right: the composite inherits $g$'s domain, so $f(g(x)) = x$ only on $[0,\infty)$.
• Wrong: $x^{100}$ grows faster than $2^{x}$ — just check $x=10$. Right: the two graphs cross where $100\ln x = x\ln 2$, at $x \approx 1.007$ and again at $x \approx 996$; $x^{100}$ is larger only strictly between those, and $2^{x}$ wins forever after $x \approx 996$. Growth statements are claims about the long run only.