The fundamental objects of calculus — rules that map inputs to outputs.
A function \(f : A \to B\) is a rule that assigns to each element \(x \in A\) exactly one element \(f(x) \in B\).
| Type | General Form | Example | Domain |
|---|---|---|---|
| Polynomial | \(a_n x^n + \cdots + a_0\) | \(3x^2 - 2x + 1\) | \(\mathbb{R}\) |
| Rational | \(p(x)/q(x)\) | \(\dfrac{x+1}{x^2-4}\) | \(q(x)\neq 0\) |
| Trigonometric | \(\sin x,\ \cos x,\ \tan x\) | \(\sin(2x+1)\) | Varies |
| Exponential | \(a^x,\ a>0\) | \(e^x,\ 2^x\) | \(\mathbb{R}\) |
| Logarithmic | \(\log_a x\) | \(\ln x,\ \log_{10}x\) | \((0,\infty)\) |
| Root | \(\sqrt[n]{x}\) | \(\sqrt{x},\ x^{1/3}\) | \([0,\infty)\) if \(n\) even |
The composition \((f \circ g)(x) = f\!\left(g(x)\right)\). The output of \(g\) becomes the input of \(f\).
Example: If \(f(x) = x^2\) and \(g(x) = x+1\), then \((f\circ g)(x) = (x+1)^2\).
\(f^{-1}\) satisfies \(f\!\left(f^{-1}(x)\right)=x\) and \(f^{-1}\!\left(f(x)\right)=x\). An inverse exists when \(f\) is one-to-one (passes the horizontal line test). Graphically, \(y = f^{-1}(x)\) is the reflection of \(y=f(x)\) across the line \(y=x\).
Find the domain of \(\displaystyle f(x) = \frac{\sqrt{x-2}}{x^2 - 9}\).
Step 1 — Numerator: \(\sqrt{x-2}\) requires \(x - 2 \geq 0 \Rightarrow x \geq 2\).
Step 2 — Denominator: \(x^2 - 9 \neq 0 \Rightarrow x \neq \pm 3\). Since domain is \(x\geq 2\), we exclude \(x=3\).
Answer: Domain \(= [2,\,3) \cup (3,\,\infty)\)
Understanding the behavior of a function as its input approaches a value — even if the function is undefined there.
We write \(\displaystyle\lim_{x \to a} f(x) = L\) if \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(a\) (but \(x \neq a\)).
The value of \(f\) at \(a\) is irrelevant — only the behavior near \(a\) matters.
\(\displaystyle\lim_{x\to a^-} f(x) = L\) — left-hand limit: \(x\) approaches \(a\) from below
\(\displaystyle\lim_{x\to a^+} f(x) = L\) — right-hand limit: \(x\) approaches \(a\) from above
\(\displaystyle\lim_{x\to a} f(x) = L\) exists if and only if both one-sided limits exist and are equal.
If \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\), then all of the following hold:
| Law | Statement |
|---|---|
| Sum | \(\lim[f+g] = L+M\) |
| Difference | \(\lim[f-g] = L-M\) |
| Product | \(\lim[f\cdot g] = L\cdot M\) |
| Quotient | \(\lim[f/g] = L/M\quad\) (provided \(M\neq 0\)) |
| Power | \(\lim[f^n] = L^n\) |
| Root | \(\lim\sqrt[n]{f} = \sqrt[n]{L}\quad\) (if \(L>0\) for even \(n\)) |
| Constant multiple | \(\lim[c\cdot f] = c\cdot L\) |
Evaluate \(\displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2}\).
Direct substitution gives \(\dfrac{0}{0}\) — indeterminate. Factor the numerator:
\[\lim_{x\to 2}\frac{x^2-4}{x-2} = \lim_{x\to 2}\frac{(x-2)(x+2)}{x-2} = \lim_{x\to 2}(x+2) = \boxed{4}\]
See how function values approach the limit as \(x \to 2\) for \(f(x)=\dfrac{x^2-4}{x-2}\):
| \(x\) (from left) | \(f(x)\) | \(x\) (from right) | \(f(x)\) | |
|---|---|---|---|---|
| 1.9 | 3.9 | 2.1 | 4.1 | |
| 1.99 | 3.99 | 2.01 | 4.01 | |
| 1.999 | 3.999 | 2.001 | 4.001 | |
| → 2 | Limit = 4 | 2 ← | ||
\[\lim_{x\to 0}\frac{\sin x}{x} = 1 \qquad \lim_{x\to 0}\frac{1-\cos x}{x} = 0 \qquad \lim_{x\to\infty}\!\left(1+\frac{1}{x}\right)^{\!x} = e\]
A powerful technique using derivatives to resolve indeterminate limit forms.
Suppose \(\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}\) yields the form \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\), and \(f,g\) are differentiable near \(a\) (with \(g'(x)\neq 0\) near \(a\)). Then:
\[\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\]
provided the right-hand limit exists (finite or \(\pm\infty\)).
Always verify the form is \(0/0\) or \(\infty/\infty\) before applying L'Hôpital's rule. Applying it to a non-indeterminate form gives a wrong answer.
| Form | Direct action |
|---|---|
| \(0/0\) | Apply L'Hôpital directly |
| \(\infty/\infty\) | Apply L'Hôpital directly |
| \(0 \cdot \infty\) | Rewrite as \(\dfrac{0}{1/\infty}\) or \(\dfrac{\infty}{1/0}\), then apply L'Hôpital |
| \(\infty - \infty\) | Combine over common denominator, then apply L'Hôpital |
| \(1^\infty,\; 0^0,\; \infty^0\) | Write \(y=f^g\), take \(\ln y = g\ln f\), evaluate limit of \(g\ln f\) (now \(0\cdot\infty\)), then \(e^{\text{result}}\) |
\[\lim_{x\to 0}\frac{\sin x}{x} \;\stackrel{\text{L'H}}{=}\; \lim_{x\to 0}\frac{\cos x}{1} = \cos 0 = 1\]
\[\lim_{x\to\infty}\frac{\ln x}{x} \;\stackrel{\text{L'H}}{=}\; \lim_{x\to\infty}\frac{1/x}{1} = \lim_{x\to\infty}\frac{1}{x} = 0\]
\[\lim_{x\to 0}\frac{1-\cos x}{x^2} \;\stackrel{\text{L'H}}{=}\; \lim_{x\to 0}\frac{\sin x}{2x} \;\stackrel{\text{L'H}}{=}\; \lim_{x\to 0}\frac{\cos x}{2} = \frac{1}{2}\]
Find \(\displaystyle\lim_{x\to 0^+} x^x\). Let \(y = x^x\), so \(\ln y = x\ln x = \dfrac{\ln x}{1/x}\) which is \(\dfrac{-\infty}{\infty}\).
\[\lim_{x\to 0^+}\frac{\ln x}{1/x} \stackrel{\text{L'H}}{=} \lim_{x\to 0^+}\frac{1/x}{-1/x^2} = \lim_{x\to 0^+}(-x) = 0\]
So \(\ln y \to 0\), hence \(y = x^x \to e^0 = \mathbf{1}\).
Lines that a graph approaches — describing long-run or blow-up behavior.
The vertical line \(x = a\) is a vertical asymptote of \(f\) if any of the following hold:
\[\lim_{x\to a^+}f(x) = \pm\infty \quad\text{or}\quad \lim_{x\to a^-}f(x) = \pm\infty\]
For rational functions: look for zeros of the denominator where the numerator is non-zero.
The horizontal line \(y = L\) is a horizontal asymptote if:
\[\lim_{x\to+\infty}f(x) = L \quad\text{or}\quad \lim_{x\to-\infty}f(x) = L\]
A function can have at most two distinct horizontal asymptotes.
For \(f(x)=\dfrac{a_n x^n+\cdots}{b_m x^m+\cdots}\), compare degrees \(n\) and \(m\):
| Degree condition | Horizontal asymptote |
|---|---|
| \(n < m\) | \(y = 0\) |
| \(n = m\) | \(y = \dfrac{a_n}{b_m}\) (ratio of leading coefficients) |
| \(n > m\) | No horizontal asymptote (oblique/slant asymptote instead when \(n=m+1\)) |
Find all asymptotes of \(\displaystyle f(x) = \frac{3x^2}{x^2-4}\).
Vertical: Set \(x^2-4=0 \Rightarrow x=\pm 2\). Numerator at \(x=\pm 2\): \(3(4)=12\neq 0\). So \(x=2\) and \(x=-2\) are vertical asymptotes.
Horizontal: Degrees are equal (\(n=m=2\)), so \(y = \dfrac{3}{1} = \mathbf{3}\).
Find the oblique asymptote of \(\displaystyle f(x) = \frac{x^2+1}{x-1}\).
Perform polynomial long division: \(x^2+1 = (x-1)(x+1) + 2\), so \(f(x)=x+1+\dfrac{2}{x-1}\).
As \(x\to\pm\infty\), \(\dfrac{2}{x-1}\to 0\), so the oblique asymptote is \(\mathbf{y = x+1}\).
Evaluate difficult limits by trapping a function between two simpler bounds.
If, for all \(x\) near \(a\) (except possibly at \(a\) itself):
\[g(x) \leq f(x) \leq h(x)\]
and if \(\;\displaystyle\lim_{x\to a} g(x) = \lim_{x\to a} h(x) = L\), then:
\[\lim_{x\to a} f(x) = L\]
Think of it as a sandwich: if the bread (top and bottom) squeezes to the same point \(L\), whatever is inside is forced to meet at \(L\) too. No choice!
The key tool is usually: \(-1 \leq \sin(\text{anything}) \leq 1\) and \(-1 \leq \cos(\text{anything}) \leq 1\).
We know \(-1 \leq \sin(1/x) \leq 1\) for all \(x \neq 0\).
Multiply through by \(x^2 \geq 0\):
\[-x^2 \leq x^2\sin(1/x) \leq x^2\]
Since \(\lim_{x\to 0}(-x^2) = 0\) and \(\lim_{x\to 0}x^2 = 0\), by the Squeeze Theorem:
\[\lim_{x\to 0} x^2\sin(1/x) = \boxed{0}\]
Using unit circle geometry, for \(0 < x < \pi/2\) one can show:
\[\cos x \leq \frac{\sin x}{x} \leq 1\]
Since \(\lim_{x\to 0^+}\cos x = 1\) and \(\lim_{x\to 0^+}1 = 1\), the Squeeze Theorem gives \(\lim_{x\to 0^+}\dfrac{\sin x}{x} = 1\). By symmetry (the function is even), the two-sided limit equals 1.
A function is continuous when its graph has no gaps, jumps, or holes — you can draw it without lifting your pen.
\(f\) is continuous at \(x = a\) if all three hold:
| Type | What fails | Appearance | Example |
|---|---|---|---|
| Removable | Condition 3 fails (or 1 fails but limit exists) | Hole in the graph | \(\dfrac{x^2-1}{x-1}\) at \(x=1\) |
| Jump | Condition 2 fails: one-sided limits differ | Vertical jump | \(\text{sgn}(x)\) at \(x=0\) |
| Infinite | Limit is \(\pm\infty\) | Vertical asymptote | \(1/x\) at \(x=0\) |
| Oscillatory | Limit doesn't exist (oscillates infinitely) | Oscillates wildly | \(\sin(1/x)\) at \(x=0\) |
If \(f\) is continuous on \([a,b]\) and \(N\) is any number strictly between \(f(a)\) and \(f(b)\), then there exists at least one \(c \in (a,b)\) with \(f(c) = N\).
Root-finding application: If \(f(a)\) and \(f(b)\) have opposite signs, then \(f\) has at least one root in \((a,b)\).
Show \(f(x)=x^3 - x - 1\) has a root in \((1,2)\).
\(f(1)=1-1-1=-1 < 0\) and \(f(2)=8-2-1=5 > 0\). Since \(f\) is continuous (polynomial) and changes sign, IVT guarantees a root in \((1,2)\).
On their entire domains: polynomials, rational functions, \(\sin x\), \(\cos x\), \(e^x\), \(\ln x\), \(|x|\), \(\sqrt[n]{x}\).
Continuity is preserved under \(+,\, -,\, \times,\, \div\) (denominator \(\neq 0\)), and composition.
The derivative captures the instantaneous rate of change — the slope of the tangent line.
The derivative of \(f\) at \(x = a\) is:
\[f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\]
If this limit exists, \(f\) is said to be differentiable at \(a\). Geometrically, \(f'(a)\) is the slope of the tangent line to \(y=f(x)\) at \(x=a\).
Differentiable at \(a\) \(\Longrightarrow\) Continuous at \(a\)
But the converse is false: continuity does NOT guarantee differentiability.
| Situation | Reason | Example |
|---|---|---|
| Sharp corner / kink | Left and right derivatives differ | \(|x|\) at \(x=0\): \(f'(0^-)=-1,\; f'(0^+)=+1\) |
| Vertical tangent | Slope is \(\infty\) (limit DNE) | \(\sqrt[3]{x}\) at \(x=0\) |
| Discontinuity | Not continuous \(\Rightarrow\) not differentiable | Any jump discontinuity |
| Cusp | Slopes diverge to \(\pm\infty\) from both sides | \(x^{2/3}\) at \(x=0\) |
Find \(f'(x)\) for \(f(x) = x^2\).
\[f'(x) = \lim_{h\to 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h} = \lim_{h\to 0}\frac{2xh+h^2}{h} = \lim_{h\to 0}(2x+h) = 2x\]
Find \(f'(x)\) for \(f(x) = \dfrac{1}{x}\).
\[f'(x) = \lim_{h\to 0}\frac{\frac{1}{x+h}-\frac{1}{x}}{h} = \lim_{h\to 0}\frac{\frac{x-(x+h)}{x(x+h)}}{h} = \lim_{h\to 0}\frac{-h}{h\cdot x(x+h)} = \lim_{h\to 0}\frac{-1}{x(x+h)} = \frac{-1}{x^2}\]
Drag the slider to move the point \(P\) along \(y=x^2\) and see the tangent line.
The systematic rules for finding derivatives — no limit calculation needed.
| Rule | Formula |
|---|---|
| Constant | \(\dfrac{d}{dx}[c] = 0\) |
| Power Rule | \(\dfrac{d}{dx}[x^n] = nx^{n-1}\quad\) (for any real \(n\)) |
| Constant Multiple | \(\dfrac{d}{dx}[c\cdot f] = c\cdot f'\) |
| Sum / Difference | \(\dfrac{d}{dx}[f\pm g] = f' \pm g'\) |
\[\frac{d}{dx}[f\cdot g] = f'\cdot g + f\cdot g'\]
Memory aid: "derivative of first × second + first × derivative of second"
\[\frac{d}{dx}\!\left[\frac{f}{g}\right] = \frac{f'g - fg'}{g^2}\]
Memory aid: "low d-high minus high d-low, over low squared" (where "low" = \(g\), "high" = \(f\))
If \(y = f(g(x))\), then:
\[\frac{dy}{dx} = f'(g(x))\cdot g'(x)\]
Strategy: Identify the outer function and inner function. Differentiate the outer (leaving the inner untouched), then multiply by the derivative of the inner.
| Function | Derivative | Function | Derivative |
|---|---|---|---|
| \(\sin x\) | \(\cos x\) | \(\csc x\) | \(-\csc x\cot x\) |
| \(\cos x\) | \(-\sin x\) | \(e^x\) | \(e^x\) |
| \(\tan x\) | \(\sec^2 x\) | \(a^x\) | \(a^x\ln a\) |
| \(\sec x\) | \(\sec x\tan x\) | \(\ln x\) | \(1/x\) |
| \(\cot x\) | \(-\csc^2 x\) | \(\log_a x\) | \(1/(x\ln a)\) |
Differentiate \(y = \sin(3x^2+1)\).
Outer: \(\sin(\cdot)\); inner: \(3x^2+1\).
\[y' = \cos(3x^2+1)\cdot 6x = 6x\cos(3x^2+1)\]
Differentiate \(y = x^2 e^{\sin x}\).
\[y' = 2x\cdot e^{\sin x} + x^2\cdot e^{\sin x}\cdot\cos x = xe^{\sin x}(2+x\cos x)\]
Differentiate \(y = \dfrac{\ln x}{x^2}\).
\[y' = \frac{(1/x)\cdot x^2 - \ln x\cdot 2x}{x^4} = \frac{x - 2x\ln x}{x^4} = \frac{1-2\ln x}{x^3}\]
When \(y\) is defined implicitly as a function of \(x\), differentiate both sides with respect to \(x\), using the chain rule on any \(y\)-terms (giving a factor \(dy/dx\)), then solve for \(dy/dx\).
Example: Find \(dy/dx\) for \(x^2 + y^2 = 25\).
Differentiate both sides: \(2x + 2y\dfrac{dy}{dx} = 0 \;\Rightarrow\; \dfrac{dy}{dx} = -\dfrac{x}{y}\).
Using derivatives to decode the complete shape of a function's graph.
On an interval where \(f'(x) > 0\): \(f\) is increasing (graph rises left to right).
On an interval where \(f'(x) < 0\): \(f\) is decreasing (graph falls left to right).
Where \(f'(x) = 0\): function is momentarily flat — a potential turning point.
\(x=c\) is a critical point of \(f\) if \(f'(c)=0\) or \(f'(c)\) does not exist.
Fermat's Theorem: If \(f\) has a local extremum at \(c\) and \(f'(c)\) exists, then \(f'(c)=0\). So all extrema lie at critical points (but not all critical points are extrema!).
At a critical point \(c\):
Where \(f''(x) > 0\): graph is concave up — shaped like \(\cup\), slope is increasing.
Where \(f''(x) < 0\): graph is concave down — shaped like \(\cap\), slope is decreasing.
An inflection point is where concavity changes (requires \(f''(c)=0\) or DNE, and a sign change).
At a critical point \(c\) where \(f'(c)=0\):
Fully analyze \(f(x) = x^3 - 3x^2 - 9x + 5\).
Step 1 — Critical points:
\(f'(x) = 3x^2-6x-9 = 3(x^2-2x-3) = 3(x-3)(x+1)\)
Critical points: \(x=-1\) and \(x=3\).
Step 2 — Increasing/Decreasing (sign chart for \(f'\)):
| Interval | Sign of \(f'\) | Behavior |
|---|---|---|
| \((-\infty,-1)\) | \(+\) | Increasing |
| \((-1,3)\) | \(-\) | Decreasing |
| \((3,\infty)\) | \(+\) | Increasing |
Local max at \(x=-1\): \(f(-1)=10\). Local min at \(x=3\): \(f(3)=-22\).
Step 3 — Concavity:
\(f''(x) = 6x-6 = 6(x-1)\)
Concave down on \((-\infty,1)\), concave up on \((1,\infty)\). Inflection point at \(x=1\): \(f(1)=-6\).
Local max at (−1, 10) | Local min at (3, −22) | Inflection at (1, −6)
Polynomial approximations and fundamental theorems connecting function values to derivatives.
\[P_n(x) = \sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n\]
This polynomial matches \(f\) and its first \(n\) derivatives at \(x=a\).
A Taylor polynomial centered at \(a=0\):
\[P_n(x) = f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\cdots+\frac{f^{(n)}(0)}{n!}x^n\]
| Function | Maclaurin Series | Radius of Convergence |
|---|---|---|
| \(e^x\) | \(\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots\) | All \(x\) |
| \(\sin x\) | \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!} = x-\frac{x^3}{6}+\frac{x^5}{120}-\cdots\) | All \(x\) |
| \(\cos x\) | \(\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!} = 1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\) | All \(x\) |
| \(\ln(1+x)\) | \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n} = x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\) | \(|x|\leq 1\), \(x\neq -1\) |
| \(\dfrac{1}{1-x}\) | \(\displaystyle\sum_{n=0}^{\infty}x^n = 1+x+x^2+x^3+\cdots\) | \(|x|<1\) |
For \(f(x)=e^x\), all derivatives are \(e^x\), so \(f^{(k)}(0)=1\) for all \(k\).
\[P_4(x) = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}\]
Approximation check: \(e^1 \approx P_4(1) = 1+1+0.5+0.167+0.042 = 2.708\) (actual: \(2.718\ldots\))
If \(f\) satisfies:
Then there exists at least one \(c\in(a,b)\) such that \(f'(c)=0\).
Geometric meaning: A smooth curve that starts and ends at the same height must have at least one horizontal tangent line somewhere in between.
If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists at least one \(c\in(a,b)\) such that:
\[f'(c) = \frac{f(b)-f(a)}{b-a}\]
Geometric meaning: There is a point on the curve where the tangent line is parallel to the secant line joining \((a,f(a))\) to \((b,f(b))\).
Rolle's theorem is a special case of MVT: if \(f(a)=f(b)\), the right side becomes \(\dfrac{f(b)-f(a)}{b-a}=0\), so MVT gives \(f'(c)=0\) — exactly Rolle's conclusion.
Both require continuity on the closed interval AND differentiability on the open interval.
Verify the MVT for \(f(x)=x^2\) on \([1,3]\) and find the point \(c\).
MVT average rate of change: \(\dfrac{f(3)-f(1)}{3-1}=\dfrac{9-1}{2}=4\).
Set \(f'(c)=2c=4\Rightarrow c=2\). Indeed \(c=2\in(1,3)\). ✓
The tangent at \(x=2\) (slope 4) is parallel to the secant from \((1,1)\) to \((3,9)\).
Verify Rolle's theorem for \(f(x)=x^2-4x+3\) on \([1,3]\).
Check: \(f(1)=1-4+3=0\) and \(f(3)=9-12+3=0\). So \(f(1)=f(3)\). ✓
\(f\) is a polynomial — continuous and differentiable everywhere. ✓
\(f'(x)=2x-4=0\Rightarrow x=2\). Indeed \(c=2\in(1,3)\). ✓